basic analysis iv: measure theory and...

Basic Analysis IV: Measure Theory and Integration

The giant squids are interested in advanced mathematicaltraining and Jim is happy to help!

James K. PetersonDepartment of Mathematical Sciences

Clemson Universityemail: [email protected]© James K. Peterson First Edition

Gneural Gnome Press

Version 07.10.2018 : Compiled July 10, 2018

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Dedication I dedicate this work to all of my students who have learning these ideas of analysisthrough my courses. I have learned as much from them as I hope they have from me. I am a firmbeliever that all my students are capable of excellence and that the only path to excellence is throughdiscipline and study. I am always proud of my students for doing so well on this journey. I hopethese notes in turn make you proud of my efforts.

Abstract This book introduces graduate students in mathematics concepts from measure theoryand also, continues their training inthe abstract way of looking at the world. We feel that is a mostimportant skill to have when your lifes work will involve quantitative modeling to gain insight intothe real world.

Acknowledgements We want to acknowledge the great debt we have to our wife, Pauli, for herpatience in dealing with those vacant stares and the long hours spent in typing and thinking. You arethe love of my life.

The cover for this book is an original painting by me done in July 2017. It shows the momentwhen the giant squids reached out to me to learn advanced mathematics.

History Based On:Handwritten Notes For Measure and IntegrationMATH 82201995 - 2006Spring 2009

Table Of Contents

I Introductory Matter 1

1 Introduction 31.1 The Analysis Courses . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Senior Level Analysis . . . . . . . . . . . . . . . . . 41.1.2 The Graduate Analysis Courses . . . . . . . . . . . . 41.1.3 More Advanced Courses . . . . . . . . . . . . . . . . 7

1.2 Teaching The Measure and Integration Course . . . . . . . . 8

II Classical Riemann Integration 11

2 Riemann Overview 132.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.3.1 A Riemann Sum Example . . . . . . . . . . . . . . . 162.3.2 The Riemann Integral As A Limit . . . . . . . . . . . 162.3.3 The Fundamental Theorem Of Calculus . . . . . . . . 192.3.4 The Cauchy Fundamental Theorem Of Calculus . . . . 222.3.5 Applications . . . . . . . . . . . . . . . . . . . . . . 242.3.6 Simple Substitution Techniques . . . . . . . . . . . . 25

2.4 Handling Jumps . . . . . . . . . . . . . . . . . . . . . . . . 282.4.1 Removable Discontinuity . . . . . . . . . . . . . . . . 282.4.2 Jump Discontinuity . . . . . . . . . . . . . . . . . . . 302.4.3 Homework . . . . . . . . . . . . . . . . . . . . . . . 31

3 Bounded Variation 333.1 Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.1.1 Homework . . . . . . . . . . . . . . . . . . . . . . . 353.2 Monotone . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.2.1 Worked Out Example . . . . . . . . . . . . . . . . . . 423.2.2 Homework . . . . . . . . . . . . . . . . . . . . . . . 44

3.3 Bounded Variation . . . . . . . . . . . . . . . . . . . . . . . 453.3.1 Homework . . . . . . . . . . . . . . . . . . . . . . . 49

3.4 Total Variation . . . . . . . . . . . . . . . . . . . . . . . . . 493.5 Continuous Also . . . . . . . . . . . . . . . . . . . . . . . . 52

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4 Riemann Integration 554.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.2 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.3 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.4 Riemann Integrable? . . . . . . . . . . . . . . . . . . . . . . 704.5 More Properties . . . . . . . . . . . . . . . . . . . . . . . . 714.6 Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . 74

4.6.1 Homework . . . . . . . . . . . . . . . . . . . . . . . 804.7 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . 804.8 Same Integral? . . . . . . . . . . . . . . . . . . . . . . . . . 83

5 Further Riemann Results 875.1 Limit Interchange . . . . . . . . . . . . . . . . . . . . . . . 875.2 Riemann Integrable? . . . . . . . . . . . . . . . . . . . . . . 925.3 Content Zero . . . . . . . . . . . . . . . . . . . . . . . . . . 93

III Riemann - Stieljes Integrals 99

6 Riemann-Stieljes 1016.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 1026.2 Step Integrators . . . . . . . . . . . . . . . . . . . . . . . . 1046.3 Monotone Integrators . . . . . . . . . . . . . . . . . . . . . 1086.4 Equivalence Theorem . . . . . . . . . . . . . . . . . . . . . 1106.5 Further Properties . . . . . . . . . . . . . . . . . . . . . . . 1116.6 Bounded Variation Integrators . . . . . . . . . . . . . . . . . 113

7 Further Riemann-Stieljes 1177.1 Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . 1177.2 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1207.3 Computations . . . . . . . . . . . . . . . . . . . . . . . . . 1237.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

IV Abstract Measure Theory One 131

8 Measurability 1338.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1348.2 Borel Sigma Algebra . . . . . . . . . . . . . . . . . . . . . . 135

8.2.1 Homework . . . . . . . . . . . . . . . . . . . . . . . 1368.3 Extended Borel Sigma Algebra . . . . . . . . . . . . . . . . 1378.4 Measurable Functions . . . . . . . . . . . . . . . . . . . . . 138

8.4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . 1408.5 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 1418.6 Extended Valued . . . . . . . . . . . . . . . . . . . . . . . . 1438.7 Extended Properties . . . . . . . . . . . . . . . . . . . . . . 1458.8 Continuous Compositions . . . . . . . . . . . . . . . . . . . 148

8.8.1 The Composition With Finite Measurable Functions . 1488.8.2 The Approximation Of Non-negative Mea-

surable Functions . . . . . . . . . . . . . . . . . . . . 1498.8.3 Continuous Functions of Extended Valued

Measurable Functions . . . . . . . . . . . . . . . . . 150

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8.9 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

9 Abstract Integration 1539.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 1559.2 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 1609.3 Equality a.e. . . . . . . . . . . . . . . . . . . . . . . . . . . 1649.4 Convergence Theorems . . . . . . . . . . . . . . . . . . . . 1679.5 Extended Integrands . . . . . . . . . . . . . . . . . . . . . . 1749.6 Summable Properties . . . . . . . . . . . . . . . . . . . . . 1789.7 The DCT . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1819.8 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . 1839.9 Alternate Integration . . . . . . . . . . . . . . . . . . . . . . 184

9.9.1 Homework . . . . . . . . . . . . . . . . . . . . . . . 190

10 The Lp Spaces 19110.1 The General Lp spaces . . . . . . . . . . . . . . . . . . . . . 19510.2 The World Of Counting Measure . . . . . . . . . . . . . . . 20410.3 Essentially Bounded Functions . . . . . . . . . . . . . . . . 20510.4 The Hilbert Space L2 . . . . . . . . . . . . . . . . . . . . . 21110.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

V Constructing Measures 213

11 Building Measures 21511.1 Via Outer Measure . . . . . . . . . . . . . . . . . . . . . . . 21511.2 Via Metric Outer Measure . . . . . . . . . . . . . . . . . . . 22111.3 Building Outer Measure . . . . . . . . . . . . . . . . . . . . 22611.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23011.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

12 Lebesgue Measure 23512.1 Outer Measure . . . . . . . . . . . . . . . . . . . . . . . . . 23512.2 LOM Is MOM . . . . . . . . . . . . . . . . . . . . . . . . . 24512.3 Approximation Results . . . . . . . . . . . . . . . . . . . . 249

12.3.1 Approximating Measurable Sets . . . . . . . . . . . . 24912.3.2 Approximating Measurable Functions . . . . . . . . . 252

12.4 Non Measurable Sets . . . . . . . . . . . . . . . . . . . . . 25512.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 257

12.5 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 25712.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 260

13 Cantor Sets 26113.1 Generalized . . . . . . . . . . . . . . . . . . . . . . . . . . 26113.2 Representation . . . . . . . . . . . . . . . . . . . . . . . . . 26313.3 Cantor Function . . . . . . . . . . . . . . . . . . . . . . . . 26513.4 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . 266

14 Lebesgue Stieljes Measure 26914.1 Lebesgue-Stieljes . . . . . . . . . . . . . . . . . . . . . . . 27014.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 27614.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

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VI Abstract Measure Theory Two 281

15 Convergence Modes 28315.1 Extracting Subsequences . . . . . . . . . . . . . . . . . . . . 28415.2 Egoroff’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 29215.3 Vitali’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 29415.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29915.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

16 Decomposing Measures 30316.1 Jordan Decomposition . . . . . . . . . . . . . . . . . . . . . 30316.2 Hahn Decomposition . . . . . . . . . . . . . . . . . . . . . . 30716.3 Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30916.4 Absolute Continuity . . . . . . . . . . . . . . . . . . . . . . 31216.5 Radon-Nikodym . . . . . . . . . . . . . . . . . . . . . . . . 31416.6 Lebesgue Decomposition . . . . . . . . . . . . . . . . . . . 32116.7 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . 325

17 Connections To Riemann Integration 327

18 Differentiation 33118.1 Absolutely Continuous Functions . . . . . . . . . . . . . . . 33118.2 LS and AC . . . . . . . . . . . . . . . . . . . . . . . . . . . 33218.3 Bounded Variation Derivatives . . . . . . . . . . . . . . . . . 333

VII Summing It All Up 345

19 Summing It All Up 347

VIII References 349

IX Detailed Indices 353

X Glossary Of Terms 361

XI Appendix: Undergraduate Analysis Exam-inations 363

A Advanced Calculus I 365A-1 Course Structure . . . . . . . . . . . . . . . . . . . . . . . . 365A-2 Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . 365A-3 Exams Version A . . . . . . . . . . . . . . . . . . . . . . . . 367

A-3.1 Exam 1A . . . . . . . . . . . . . . . . . . . . . . . . 367A-3.2 Exam 2A . . . . . . . . . . . . . . . . . . . . . . . . 368A-3.3 Exam 3A . . . . . . . . . . . . . . . . . . . . . . . . 369A-3.4 Final A . . . . . . . . . . . . . . . . . . . . . . . . . 370

A-4 Exams Version B . . . . . . . . . . . . . . . . . . . . . . . . 371A-4.1 Exam 1B . . . . . . . . . . . . . . . . . . . . . . . . 371

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A-4.2 Exam 2B . . . . . . . . . . . . . . . . . . . . . . . . 372A-4.3 Exam 3B . . . . . . . . . . . . . . . . . . . . . . . . 374A-4.4 Final B . . . . . . . . . . . . . . . . . . . . . . . . . 374

A-5 Exams Version C . . . . . . . . . . . . . . . . . . . . . . . . 376A-5.1 Exam 1C . . . . . . . . . . . . . . . . . . . . . . . . 376A-5.2 Exam 2C . . . . . . . . . . . . . . . . . . . . . . . . 377A-5.3 Exam 3C . . . . . . . . . . . . . . . . . . . . . . . . 378A-5.4 Final C . . . . . . . . . . . . . . . . . . . . . . . . . 379

B Advanced Calculus II 383B-1 MTHSC 454 . . . . . . . . . . . . . . . . . . . . . . . . . . 383B-2 Course Structure . . . . . . . . . . . . . . . . . . . . . . . . 383B-3 Exams Version A . . . . . . . . . . . . . . . . . . . . . . . . 386

B-3.1 Exam 1A . . . . . . . . . . . . . . . . . . . . . . . . 386B-3.2 Exam 2A . . . . . . . . . . . . . . . . . . . . . . . . 387B-3.3 Exam 3A . . . . . . . . . . . . . . . . . . . . . . . . 388B-3.4 Final A . . . . . . . . . . . . . . . . . . . . . . . . . 389

B-4 Exams Version B . . . . . . . . . . . . . . . . . . . . . . . . 391B-4.1 Exam 1B . . . . . . . . . . . . . . . . . . . . . . . . 391B-4.2 Exam 2B . . . . . . . . . . . . . . . . . . . . . . . . 392B-4.3 Exam 3B . . . . . . . . . . . . . . . . . . . . . . . . 393

XII Appendix: Linear Analysis Examinations 395

C Linear Analysis I 397C-1 Course Structure . . . . . . . . . . . . . . . . . . . . . . . . 397C-2 Exams Version A . . . . . . . . . . . . . . . . . . . . . . . . 397

C-2.1 Exam 1A . . . . . . . . . . . . . . . . . . . . . . . . 397C-2.2 Exam 2A . . . . . . . . . . . . . . . . . . . . . . . . 398C-2.3 Exam 3A . . . . . . . . . . . . . . . . . . . . . . . . 400

C-3 Exams Version B . . . . . . . . . . . . . . . . . . . . . . . . 401C-3.1 Exam 1B . . . . . . . . . . . . . . . . . . . . . . . . 401C-3.2 Exam 2B . . . . . . . . . . . . . . . . . . . . . . . . 402C-3.3 Final B . . . . . . . . . . . . . . . . . . . . . . . . . 403

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Part I

Introductory Matter

1

Chapter 1

Introduction

We believe that all students who are seriously interested in mathematics at the Master’s and Doctorallevel should have a passion for analysis even if it is not the primary focus of their own researchinterests. So you should all understand that my own passion for the subject will shine though in thenotes that follow! And, it goes without saying that we assume that you are all mature mathematicallyand eager and interested in the material! Now, the present text focuses on the topics of Measure andIntegration from a very abstract point of view, but it is very helpful to place this course into its propercontext. Also, for those of you who are preparing to take the qualifying examination in analysis, theoverview below will help you see why all this material fits together into a very interesting web ofideas.

1.1 The Analysis CoursesIn outline form, the classical material on analysis would cover the material using textbooks equivalentto the ones listed below:

(A): Undergraduate Analysis, text Advanced Calculus: An Introduction to Analysis, by WatsonFulks. Here these are MATH 4530 and MATH 4540. Our take on this material is (Peterson(6) 2019). Additional material, not covered in courses now but very useful is discussed in(Peterson (8) 2019).

(B): Introduction to Abstract Spaces, text Introduction to Functional Analysis and Applications,by Ervin Kreyszig. Here this is MATH 8210. Our take on this material is (Peterson (7) 2019).

(C): Measure Theory and Abstract Integration, texts General Theory of Functions and Integra-tion, by Angus Taylor and Real Analysis, by Royden are classics but actually very hard toread and absorb. Our take on this material is the text you are currently reading. Here this isMATH 8220.

(D:) Topology and Functional Analysis. A classical text is Topology and Analysis by Simmons.Our take on this material is (Peterson (4) 2019).

In addition, a nice book that organizes the many interesting examples and counterexamples in thisarea is good to have on your shelf. We recommend the text Counterexamples in Analysis by Gel-baum and Olmstead. There are thus essentially five courses required to teach you enough of theconcepts of mathematical analysis to enable you to read technical literature (such as engineering,control, physics, mathematics, statistics and so forth) at the beginning research level. Here are somemore details about these courses.

3

4 CHAPTER 1. INTRODUCTION

1.1.1 Senior Level AnalysisTypically, this is a full two semester sequence that discusses thoroughly what we would call theanalysis of functions of a real variable. At Clemson, this is the sequence MTHSC 453-454. This twosemester sequence covers the following:

Advanced Calculus I: Here MATH 4530: This course studies sequences and functions whose do-main is simply the real line. There are, of course, many complicated ideas, but everything wedo here involves things that act on real numbers to produce real numbers. If we call these thingsthat act on other things, OPERATORS, we see that this course is really about real–valued op-erators on real numbers. This course invests a lot of time in learning how to be precise withthe notion of convergence of sequences of objects, that happen to be real numbers, to othernumbers.

1. Basic Logic, Inequalities for Real Numbers, Functions

2. Sequences of Real Numbers, Convergence of Sequences

3. Subsequences and the Bolzano–Weierstrass Theorem

4. Cauchy Sequences

5. Continuity of Functions

6. Consequences of Continuity

7. Uniform Continuity

8. Differentiability of Functions

9. Consequences of Differentiability

10. Taylor Series Approximations

Advanced Calculus II: MATH 4540: In this course, we rapidly become more abstract. First, wedevelop carefully the concept of the Riemann Integral. We show that although differentiation isintellectually quite a different type of limit process, it is intimately connected with the Riemannintegral. Also, for the first time, we begin to explore the idea that we could have sequencesof objects other than real numbers. We study carefully their convergence properties. We learnabout two fundamental concepts: pointwise and uniform convergence of sequences of objectscalled functions. We are beginning to see the need to think about sets of objects, such asfunctions, and how to define the notions of convergence and so forth in this setting.

1. The Riemann Integral

2. Sequences of Functions

3. Uniform Convergence of Sequence of Functions

4. Series of Functions

Our version of this course, (Peterson (6) 2019) adds two variable calculus, convex analysis in < anda nice discussion of Fourier Series. It turns out many of our mathematics majors do not take anadvanced engineering mathematics class, so they are not necessarily exposed to Fourier Series tools.And there is a lot of nice mathematics in that!

1.1.2 The Graduate Analysis CoursesThere are three basic courses at Clemson. First, linear analysis (MATH 8210), then measure andintegration (MATH 8220) and finally, functional analysis (MATH 9270). MATH 8210 is the core

1.1. THE ANALYSIS COURSES 5

analysis course and all Masters students at Clemson University must take it. Also, at Clemson Uni-versity, MATH 8210 and MATH 8220 form the two courses which we test prospective Ph.D. studentson as part of the analysis preliminary examination. The content of these courses, also must fit withina web of other responsibilities. Many students are typically weak in abstraction coming in, so ifwe teach the material too fast, we lose them. Now if 20 students take MATH 8210, usually 15 or75% are already committed to an M.S. program which emphasizes Operations Research, Statistics,Algebra/ Combinatorics or Computation in addition to applied Analysis. Hence, currently, there areonly about 5 students in MATH 8210 who might be interested in an M.S. specialization in analysis.The other students typically either don’t like analysis at all and are only there because they have to beor they like analysis but it is part of their studies in number theory, partial differential equations forthe Computation area and so forth. Either way, the students will not continue to study analysis for adegree specialization. However, we think it is important for all students to both know and appreciatethis material. Traditionally, there are several ways to go. The Cynical Approach: Nothing you cando will make students who don’t like analysis change their mind. So teach the material hard and fastand target the 2 - 3 students who can benefit. The rest will come along for the ride and leave thecourse convinced that analysis is just like they thought – too hard and too complicated. If you do thisapproach, you can pick about any book you like. Most books for our students are too abstract and soare very hard for them to read. But the 2 -3 students who can benefit from material at this level, willbe happy with the book. We admit this is not our style although some think it is a good way to find thereally bright analysis students. We prefer the alternate Enthusiastic “maybe I can get them interestedanyway” Approach: The instructor scours the available literature in order to make up notes to leadthe students “gently” into the required abstract way of thinking. We haven’t had much luck findinga published book for this so since this is our preferred plan of action: we have have worked on notessuch as the ones you have in your hand for a long time. These notes start out handwritten and slowlymature into the typed versions. many students have suffered with the earlier versions and we thankthem for that! We believe it is important to actively try to get all the students interested but, of course,this is never completely successful. However, we still think there is great value in this approach andit is the one we have been trying for many years.

Introductory Linear Analysis: MATH 8210 Our constraints for MATH 8210 content are that weget the students adequately exposed to a more abstract way of thinking about the world. Wegenerally cover

• metric spaces.

• vector spaces with a norm.

• vector spaces with an inner product.

It doesn’t sound like much but there is a lot of material in here the students haven’t seen. Forexample, we typically focus a lot on how we are really talking about sets of objects with someadditional structure. A set plus a way to measure distance between objects gives a metric space;if we can add and scale objects, we get a vector space; if we have a vector space and add thestructure that allows us to project a vector to a subspace, we get an inner product space. Wealso mention we could have a set of objects and define one operation to obtain a group or if wedefine a special collection of sets we call open, we get a topological space and so forth. If wework hard, we can help open their minds to the fact that each of the many sub disciplines in theMathematical Sciences focuses on special structure we add to a set to help us solve problemsin that arena.

There are lots of ways to cover the important material in these topic areas and even many waysto decide on exactly what is important from metric, normed and inner product spaces. So thereis that kind of freedom, but not so much freedom that you can decide to drop say, inner productspaces. For example, we could use Sturm Liouville systems as an example when the discussion


turns to eigenvalues of operators. It is nice to use projection theorems in an inner productsetting as a big finishing application, but remember the students are weak in background, e.g.,their knowledge of ordinary differential equations and Calculus in <n is normally weak. Sowe are limited in our coverage of the completeness of an orthonormal sequence in an innerproduct space in many respects. If you look carefully at that material, you need to cover someelementary versions of the Hahn Banach theorem to do it right. However, we run out of time tocover such advanced topics.. The trade off seems to be between thorough coverage of a smallnumber of topics or rapid coverage of many topics superficially. We like the former approachmyself, but it can be done other ways. We believe this course is about teaching the studentsabout the abstract way of thinking about problems and hence, we feel there is great value inteaching very, very carefully the basics of this material.

Also, this MATH 8210 material is a nice prerequisite for partial differential equations andordinary differential equations as well as statistics and probability courses.

This course takes a huge amount of time for lecture preparation and student interaction in youroffice, so when we teach this material, we slow down in our research output!

In more detail, in MATH 8210, we now begin to rephrase all of our knowledge about conver-gence of sequence of objects in a much more general setting.

1. Metric Spaces: A set of objects and a way of measuring distance between objects whichsatisfies certain special properties. This function is called a metric and its propertieswere chosen to mimic the properties that the absolute value function has on the real line.We learn to understand convergence of objects in a general metric space. It is reallyimportant to note that there is NO additional structure imposed on this set of objects; nolinear structure (i.e. vector space structure), no notion of a special set of elements calleda basis which we can use to represent arbitrary elements of the set. The metric in a sensegeneralizes the notion of distance between numbers. We can’t really measure the size ofan object by itself, so we do not yet have a way of generalizing the idea of size or length.

A fundamentally important concept now emerges: the notion of completeness and how itis related to our choice of metric on a set of objects. We learn a clever way of constructingan abstract representation of the completion of any metric space, but at this time, we haveno practical way of seeing this representation.

2. Normed Spaces: We add linear structure to the set of objects and a way of measuringthe magnitude of an object; that is, there is now an operation we think of as additionand another operation which allows us to scale objects and a special function called anorm whose value for a given object can be thought of as the object’s magnitude. Wethen develop what we mean by convergence in this setting. Since we have a vector spacestructure, we can now begin to talk about a special subset of objects called a basis whichcan be used to find a useful way of representing an arbitrary object in the space.

Another most important concept now emerges: the cardinality of this basis may be finiteor infinite. We begin to explore the consequences of a space being finite versus infinitedimensional.

3. Inner Product Spaces: To a set of objects with vector space structure, we add a functioncalled an inner product which generalizes the notion of dot product of vectors. This hasthe extremely important consequence of allowing the inner product of two objects to zeroeven though the objects are not the same. Hence, we can develop an abstract notion ofthe orthogonality of two objects. This leads to the idea of a basis for the set of objects inwhich all the elements are mutually orthogonal. We then finally can learn how to buildrepresentations of arbitrary objects efficiently.

1.1. THE ANALYSIS COURSES 7

4. Completions: We learn how to complete an arbitrary metric, normed or inner productspace in an abstract way, but we know very little about the practical representations ofsuch completions.

5. Linear Operators: We study a little about functions whose domain is one set of objectsand whose range is another. These functions are typically called operators. We learn alittle about them here.

6. Linear Functionals: We begin to learn the special role that real-valued functions actingon objects play in analysis. These types of functions are called linear functionals andlearning how to characterize them is the first step in learning how to use them. We justbarely begin to learn about this here.

We have implemented this approach in (Peterson (7) 2019).

Measure Theory: MATH 8220: This course generalizes the notion of integration to a very abstractsetting. The set of notes you are reading is a textbook for this material. Roughly speaking, wefirst realize that the Riemann integral is a linear mapping from the space of bounded real valuedfunctions on a compact interval into the reals which has a number of interesting properties. Wethen study how we can generalize such mappings so that they can be applied to arbitrary setsX , a special collection of subsets of X called a sigma-algebra and a new type of mappingcalled a measure which on < generalizes our usual notion of the length of an interval. In thisclass, we discuss the following:

1. The Riemann Integral

2. Measures on a sigma-algebra S in the set X and integration with respect to the measure.

3. Measures specialized to sigma-algebras on the set <n and integrations with respect tothese measures. The canonical example of this is Lebesgue measure on <n.

4. Differentiation and Integration in these abstract setting and their connections.

This is what we cover in the notes you are reading now and some other things!

1.1.3 More Advanced CoursesIt is also recommended that students consider taking a course in what is called Functional Analysis.Here that is called MATH 9270. While not part of the qualifying examination, in this course, we canfinally develop in a careful way the necessary tools to work with linear operators, weak convergenceand so forth. This is a huge area of mathematics, so there are many possible ways to design anintroductory course. A typical such course would cover:

1. The Open Mapping and Closed Graph Theorem.

2. An Introduction to General Operator Theory.

3. Topological Vector Spaces and Distributions.

4. An Introduction to the Spectral Theory of Linear Operators; this is the study of the eigenvaluesand eigenobjects for a given linear operator–lots of applications here!

5. Some advanced topic using these ideas: possibilities include

(a) Existence Theory of Boundary Value Problems.

(b) Existence Theory for Integral Equations.

(c) Existence Theory in Control.


1.2 Teaching The Measure and Integration CourseSo now that you have seen how the analysis courses all fit together, it is time for the main course. Soroll up your sleeves and prepare to work! Let’s start with a few more details on what this course onMeasure and Integration will cover.

In this course, we assume mathematical maturity and we tend to follow the The Enthusiastic“maybe I can get them interested anyway” Approach in lecturing (so, be warned)! It is difficult todecide where to start in this course. There is usually a reasonable fraction of you who have neverseen an adequate treatment of Riemann Integration. For example, not everyone may have seen theequivalent of MTHSC 454 where Riemann integration is carefully discussed. We therefore haveseveral versions of this course. We have divided the material into blocks as follows: We believe thereare a lot of advantages in treating integration abstractly. So, if we covered the Lebesgue integral on< right away, we can take advantage of a lot of the special structure < has which we don’t have ingeneral. It is better for long term intellectual development to see measure and integration approachedwithout using such special structure. Also, all of the standard theorems we want to do are just as easyto prove in the abstract setting, so why specialize to <? So we tend to do abstract measure stuff first.The core material for Block 1 is as follows:

1. abstract measure ν on a sigma - algebra S of subsets of a universeX .

2. measurable functions with respect to a measure ν; these are also called random variables whenν is a probability measure.

3. integration∫fdν

4. convergence results: monotone convergence theorem, dominated convergence theorem etc.

Then we develop the Lebesgue Integral in <n via outer measures as the great example of a nontrivialmeasure. So Block 2 of material is thus

1. outer measures in <n

2. Caratheodory conditions for measurable sets

3. construction of the Lebesgue sigma algebra

4. connections to Borel sets

To fill out the course, we pick topics from the following

1. Riemann and Riemann - Stieljes integration. This would go before Block 1 if we do it. Call itblock Riemann.

2. Decomposition of measures – I love this material so this is after Block 2. Call it block Decom-position.

3. Connection to Riemann integration via absolute continuity of functions. this is actually hardstuff and takes about 3 weeks to cover nicely. Call it Block Riemann and Lebesgue. If thisis done without Block Riemann, you have to do a quick review of Riemann stuff so they canfollow the proofs.

4. Fubini type theorems. This would go after Block 2. Call this Block Fubini.

5. Differentiation via the Vitali approach. This is pretty hard too. Call this Differentiation.

6. Treatment of the usual Lp spaces. Call this Block Lp.

1.2. TEACHING THE MEASURE AND INTEGRATION COURSE 9

7. More convergence stuff like convergence in measure, Lp convergence implies convergence ofa subsequence pointwise etc. These are hard theorems and to do them right requires a lot oftime. Call this More Convergence.

We have taught this in at least the following ways. And always, lots of homework and projects,as we believe only hands on work really makes this stuff sink in.

Way 1: Block Riemann, Block 1, Block 2 and Block Decomposition.

Way 2: Block 1, Block 2, Block Decomposition and Block Riemann and Lebesgue.

Way 3: Block 1, Block 2, Block Decomposition and Differentiation.

Way 4: Block 1, Block 2, Block Lp, Block More Convergence and Block Decomposition.

Way 5: Block 1, Block 2, Block Fubini, Block More Convergence and Block Decomposition.

So as you can see it will be an interesting ride!

So we have a series of five books on analysis, one of which you are reading. Each course has aweb site for it which contains presentations and other material so feel free to download and use anymaterial on those sites. These courses are

• First Half of Primer One for Analysis: MATH 4530 here. The web site is at The Primer OnAnalysis I: Part One Home Page .

• Second Half of Primer One for Analysis: MATH 4540 here. The web site is at The PrimerOn Analysis I: Part Two Home Page .

• Second Primer on Analysis: this is material we do not teach: essentially calculus in <n ideas.The web site is at The Primer On Analysis II Home Page .

• Third Primer on Analysis: MATH 8210 here. The web site is at The Primer On Analysis III:Linear Analysis Home Page .

• Fourth Primer on Analysis: MATH 8220 here. The web site is at The Primer On AnalysisIV: Measure Theory Home Page .

• Fifth Primer on Analysis: MATH 9270 here. The web site is at The Primer On Analysis V:Functional Analysis and Topology Home Page .

All of these books will be published in 2019 as the set (Peterson (6) 2019), (Peterson (8) 2019),(Peterson (7) 2019), (Peterson (5) 2019) and (Peterson (4) 2019).

www.ces.clemson.edu/~petersj/AdvancedCalcOne.html

www.ces.clemson.edu/~petersj/AdvancedCalcOne.html

www.ces.clemson.edu/~petersj/AdvancedCalcTwo.html

www.ces.clemson.edu/~petersj/AdvancedCalcTwo.html

www.ces.clemson.edu/~petersj/BasicAnalysisTwo.html

www.ces.clemson.edu/~petersj/BasicAnalysisThree.html

www.ces.clemson.edu/~petersj/BasicAnalysisThree.html

www.ces.clemson.edu/~petersj/BasicAnalysisFour.html

www.ces.clemson.edu/~petersj/BasicAnalysisFour.html

www.ces.clemson.edu/~petersj/BasicAnalysisFive.html

www.ces.clemson.edu/~petersj/BasicAnalysisFive.html

Part II

Classical Riemann Integration

11

Chapter 2

An Overview Of RiemannIntegration

In this Chapter, we will give you a quick overview of Riemann integration. There are few real proofsbut it is useful to have a quick tour before we get on with the job of extending this material to amore abstract setting. Much of this material can be found in a good Calculus book although the moreadvanced stuff requires that you look at a book on beginning real analysis such as (Fulks (3) 1978).

2.1 Continuity

A function f defined on an interval [a, b] where a and b are finite numbers can be quite strange. Forexample, here is a legitimate function f defined on [−1, 1]:

f(t) =

1 if t is a rational number−1 if t is an irrational number (2.1)

We will assume you know what a rational and irrational number is! Now, this function is horriblyodd: it is not possible to graph it at all. But we can understand it intellectually. This function isnot differentiable or continuous at any points! We naturally want to study and use functions that aremuch better behaved than this. Since continuity and differentiability are pointwise concepts, to doanything useful for biological modeling, we usually want functions that satisfy the requirements forcontinuity and differentiability at all points in an entire interval. For example, the function f definedby

f(t) = 2t3 + 32 t + 16

is defined at all real numbers t and even is continuous and differentiable at each such t. Recall, thedefinition of [Continuity??] of a function f at a point p in its domain.

Definition 2.1.1 Continuity Of A Function At A Point: ε − δ Version

f is said to be continuous at a point p in its domain if given any tolerance ε there is arestriction δ on the values of t so that | f(t) − f(p) |< ε if t is in the domain of f and tsatisfies | t − p |< δ.

13

14 CHAPTER 2. RIEMANN OVERVIEW

Now, this definition is written in the very formal language of mathematics. We require such pre-cision so that we can be absolutely clear as to what we mean. However, if we are willing to burysome of this detail, this can be rephrased as

Definition 2.1.2 Continuity Of A Function At A Point: Limit Version

f is said to be continuous at a point p in its domain if several conditions hold:

1. f is actually defined at p

2. The limit as t approaches p of f exists

3. The value of the limit above matches the value f(p).

This is usually stated more succinctly as f(p) exists and limt→ p f(t) = f(p), but both ways ofsaying it mean the same. You should be able to understand that our polynomial f(t) = 2t3 +32t+16is continuous at all t values using these definitions. That is, you should have studied the ideas of limitsand continuity at this level of abstraction in your past exposure to this material.

If a function is continuous at a point p, the next question we can ask is about its [Differentiabil-ity??].

2.2 DifferentiabilityRecall the definition of differentiability of a function f at a point p.

Definition 2.2.1 Differentiability of A Function At A Point

f is said to be differentiable at a point p in its domain if the limit as t approaches p, t 6= p, ofthe quotients f(t)− f(p)

t− p exists. When this limit exists, the value of this limit is denoted by a

number of possible symbols: f ′(p) or dfdt (p). This can also be phrased in terms of the right

and left hand limits f ′(p+) = limt→ p+f(t)− f(p)

t− p f and f ′(p−) = limt→ p−f(t)− f(p)

t− p f .If both exist and match at p, then f ′(p) exists and the value of the derivative is the commonvalue.

A fundamental consequence of the existence of a derivative of a function at a point t is that it mustalso be continuous there. Remember that as it will be important in a lot of things later. We state thisas Theorem 2.2.1

Theorem 2.2.1 Differentiability Implies Continuity

Let f be a function which is differentiable at a point t in its domain. Then f is alsocontinuous at t.

You should have discussed this idea carefully in your previous classes so that you know about secantlines and the way they approximate the value of the limit in Definition 2.2.1 when it exists.

2.3 IntegrationYou should also have been exposed to the idea of the integration of a function f . There are twointellectually separate ideas here:

2.3. INTEGRATION 15

1. The idea of a [Primitive??] or antiderivative of a function f . This is any function F which isdifferentiable and satisfies F ′(t) = f(t) at all points in the domain of f . Normally, the domainof f is a finite interval of the form [a, b], although it could also be an infinite interval like all of< or [1,∞) and so on. Note that an antiderivative does not require any understanding of theprocess of Riemann integration at all – only what differentiation is!

2. The idea of the Riemann integral of a function. You should have been exposed to this in yourfirst Calculus course and perhaps a bit more rigorously in your undergraduate second semesteranalysis course.

Let’s review what Riemann Integration involves. First, we start with a bounded function f on afinite interval [a, b]. This kind of function f need not be continuous! Then select a finite number ofpoints from the interval [a, b], x0, x1, , . . . , xn−1, xn. We don’t know how many points thereare, so a different selection from the interval would possibly gives us more or less points. But forconvenience, we will just call the last point xn and the first point x0. These points are not arbitrary –x0 is always a, xn is always b and they are ordered like this:

x0 = a < x1 < x2 < . . . < xn−1 < xn = b

The collection of points from the interval [a, b] is called a Partition of [a, b] and is denoted bysome letter – here we will use the letter π. So if we say π is a partition of [a, b], we know it willhave n+ 1 points in it, they will be labeled from x0 to xn and they will be ordered left to right withstrict inequalities. But, we will not know what value the positive integer n actually is. The simplestPartition π is the two point partition a, b. Note these things also:

1. Each partition of n+ 1 points determines n subintervals of [a, b]

2. The lengths of these subintervals always adds up to the length of [a, b] itself, b− a.

3. These subintervals can be represented as

[x0, x1], [x1, x2], . . . , [xn−1, xn]

or more abstractly as [xi, xi+1] where the index i ranges from 0 to n− 1.

4. The length of each subinterval is xi+1 − xi for the indices i in the range 0 to n− 1.

Now from each subinterval [xi, xi+1] determined by the Partition π, select any point you wantand call it si. This will give us the points s0 from [x0, x1], s1 from [x1, x2] and so on up to the lastpoint, sn−1 from [xn−1, xn]. At each of these points, we can evaluate the function f to get the valuef(sj). Call these points an Evaluation Set for the partition π. Let’s denote such an evaluation setby the letter σ. Note there are many such evaluation sets that can be chosen from a given partitionπ. We will leave it up to you to remember that when we use the symbol σ, you must remember it isassociated with some partition.

If the function f was nice enough to be positive always and continuous, then the product f(si) ×(xi+1 − xi) can be interpreted as the area of a rectangle. Then, if we add up all these rectangle areaswe get a sum which is useful enough to be given a special name: the Riemann sum for the functionf associated with the Partition π and our choice of evaluation set σ = s0, . . . , sn−1. This sumis represented by the symbol S(f,π,σ) where the things inside the parenthesis are there to remindus that this sum depends on our choice of the function f , the partition π and the evaluations set σ.So formally, we have the definition


Definition 2.3.1 Riemann Sum

The Riemann sum for the bounded function f , the partition π and the evaluation set σ =s0, . . . , sn−1 from πx0, x1, , . . . , xn−1, xn is defined by

S(f,π,σ) =

n−1∑i=0

f(si) (xi+1 − xi)

It is pretty misleading to write the Riemann sum this way as it can make us think that the nis always the same when in fact it can change value each time we select a different partitionπ. So many of us write the definition this way instead

S(f,π,σ) =∑i ∈ π

f(si) (xi+1 − xi) =∑π

f(si) (xi+1 − xi)

and we just remember that the choice of π will determine the size of n.

2.3.1 A Riemann Sum ExampleLet’s look at an example of all this. In Figure 2.1, we see the graph of a typical function which isalways positive on some finite interval [a, b]

(a, f(a))(b, f(b))

a b

A generic curve f on the interval [a, b]which is always positive. Note the areaunder this curve is the shaded region.

Figure 2.1: The Area Under The Curve f

Next, let’s set the interval to be [1, 6] and compute the Riemann Sum for a particular choice ofPartition π and evaluation set π. This is shown in Figure 2.2.

We can also interpret the Riemann sum as an approximation to the area under the curve as shownin Figure 2.1. This is shown in Figure 2.3.

2.3.2 The Riemann Integral As A LimitWe can construct many different Riemann Sums for a given function f . To define the [Riemann˙Integral??]of f , we only need a few more things:

2.3. INTEGRATION 17

(1, f(1)) (6, f(6))

1 6

The partition is π =1.0, 1.5, 2.6, 3.8, 4.3, 5.6, 6.0. Hence, wehave subinterval lengths of x1 − x0 = 0.5,x2 − x1 = 1.1, x3 − x2 = 1.2,x4 − x3 = 0.5, x5 − x4 = 1.3 andx6 − x5 = 0.4, giving || P ||= 1.3. Thus,

S(f,π,σ) =

5∑i=0

f(si) (xi+1 − xi)

For the evaluation set σ = 1.1, 1.8, 3.0, 4.1, 5.3, 5.8 shown in red in Figure 2.2, we wouldfind the Riemann sum is

S(f,π,σ) = f(1.1)× 0.5

+ f(1.8)× 1.1

+ f(3.0)× 1.2

+ f(4.1)× 0.5

+ f(5.3)× 1.3

+ f(5.8)× 0.4

Of course, since our picture shows a generic f , we can’t actually put in the function valuesf(si)!

Figure 2.2: A Simple Riemann Sum

1. Each partition π has a maximum subinterval length – let’s use the symbol || π || to denote thislength. We read the symbol || π || as the norm or gauge of π.

2. Each partition π and evaluation set σ determines the number S(f,π,σ) by a simple calcula-tion.

3. So if we took a collection of partitions π1, π2 and so on with associated evaluation sets σ1, σ2

etc., we would construct a sequence of real numbers S(f,π1,σ1), S(f,π2,σ2), . . . , , S(f,πn,σn), . . . , .Let’s assume the norm of the partition πn gets smaller all the time; i.e. limn→∞ || πn ||= 0.We could then ask if this sequence of numbers converges to something.

What if the sequence of Riemann sums we construct above converged to the same number I nomatter what sequence of partitions whose norm goes to zero and associated evaluation sets we chose?Then, we would have that the value of this limit is independent of the choices above. This is indeedwhat we mean by the Riemann Integral of f on the interval [a, b].


(1, f(1)) (6, f(6))

1 6

The partition is π =1.0, 1.5, 2.6, 3.8, 4.3, 5.6, 6.0.

Figure 2.3: The Riemann Sum As An Approximate Area

Definition 2.3.2 Riemann Integrability Of A Bounded Function

Let f be a bounded function on the finite interval [a, b]. if there is a number I so that

limn→∞

S(f,πn,σn) = I

no matter what sequence of partitions πn with associated sequence of evaluation setsσn we choose as long as limn→∞ || πn ||= 0, we will say that the Riemann Integralof f on [a, b] exists and equals the value I .

The value I is dependent on the choice of f and interval [a, b]. So we often denote this value byI(f, [a, b]) or more simply as, I(f, a, b). Historically, the idea of the Riemann integral was developedusing area approximation as an application, so the summing nature of the Riemann Sum was denotedby the 16th century letter S which resembled an elongated or stretched letter S which looked likewhat we call the integral sign

∫. Hence, the common notation for the Riemann Integral of f on [a, b],

when this value exists, is∫ baf . We usually want to remember what the independent variable of f is

also and we want to remind ourselves that this value is obtained as we let the norm of the partitionsgo to zero. The symbol dx for the independent variable x is used as a reminder that xi+1 − xiis going to zero as the norm of the partitions goes to zero. So it has been very convenient to addto the symbol

∫ baf this information and use the augmented symbol

∫ baf(x) dx instead. Hence, if

the independent variable was t instead of x, we would use∫ baf(t) dt. Since for a function f , the

name we give to the independent variable is a matter of personal choice, we see that the choice ofvariable name we use in the symbol

∫ baf(t) dt is very arbitrary. Hence, it is common to refer to the

independent variable we use in the symbol∫ baf(t) dt as the dummy variable of integration.

We need a few more facts. We shall prove later the following things are true about the RiemannIntegral of a bounded function. First, we know when a bounded function actually has a Riemannintegral from Theorem 2.3.1.

2.3. INTEGRATION 19

Theorem 2.3.1 Existence Of The Riemann Integral

Let f be a bounded function on the finite interval [a, b]. Then the Riemann integral of f on[a, b],

∫ baf(t)dt exists if

1. f is continuous on [a, b]

2. f is continuous except at a finite number of points on [a, b].

Further, if f and g are both Riemann integrable on [a, b] and they match at all but a finitenumber of points, then their Riemann integrals match; i.e.

∫ baf(t)dt equals

∫ bag(t)dt.

The function given by Equation 2.1 is bounded but continuous nowhere on [−1, 1] and it is indeedpossible to prove it does not have a Riemann integral on that interval. However, most of the functionswe want to work with do have a lot of smoothness, i.e. continuity and even differentiability on theintervals we are interested in. Hence, Theorem 2.3.1 will apply. Here are some examples:

1. If f(t) is t2 on the interval [−2, 4], then∫ 4

−2f(t)dt does exist as f is continuous on this

interval.

2. If g was defined by

g(t) =

t2 −2 ≤ t < 1 and 1 < t ≤ 45 t = 1

we see g is not continuous at only one point and so it is Riemann integrable on [−2, 4]. More-over, since f and g are both integrable and match at all but one point, their Riemann integralsare equal.

However, with that said, in this course, we want to relax the smoothness requirements on thefunctions f we work with and define a more general type of integral for this less restricted class offunctions.

2.3.3 The Fundamental Theorem Of Calculus

There is a big connection between the idea of the antiderivative of a function f and its Riemannintegral. For a positive function f on the finite interval [a, b], we can construct the area under thecurve function F (x) =

∫ xaf(t)dt where for convenience we choose an x in the open interval (a, b).

We show F (x) and F (x + h) for a small positive h in Figure 2.4. Let’s look at the difference inthese areas:

F (x + h) − F (x) =

∫ x+h

a

f(t) dt −∫ x

a

f(t) dt

=

∫ x

a

f(t) dt +

∫ x+h

x

f(t) dt −∫ x

a

f(t) dt

=

∫ x+h

x

f(t) dt


where we have used standard properties of the Riemann integral to write the first integral as twopieces and then do a subtraction. Now divide this difference by the change in x which is h. We find

F (x + h) − F (x)

h=

1

h

∫ x+h

x

f(t) dt (2.2)

The difference in area,∫ x+h

xf(t) dt, is the second shaded area in Figure 2.4. Clearly, we have

F (x + h) − F (x) =

∫ x+h

x

f(t) dt (2.3)

We know that f is bounded on [a, b]; hence, there is a number B so that f(t) ≤ B for all t in [a, b].Thus, using Equation 2.3, we see

0 ≤ F (x + h) − F (x) ≤∫ x+h

x

B dt = B h (2.4)

From this we have

0 ≤ limh→ 0

(F (x + h) − F (x)) ≤ limh→ 0

B h

= 0

We conclude that F is continuous at each x in [a, b] as

limh→ 0

(F (x + h) − F (x)) = 0

It seems that the new function F we construct by integrating the function f in this manner, alwaysbuilds a new function that is continuous. Is F differentiable at x? If f is continuous at x, then givena positive ε, there is a positive δ so that

f(x)− ε < f(t) < f(x) + ε if x− δ < t < x+ δ

and t is in [a, b]. So, if h is less than δ, we have

1

h

∫ x+h

x

(f(x)− ε) <F (x + h) − F (x)

h=

1

h

∫ x+h

x

f(t) dt <1

h

∫ x+h

x

(f(x) + ε)

This is easily evaluated to give

f(x)− ε < F (x + h) − F (x)

h=

∫ x+h

x

f(t) dt < f(x) + ε

if h is less than δ. This shows that

limh→ 0+

F (x + h) − F (x)

h= f(x)

2.3. INTEGRATION 21

You should be able to believe that a similar argument would work for negative values of h: i.e.,

limh→ 0−

F (x + h) − F (x)

h= f(x)

This tells us that F ′(x) exists and equals f(x) as long as f is continuous at x as

F ′(x+) = limh→ 0+

F (x + h) − F (x)

h= f(x)

F ′(x−) = limh→ 0−

F (x + h) − F (x)

h= f(x)

This relationship is called [Fundamental˙Theorem˙Calculus??]. The same sort of argument worksfor x equals a or b but we only need to look at the derivative from one side. We will prove this sort oftheorem using fairly relaxed assumptions on f for the interval [a, b] in the later Chapters. Even if wejust consider the world of Riemann Integration, we only need to assume that f is Riemann Integrableon [a, b] which allows for jumps in the function.

Theorem 2.3.2 Fundamental Theorem Of Calculus

Let f be Riemann Integrable on [a, b]. Then the function F defined on [a, b] by F (x) =∫ xaf(t) dt satisfies

1. F is continuous on all of [a, b]

2. F is differentiable at each point x in [a, b] where f is continuous and F ′(x) = f(x).

Using the same f as before, suppose G was defined on [a, b] as follows

G(x) =

∫ b

x

f(t) dt.

Note that

F (x) + G(x) =

∫ x

a

f(t) dt +

∫ b

x

f(t) dt

=

∫ b

a

f(t) dt.

Since the Fundamental Theorem of Calculus tells us F is differentiable, we seeG(x) =∫ baf(t)dt−

F (x) must also be differentiable. It follows that

G′(x) = − F ′(x) = −f(x).

Let’s state this as a variant of the Fundamental Theorem of Calculus, the Reversed FundamentalTheorem of Calculus so to speak.


(a, f(a))

(b, f(b))

a b

x x + h

F (x) F (x + h)

A generic curve f on theinterval [a, b] which isalways positive. We letF (x) be the area under thiscurve from a to x. Thisis indicated by the shadedregion.

Figure 2.4: The Function F (x)

Theorem 2.3.3 Fundamental Theorem Of Calculus Reversed

Let f be Riemann Integrable on [a, b]. Then the function F defined on [a, b] by F (x) =∫ bxf(t) dt satisfies

1. F is continuous on all of [a, b]

2. F is differentiable at each point x in [a, b] where f is continuous and F ′(x) =−f(x).

2.3.4 The Cauchy Fundamental Theorem Of Calculus

We can use the Fundamental Theorem of Calculus to learn how to evaluate many Riemann integrals.Let G be an antiderivative of the function f on [a, b]. Then, by definition, G′(x) = f(x) and so weknow G is continuous at each x. But we still don’t know that f itself is continuous. However, if weassume f is continuous, then if we define F on [a, b] by

2.3. INTEGRATION 23

F (x) = f(a) +

∫ x

a

f(t) dt,

the Fundamental Theorem of Calculus, Theorem 2.3.2, is applicable. Thus, F ′(x) = f(x) at eachpoint. But that means F ′ = G′ = f at each point. Functions whose derivatives are the same mustdiffer by a constant. Call this constant C. We thus have F (x) = G(x) + C. So, we have

F (b) = f(a) +

∫ b

a

f(t)dt = G(b) + C

F (a) = f(a) +

∫ a

a

f(t)dt = G(a) + C

But∫ aaf(t) dt is zero, so we conclude after some rewriting

G(b) = f(a) +

∫ b

a

f(t)dt + C

G(a) = f(a) + C

And after subtracting, we find the important result

G(b) − G(a) =

∫ b

a

f(t)dt

This is huge! This is what tells us how to integrate many functions. For example, if f(t) = t3,we can guess the antiderivatives have the form t4/4 + C for an arbitrary constant C. Thus, sincef(t) = t3 is continuous, the result above applies. We can therefore calculate Riemann integrals likethese:

1. ∫ 3

1

t3 dt =t4

4

∣∣∣∣31

=34

4− 14

4

=80

4

2. ∫ 4

−2

t3 dt =t4

4

∣∣∣∣4−2

=44

4− (−2)4

4

=256

4− 16

4

=240

4


Let’s formalize this as a theorem called the [Cauchy˙FTC??]. All we really need to prove thisresult is that f is Riemann integrable on [a, b], which is true if our function f is continuous.

Theorem 2.3.4 Cauchy Fundamental Theorem Of Calculus

Let G be any antiderivative of the Riemann integrable function f on the interval [a, b].Then G(b) − G(a) =

∫ baf(t) dt.

2.3.5 Applications

With the Cauchy Fundamental Theorem of Calculus under our belt, we can guess a lot of antideriva-tives and from that know how to evaluate many Riemann integrals. Let’s get started.

1. It is easy to guess the antiderivative of a power of t as we have already mentioned. We knowthe antiderivative of the following are easy to figure out:

(a) If f(t) = t5, then the antiderivative of f is any function of the form F (t) = t6/6 + Cwhere C can be any constant.

(b) If f(t) = t−5, it is still easy to guess the antiderivative which is F (t) = t−4/(−4) + C,where C is an arbitrary constant.

The common symbol for the antiderivative of f has evolved to be∫f because of the close

connection between the antiderivative of f and the Riemann integral of f which is given inthe Cauchy Fundamental Theorem of Calculus, Theorem 2.3.4. The usual Riemann integral,∫ baf(t) dt of f on [a, b] computes a definite value – hence, the symbol

∫ baf(t) dt is usually

referred to as the definite integral of f on [a, b] to contrast it with the family of functionsrepresented by the antiderivative

∫f . Since the antiderivatives are arbitrary up to a constant,

most of us refer to the antiderivative as the indefinite integral of f . Also, we hardly ever say“let’s find the antiderivative of f” – instead, we just say, “let’s integrate f”. We will beginusing this shorthand now! We can state these results as Theorem 2.3.5.

Theorem 2.3.5 Antiderivatives Of Simple Powers

If p is any power other than−1, then the antiderivative of f(t) = tp is F (t) = tp+1/(p+1) + C. This is also expressed as

∫tp dt = tp+1/(p+ 1) + C

2. The Riemann integral of the function f on [a, b] can also be easily computed. We state thisTheorem 2.3.6

Theorem 2.3.6 Definite Integrals Of Simple Powers

If p is any power other than −1, then the definite integral of f(t) = tp on [a, b] is∫ batp dt = tp+1/(p+ 1)

∣∣∣∣ba

3. The simple trigonometric functions sin(t) and cos(t) also have straightforward antiderivativesas shown in Theorem 2.3.7.

Theorem 2.3.7 Antiderivatives of Simple Trigonometric Functions

2.3. INTEGRATION 25

(a) The antiderivative of sin(t) equals − cos(t) + C

(b) The antiderivative of cos(t) equals sin(t) + C

4. The definite integrals of the sin and cos functions are then:

Theorem 2.3.8 Definite Integrals Of Simple Trigonometric Functions

(a)∫ ba

sin(t) dt is − cos(t)

∣∣∣∣ba

(b)∫ ba

cos(t) is sin(t)

∣∣∣∣ba

2.3.6 Simple Substitution TechniquesWe can use the tools above to figure out how to integrate many functions that seem complicated butinstead are just disguised versions of simple power function integrations. Let’s go through some ingreat detail.

Exercise 2.3.1 Compute∫

(t2 + 1) 2t dt

Solution When you look at this integral, you should train yourself to see the simpler integral∫u du

where u(t) = t2 + 1. Here are the steps:

1. We make the change of variable u(t) = t2 + 1. Now differentiate both sides to see u′(t) = 2t.Thus, we have ∫

(t2 + 1) 2t dt =

∫u(t) u′(t) dt

2. Now recall the chain rule for powers of functions, we know((u(t))2

)′(t) = 2 u(t) u′(t)

Thus,

u(t) u′(t) =1

2

((u(t))2

)′(t)

This then tells us that ∫(t2 + 1) 2t dt =

∫u(t) u′(t) dt

=

∫1

2

((u(t))2

)′(t)dt

Now, the notation∫ (

(u(t))2)′

(t)dt is just our way of asking for the antiderivative of thefunction behind the integral sign. Here, that function is (u2)′. This antiderivative is, of course,just u2! Plugging that into the original problem, we find∫

(t2 + 1) 2t dt =

∫u(t) u′(t) dt


=

∫1

2

((u(t))2

)′(t)dt

=1

2u2(t) + C

=1

2(t2 + 1)2 + C

Whew!! That was awfully complicated looking. Let’s do it again in a bit more streamlinedfashion. Note all of the steps we go through below are the same as the longer version above, but sincewe write less detail down, it is much more compact. You need to get very good at understanding anddoing all these steps!! Here is the second version:

Solution

1. We make the change of variable u(t) = t2 + 1. But we write this more simply as u = t2 + 1so that the dependence of u on t is implied rather than explicitly stated. This simplifies ournotation already! Now differentiate both sides to see u′(t) = 2t. We will write this asdu = 2t dt, again hiding the t variable, using the fact that dudt = 2t can be written in itsdifferential form (you should have seen this idea in your first Calculus course). Thus, we have∫

(t2 + 1) 2t dt =

∫u du

2. The antiderivative of u is u2/2 + C and so we have∫(t2 + 1) 2t dt =

∫u du

=1

2u2 + C

=1

2(t2 + 1)2 + C

Now let’s try one a bit harder:


(t2 + 1)3 4tdt

Solution When you look at this integral, again you should train yourself to see the simpler integral2∫u3 du where u(t) = t2 + 1. Here are the steps: first, the detailed version

1. We make the change of variable u(t) = t2 + 1. Now differentiate both sides to see u′(t) = 2t.Thus, we have ∫

(t2 + 1)3 4tdt = 2

∫u3(t) u′(t) dt

2. Now recall the chain rule for powers of functions, we know((u(t))4

)′(t) = 4 u3(t) u′(t)

Thus,

2 u3(t) u′(t) = 21

4

((u(t))4

)′(t)

2.3. INTEGRATION 27

This then tells us that ∫(t2 + 1)3 4dt = 2

∫u3(t) u′(t) dt

=

∫1

2

((u(t))4

)′(t)dt

Now, the notation∫ (

(u(t))4)′

(t)dt is just our way of asking for the antiderivative of thefunction behind the integral sign. Here, that function is (u4)′. This antiderivative is, of course,just u4! Plugging that into the original problem, we find∫

(t2 + 1)3 4dt = 2

∫u3(t) u′(t) dt

=1

2u4(t) + C

=1

2(t2 + 1)4 + C

Again, this was awfully complicated looking. the streamlined version is as follows:

1. We make the change of variable u(t) = t2 + 1. Now differentiate both sides to see u′(t) = 2tand write this as du = 2t dt. Thus, we have∫

(t2 + 1)3 4dt = 2

∫u3 du

2. The antiderivative of u3 is u4/4 + C and so we have∫(t2 + 1)3 4dt = 2

∫u3 du

=1

2u4 + C

=1

2(t2 + 1)4 + C

Now let’s do one the short way only.

Exercise 2.3.3 Compute∫ √

t2 + 1 3t dt.

Solution When you look at this integral, again you should train yourself to see the simpler integral3/2

∫u1/2 du where u(t) = t2 + 1. Here are the steps: we know du = 2t dt. Thus∫ √

t2 + 1 3t dt =3

2

∫u

12 du

=3

2

132

u32 + C

=3

2

2

3(t2 + 1)

32 + C


sin(t2 + 1) 5t dt.



∫sin(u) du where u(t) = t2 + 1. Here are the steps: we know du = 2t dt. Thus∫

sin(t2 + 1) 5t dt =5

2

∫sin(u) du

=5

2(− cos(u)) + C

= −5

2cos(t2 + 1) + C

Now let’s do a definite integral:

Exercise 2.3.5 Compute∫ 5

1(t2 + 2t + 1)2 (t + 1) dt.


∫u2 du where u(t) = t2 + 2t + 1. Here are the steps: we know du = (2t + 2)dt. Thus∫ 5

1

(t2 + 2t + 1)2 (t + 1) dt =1

2

∫ t=5

t=1

u2 du

where we label the bottom and top limit of the integral in terms of the t variable to remind ourselvesthat the original integration was respect to t. Then,

1

2

∫ t=5

t=1

u2 du =1

2

u3

3|t=5t=1

=1

2

1

3(t2 + 1)3

∣∣∣∣51

=1

6

((26)3 − 23

)We will prove general substitution theorems for Riemann Integrable functions later. But it is

really just an application of the chain rule!

2.4 The Riemann Integral of Functions With JumpsNow let’s look at the Riemann integral of functions which have points of discontinuity.

2.4.1 Removable DiscontinuityConsider the function f defined on [−2, 5] by

f(t) =

2t −2 ≤ t < 01 t = 0(1/5)t2 0 < t ≤ 5

Let’s calculate F (t) =∫ t−2

f(s) ds. This will have to be done in several parts because of theway f is defined.

1. On the interval [−2, 0], note that f is continuous except at one point, t = 0. Hence, f isRiemann integrable by Theorem 2.3.1. Also, the function 2t is continuous on this intervaland so is also Riemann integrable. Then since f on [−2, 0] and 2t match at all but one point

2.4. HANDLING JUMPS 29

on [−2, 0], their Riemann integrals must match. Hence, if t is in [−2, 0], we compute F asfollows:

F (t) =

∫ t

−2

f(s) ds

=

∫ t

−2

2s ds

= s2

∣∣∣∣t−2

= t2 − (−2)2 = t2 − 4

2. On the interval [0, 5], note that f is continuous except at one point, t = 0. Hence, f is Riemannintegrable by Theorem 2.3.1. Also, the function (1/5)t2 is continuous on this interval and istherefore also Riemann integrable. Then since f on [0, 5] and (1/5)t2 match at all but onepoint on [0, 5], their Riemann integrals must match. Hence, if t is in [0, 5], we compute F asfollows:

F (t) =

∫ t

−2

f(s) ds

=

∫ 0

−2

f(s) ds +

∫ t

0

f(s) ds

=

∫ 0

−2

2s ds +

∫ t

0

(1/5)s2 ds

= s2

∣∣∣∣0−2

+ (1/15)s3

∣∣∣∣t0

= −4 + t3/15

Thus, we have found that

F (t) =

t2 − 4 −2 ≤ t < 0t3/15 − 4 0 < t ≤ 5

Note, we didn’t define F at t = 0 yet. Since f is Riemann Integrable on [−2, 5], we know fromthe Fundamental Theorem of Calculus, Theorem 2.3.2, that F must be continuous. Let’s check. F isclearly continuous on either side of 0 and we note that limt→ 0− F (t) which is F (0−) is −4 whichis exactly the value of F (0+). Hence, F is indeed continuous at 0 and we can write

F (t) =

t2 − 4 −2 ≤ t ≤ 0t3/15 − 4 0 ≤ t ≤ 5

What about the differentiability of F ? The Fundamental Theorem of Calculus guarantees that F hasa derivative at each point where f is continuous and at those points F ′(t) = f(t). Hence, we knowthis is true at all t except 0. Note at those t, we find

F ′(t) =

2t −2 ≤ t < 0(1/5)t2 0 < t ≤ 5


which is exactly what we expect. Also, note F ′(0−) = 0 and F ′(0+) = 0 as well. Hence, sincethe right and left hand derivatives match, we see F ′(0) does exist and has the value 0. But this is notthe same as f(0) = 1. Note, F is not the antiderivative of f on [−2, 5] because of this mismatch.

2.4.2 Jump DiscontinuityNow consider the function f defined on [−2, 5] by

f(t) =

2t −2 ≤ t < 01 t = 02 + (1/5)t2 0 < t ≤ 5

Let’s calculate F (t) =∫ t−2

f(s) ds. Again, this will have to be done in several parts because ofthe way f is defined.

1. On the interval [−2, 0], note that f is continuous except at one point, t = 0. Hence, f isRiemann integrable by Theorem 2.3.1. Also, the function 2t is continuous on this interval andhence is also Riemann integrable. Then since f on [−2, 0] and 2t match at all but one pointon [−2, 0], their Riemann integrals must match. Hence, if t is in [−2, 0], we compute F asfollows:

F (t) =

∫ t

−2

f(s) ds

=

∫ t

−2

2s ds

= s2

∣∣∣∣t−2

= t2 − (−2)2 = t2 − 4

2. On the interval [0, 5], note that f is continuous except at one point, t = 0. Hence, f is Riemannintegrable by Theorem 2.3.1. Also, the function 2 + (1/5)t2 is continuous on this interval andso is also Riemann integrable. Then since f on [0, 5] and 2 + (1/5)t2 match at all but onepoint on [0, 5], their Riemann integrals must match. Hence, if t is in [0, 5], we compute F asfollows:

F (t) =

∫ t

−2

f(s) ds

=

∫ 0

−2

f(s) ds +

∫ t

0

f(s) ds

=

∫ 0

−2

2s ds +

∫ t

0

(2 + (1/5)s2) ds

= s2

∣∣∣∣0−2

+ (2s + (1/15)s3)

∣∣∣∣t0

= −4 + 2t + t3/15


F (t) =

t2 − 4 −2 ≤ t < 0−4 + 2t + t3/15 0 < t ≤ 5

2.4. HANDLING JUMPS 31

As before, we didn’t define F at t = 0 yet. Since f is Riemann Integrable on [−2, 5], we knowfrom the Fundamental Theorem of Calculus, Theorem 2.3.2, that F must be continuous. F is clearlycontinuous on either side of 0 and we note that limt→ 0− F (t) which is F (0−) is−4 which is exactlythe value of F (0+). Hence, F is indeed continuous at 0 and we can write

F (t) =

t2 − 4 −2 ≤ t ≤ 0−4 + 2t + t3/15 0 ≤ t ≤ 5

What about the differentiability of F ? The Fundamental Theorem of Calculus guarantees that F hasa derivative at each point where f is continuous and at those points F ′(t) = f(t). Hence, we knowthis is true at all t except 0. Note at those t, we find

F ′(t) =

2t −2 ≤ t < 02 + (1/5)t2 0 < t ≤ 5

which is exactly what we expect. However, when we look at the one sided derivatives, we findF ′(0−) = 0 and F ′(0+) = 2. Hence, since the right and left hand derivatives do not match, wesee F ′(0) does not exist. Finally, note F is not the antiderivative of f on [−2, 5] because of thismismatch.

2.4.3 HomeworkExercise 2.4.1 Compute

∫ t−3

f(s) ds for

f(t) =

3t −3 ≤ t < 06 t = 0(1/6)t2 0 < t ≤ 6

1. Graph f and F carefully labeling all interesting points.

2. Verify that F is continuous and differentiable at all points but F ′(0) does not match f(0) andso F is not the antiderivative of f on [−3, 6]

Exercise 2.4.2 Compute∫ t

0f(s) ds for

f(t) =

−2t 2 ≤ t < 512 t = 53t − 25 5 < t ≤ 10


2. Verify that F is continuous and differentiable at all points but F ′(5) does not match f(5) andso F is not the antiderivative of f on [2, 10]

Exercise 2.4.3 Compute∫ t−3

f(s) ds for

f(t) =

3t −3 ≤ t < 06 t = 0(1/6)t2 + 2 0 < t ≤ 6


2. Verify that F is continuous and differentiable at all points except 0 and so F is not the an-tiderivative of f on [−3, 6]


Exercise 2.4.4 Compute∫ t

0f(s) ds for

f(t) =

−2t 2 ≤ t < 512 t = 53t 5 < t ≤ 10


2. Verify that F is continuous and differentiable at all points except 5 and so F is not the an-tiderivative of f on [2, 10]

Chapter 3

Functions Of Bounded Variation

Now that we have seen a quick overview of what Riemann Integration entails, let’s go back and lookat it very carefully. This will enable us to extend it to a more general form of integration calledRiemann - Stieljes. From what we already know about Riemann integrals, the Riemann integral isa mapping φ which is linear and whose domain is some subspace of the vector space of all boundedfunctions. LetB[a, b] denote this vector space which is a normed linear space using the usual infinitynorm. The set of all Riemann Integrable Functions can be denoted by the symbol RI[a, b] and weknow it is a subspace of B[a, b]. We also know that the subspace C[a, b] of all continuous functionson [a, b] is contained in RI[a, b]. In fact, if PC[a, b] is the set of all functions on [a, b] that arepiecewise continuous, then PC[a, b] is also a vector subspace contained in RI[a, b]. Hence, weknow φ : RI[a, b] ⊆ B[a, b] → < is a linear functional on the subspace RI[a, b]. Also, if f is notzero, then

|∫ baf(t) dt ||| f ||∞

≤∫ ba| f(t) | dt|| f ||∞

≤∫ ba|| f ||∞ dt

|| f ||∞= b− a

Thus, we see that || φ ||op is finite and φ is a bounded linear functional on a subspace of B[a, b] if weuse the infinity norm on RI[a, b]. But of course, we can choose other norms. There are clearly manyfunctions in B[a, b] that do not fit nicely into the development process for the Riemann Integral. Solet NI[a, b] denote a new subspace of functions which contains RI[a, b]. We know that the Riemannintegral satisfies an important idea in analysis called limit interchange. That is, if a sequence offunctions fn from RI[a, b] converges in infinity norm to f that the following facts hold:

1. f is also in RI[a, b]

2. the classic limit interchange holds:

limn→∞

∫ b

a

fn(t) dt =

∫ b

a

(lim

n→∞fn(t)

)dt

We can say this more abstractly as this: if fn → f in || · ||∞ in RI[a, b], then f remains inRI[a, b] and

limn→∞

φ (fn) = φ(

limn→∞

fn

)33

34 CHAPTER 3. BOUNDED VARIATION

But if we wanted to extend φ to the larger subspaceNI[a, b] in such a way that it remained a boundedlinear functional, we would also want to know what kind of sequence convergence we should use inorder for the interchange ideas to work. There are lots of questions:

1. Do we need to impose a norm on our larger subspace NI[a, b]?

2. Can we characterize the subspace NI[a, b] in some fashion?

3. If the extension is called φ, we want to make sure that φ is exactly φ when we restrict ourattention to functions in RI[a, b]

Also, do we have to develop integration only on finite intervals [a, b] of <? How do we evenextend traditional Riemann integration to unbounded intervals of <? All of these questions will beanswered in the upcoming chapters, but first we will see how far we can go with the traditionalRiemann approach. We will also see where the Riemann integral approach breaks down and makesus start to think of more general tools so that we can get our work done.

3.1 PartitionsDefinition 3.1.1 Partition

A partition of the finite interval [a, b] is a finite collection of points, x0, . . . , xn, orderedso that a = x0 < x1 < · · · < xn = b. We denote the partition by π and call each point xia partition point.

For each j = 1, . . . , n− 1, we let ∆xj = xj+1 − xj . The collection of all finite partitions of [a, b] isdenoted Π[a, b].

Definition 3.1.2 Partition Refinements

The partition π1 = y0, . . . , ym is said to be a refinement of the partition π2 =x0, . . . , xn if every partition point xj ∈ π2 is also in π1. If this is the case, then wewrite π2 π1, and we say that π1 is finer than π2 or π2 is coarser than π1.

Definition 3.1.3 Common Refinement

Given π1, π2 ∈ Π[a, b], there is a partition π3 ∈ Π[a, b] which is formed by taking theunion of π1 and π2 and using common points only once. We call this partition the commonrefinement of π1 and π2 and denote it by π3 = π1 ∨ π2.

Comment 3.1.1 The relation is a partial ordering of Π[a, b]. It is not a total ordering, since notall partitions are comparable. There is a coarsest partition, also called the trivial partition. It isgiven by π0 = a, b. We may also consider uniform partitions of order k. Let h = (b− a)/k. Thenπ = x0 = a, x0 + h, x0 + 2h, . . . , xk−1 = x0 + (k − 1)h, xk = b.

Proposition 3.1.1 Refinements and Common Refinements

If π1, π2 ∈ Π[a, b], then π1 π2 if and only if π1 ∨ π2 = π2.

Proof 3.1.1If π1 π2, then π1 = x0, . . . , xp ⊂ y0, . . . , yq = π2. Thus, π1 ∪ π2 = π2, and we haveπ1 ∨ π2 = π2. Conversely, suppose π1 ∨ π2 = π2. By definition, every point of π1 is also a point ofπ1 ∨ π2 = π2. So, π1 π2.

3.2. MONOTONE 35

Definition 3.1.4 The Gauge or Norm of a Partition

For π ∈ Π[a, b], we define the gauge of π, denoted ‖π‖, by ‖π‖ = max∆xj : 1 ≤ j ≤ p.

3.1.1 HomeworkExercise 3.1.1 Prove that the relation is a partial ordering of Π[a, b].

Exercise 3.1.2 Fix π1 ∈ Π[a, b]. The set C(π1) = π ∈ Π[a, b] : π1 π is called the coredetermined by π1. It is the set of all partitions of [a, b] that contain (or are finer than) π1.

1. Prove that if π1 π2, then C(π2) ⊂ C(π1).

2. Prove that if ‖π1‖ < ε, then ‖π‖ < ε for all π ∈ C(π1).

3. Prove that if ‖π1‖ < ε and π2 ∈ Π[a, b], then ‖π1 ∨ π2‖ < ε.

3.2 Monotone FunctionsIn our investigations of how monotone functions behave, we will need two fundamental facts aboutinfimum and supremum of a set of numbers which are given in Lemma 3.2.1 and Lemma 3.2.2.

Lemma 3.2.1 The Infimum Tolerance Lemma

Let S be a nonempty set of numbers that is bounded below. Then given any tolerance ε,there is at least one element s in S so that

inf(S) ≤ s < inf(S) + ε

Proof 3.2.1This is an easy proof by contradiction. Assume there is some ε so that no matter what s from S wechoose, we have

s ≥ inf(S) + ε

This says that inf(S) + ε is a lower bound for S and so by definition, inf(S) must be bigger than orequal to this lower bound. But this is clearly not possible. So the assumption that such a tolerance εexists is wrong and the conclusion follows.

and

Lemma 3.2.2 The Supremum Tolerance Lemma

Let T be a nonempty set of numbers that is bounded above. Then given any tolerance ε,there is at least one element t in T so that

sup(T ) − ε < t ≤ sup(T )

Proof 3.2.2This again is an easy proof by contradiction and we include it for completeness. Assume there issome ε so that no matter what t from T we choose, we have

t ≤ sup(T ) − ε


This says that sup(T ) − ε is an upper bound for T and so by definition, sup(T ) must be less than orequal to this upper bound. But this is clearly not possible. So the assumption that such a tolerance εexists is wrong and the conclusion must follow.

We are now in a position to discuss carefully monotone functions and other functions built fromthem. We follow discussions in (Douglas (2) 1996) at various places.

Definition 3.2.1 Monotone Functions

A real-valued function f : [a, b]→ R is said to be increasing (respectively, strictly increas-ing) if x1, x2 ∈ [a, b], x1 < x2 ⇒ f(x1) ≤ f(x2) (respectively, f(x1) < f(x2)). Similardefinitions hold for decreasing and strictly decreasing functions.

Theorem 3.2.3 A Monotone Function Estimate

Let f be increasing on [a, b], and let π = x0, . . . , xp be in Π[a, b]. For any c ∈ [a, b],define

f(c+) = limx→c+

f(x) and f(c−) = limx→c−

f(x),

where we define f(a−) = f(a) and f(b+) = f(b). Then

p∑j=0

[f(x+j )− f(x−j )] ≤ f(b)− f(a).

Proof 3.2.3First, we note that f(x+) and f(x−) always exist. The proof of this is straightforward. For x ∈ (a, b],let Tx = f(y) : a ≤ y < x. Then Tx is bounded above by f(x), since f is monotone increasing.Hence, Tx has a well-defined supremum. Let ε > 0 be given. Then, using the Supremum ToleranceLemma, Lemma 3.2.2, there is a y∗ ∈ [a, x) such that supTx − ε < f(y∗) ≤ supTx. For anyy ∈ (y∗, x), we have f(y∗) ≤ f(y) since f is increasing. Thus, 0 ≤ (supTx − f(y)) ≤ (supTx −f(y∗)) < ε for y ∈ (y∗, x). Let δ = (x− y∗)/2. Then, if 0 < x− y < δ, supTx − f(y) < ε. Sinceε was arbitrary, this shows that limy→x− f(y) = supTx. The proof for f(x+) is similar, using theInfimum Tolerance Lemma, Lemma 3.2.1. You should be able to see that f(x−) is less than or equalto f(x+) for all x. We will define f(a−) = f(a) and f(b+) = f(b) since f is not defined prior to aor after b.

To prove the stated result holds, first choose an arbitrary yj ∈ (xj , xj+1) for each j = 0, . . . , p−1. Then, since f is increasing, for each j = 1, . . . , p, we have f(yj−1) ≤ f(x−j ) ≤ f(x+

j ) ≤ f(yj).Thus,

f(x+j )− f(x−j ) ≤ f(yj)− f(yj−1). (3.1)

We also have f(a) ≤ f(a+) ≤ f(y0) and f(yp−1) ≤ f(b−) ≤ f(b). Thus, it follows that

p∑j=0

(f(x+

j )− f(x−j ))

= f(x+0 )− f(x−0 ) +

p−1∑j=1

[f(x+j − f(x−j )] + f(x+

p )− f(x−p )

≤ f(a+)− f(a−) +

p−1∑j=1

[f(yj − f(yj−1)] + f(b+)− f(b−)

3.2. MONOTONE 37

using Equation 3.1 and replacing x0 by a and xp with b. We then note the sum on the right hand sidecollapses to f(yp−1)− f(y0). Finally, since f(a−) = f(a) and f(b+) = f(b), we obtain

p∑j=0

(f(x+

j )− f(x−j ))≤ f(a+)− f(a) + f(yp−1)− f(y0) + f(b)− f(b−)

≤ f(y0)− f(a) + f(yp−1)− f(b−) + f(b)− f(y0)

≤ f(b)− f(a) + f(yp−1)− f(b−).

But f(yp−1)− f(b−) ≤ 0, so

p∑j=0

(f(x+

j )− f(x−j ))≤ f(b)− f(a).

Theorem 3.2.4 A Monotone Function Has A Countable Number of Discontinuities

If f is monotone on [a, b], the set of discontinuities of f is countable.

Proof 3.2.4For concreteness, we assume f is monotone increasing. The decreasing case is shown similarly.Since f is monotone increasing, the only types of discontinuities it can have are jump discontinuities.If x ∈ [a, b] is a point of discontinuity, then the size of the jump is given by f(x+) − f(x−). DefineDk = x ∈ (a, b) : f(x+) − f(x−) > 1/k, for each integer k ≥ 1. We want to show that Dk isfinite.

Select any finite subset S ofDk and label the points in S by x1, . . . , xp with x1 < x2 < · · · < xp.If we add the point x0 = a and xp+1 = b, these points determine a partition π. Hence, by Theorem3.2.3, we know that

p∑j=1

[f(x+j )− f(x−j )] ≤

∑π

[f(x+j )− f(x−j )] ≤ f(b)− f(a).

But each jump satisfies f(x+j )− f(x−j ) > 1/k and there are a total of p such points in S. Thus, we

must have

p/k <

p∑j=1

[f(x+j )− f(x−j )] ≤ f(b)− f(a).

Hence, p/k < f(b)− f(a), implying that p < k[f(b)− f(a)]. Thus, the cardinality of S is boundedabove by the fixed constant k[f(b) − f(a)]. Let N be the first positive integer bigger than or equalto k[f(b) − f(a)]. If the cardinality of Dk were infinite, then there would be a subset T of Dk withcardinality N + 1. The argument above would then tell us that N + 1 ≤ k[f(b)− f(a)] ≤ N givinga contradiction. Thus, Dk must be a finite set. This means that D = ∪∞k=1 Dk is countable also.

Finally, if x is a point where f is not continuous, then f(x+) − f(x−) > 0. Hence, there is apositive integer k0 so that f(x+)− f(x−) > 1/k0. This means x is in Dk0 and so is in D.

Definition 3.2.2 The Discontinuity Set Of A Monotone Function


Let f be monotone increasing on [a, b]. We will let S denote the set of discontinuities of fon [a, b]. We know this set is countable by Theorem 3.2.4 so we can label it as S = xj.Define functions u and v on [a, b] by

u(x) =

0, x = af(x)− f(x−), x ∈ (a, b]

v(x) =

f(x+)− f(x), x ∈ [a, b)0, x = b

In Figure 3.1, we show a monotone increasing function with several jumps. You should be ableto compute u and v easily at these jumps.

(a, f(a))

(b, f(b))

ax4 = bx1 x2 x3x4

33.54

77.58

10

11

12

15

16

A generic curve f on theinterval [a, b] which is al-ways positive. We showfour points of discontinuityx1, x2, x3 and x4. Noteu(x1) = 0.5, u(x2) = 0.5,u(x3) = 1 and u(x4) = 1.Also, we see v(x1) = 0.5,v(x2) = 0.5, v(x3) = 1and v(x4) = 0.

Figure 3.1: The Function F (x)

3.2. MONOTONE 39

There are several very important points to make about these functions u and v which are listedbelow.

Comment 3.2.1

1. Note that u(x) is the left-hand jump of f at x ∈ (a, b] and v(x) is the right-hand jump of f atx ∈ [a, b) .

2. Both u and v are non-negative functions and u(x) + v(x) = f(x+)− f(x−) is the total jumpin f at x, for x ∈ (a, b) .

3. Moreover, f is continuous at x from the left if and only if u(x) = 0, and f is continuous fromthe right at x if and only if v(x) = 0 .

4. Finally, f is continuous on [a, b] if and only if u(x) = v(x) = 0 on [a, b] .

Now, let S0 be any finite subset of S. From Theorem 3.2.3, we have

∑x∈S0

f(x+)− f(x−) ≤ f(b)− f(a)

This implies ∑x∈S0

u(x) + v(x) ≤ f(b)− f(a)

∑x∈S0

u(x) +∑x∈S0

v(x) ≤ f(b)− f(a).

The above tells us that the set of numbers we get by evaluating this sum over finite subsets of Sis bounded above by the number f(b) − f(a). Hence,

∑nj=1 u(xj) and

∑nj=1 v(xj) are bounded

above by f(b) − f(a) for all n. Thus, these sets of numbers have a finite supremum. But u and vare non-negative functions, so these sums form monotonically increasing sequences. Hence, thesesequences converge to their supremum which we label as

∑∞j=1 u(xj) and

∑∞j=1 v(xj).

Now, consider a nonempty subset, T , of [a, b], and suppose F ⊂ S ∩ T is finite. Then, by thearguments already presented, we know that

∑xj∈F

u(xj) +∑xj∈F

v(xj) ≤ f(b)− f(a). (3.2)

This implies

∑xj∈F

u(xj) ≤ f(b)− f(a) and∑xj∈F

v(xj) ≤ f(b)− f(a).

From this, it follows that

∑xj∈S∩T

u(xj) = sup∑xj∈F

u(xj) : F ⊂ S ∩ T, F finite.

Likewise, we also have


∑xj∈S∩T

v(xj) = sup∑xj∈F

v(xj) : F ⊂ S ∩ T, F finite.

Definition 3.2.3 The Saltus Function Associated With A Monotone Function

For x, y ∈ [a, b] with x < y, define

S[x, y] = S ∩ [x, y], S[x, y) = S ∩ [x, y), S(x, y] = S ∩ (x, y] and S(x, y) = S ∩ (x, y)

Then, define the function Sf : [a, b]→ R by

Sf (x) =

f(a), x = af(a) +

∑xj∈S(a,x] u(xj) +

∑xj∈S[a,x) v(xj), a < x ≤ b

We call Sf the Saltus Function associated with the monotone increasing function f .

Intuitively, Sf (x) is the sum of all of the jumps (i.e. discontinuities) up to and including the left-handjump at x. In essence, it is a generalization of the idea of a step function.

Theorem 3.2.5 Properties of The Saltus Function

Let f : [a, b]→ R be monotone increasing. Then

1. Sf is monotone increasing on [a, b];

2. if x < y, with x, y ∈ [a, b], then 0 ≤ Sf (y)− Sf (x) ≤ f(y)− f(x);

3. Sf is continuous on Sc ∩ [a, b] where Sc is the complement of the set S.

Proof 3.2.5Suppose x < y. Then

Sf (y)− Sf (x) =∑

xj∈S(a,y]

u(xj) −∑

xj∈S(a,x]

u(xj) +∑

xj∈S[a,y)

v(xj) −∑

xj∈S[a,x)

v(xj)

=∑

xj∈S(x,y]

u(xj) +∑

xj∈S[x,y)

v(xj)

≥ 0.

This proves the first statement. Now, suppose x, y ∈ [a, b] with x < y. Let F be a subset of [a, b]that consists of a finite number of points of the form F = x0 = x, x1, . . . , xp = y, such thatx = x0 < x1 < · · · < xp = y. In other words, F is a partition of [x, y]. Then, by Equation 3.2 weknow ∑

xj∈F∩S(x,y]

u(xj) +∑

xj∈F∩S[x,y)

v(xj) ≤ f(y)− f(x).

Taking the supremum of the left-hand side over all such sets, F , we obtain∑xj∈S(x,y]

u(xj) +∑

xj∈S[x,y)

v(xj) ≤ f(y)− f(x).

3.2. MONOTONE 41

But by the remarks made in the first part of this proof, this sum is exactly Sf (y)−Sf (x). We concludethat Sf (y)− Sf (x) ≤ f(y)− f(x) as desired.

Finally, let x be a point in Sc ∩ [a, b]. Then f is continuous at x, so, given ε > 0, there is a δ > 0such that y ∈ [a, b] and |x − y| < δ ⇒ |f(x) − f(y)| < ε. But by the second part of this proof, wehave |Sf (x)− Sf (y)| ≤ |f(y)− f(x)| < ε. Thus, Sf is continuous at x.

So, why do we care about Sf? The function Sf measures, in a sense, the degree to which f failsto be continuous. If we subtract Sf from f , we would be subtracting its discontinuities, resulting ina continuous function that behaves similarly to f .

Definition 3.2.4 The Continuous Part of A Monotone Function

Define fc : [a, b]→ R by fc(x) = f(x)− Sf (x).

Theorem 3.2.6 Properties of fc

1. fc is monotone on [a, b].

2. fc is continuous also.

Proof 3.2.6The proof that fc is monotone is left to you as an exercise with this generous hint:

Hint 3.2.1 Note if x < y in [a, b], then

fc(y) − fc(x) = (f(y)− f(x)) − (Sf (y)− Sf (x)) .

The right hand side is non negative by Theorem 3.2.5.

To prove fc is continuous is a bit tricky. We will do most of the proof but leave a few parts for youto fill in.

Pick any x in [a, b) and any positive ε. Since the f(x+) exists, there is a positive δ so that0 ≤ f(y)− f(x+) < ε if x < y < x+ δ. Thus, for such y,

fc(y) − fc(x) = [f(y)− Sf (y)] − [f(x)− Sf (x)]

= f(y) −

∑xj∈S(a,y]

u(xj) +∑

xj∈S[a,y)

v(xj)

− f(x) +

∑xj∈S(a,x]

u(xj) +∑

xj∈S[a,x)

v(xj).

Recall, S(a, y] = S(a, x] ∪ S(x, y] and S[a, y) = S[a, x) ∪ S[x, y). So,

fc(y) − fc(x) = f(y) −

∑xj∈S(x,y]

u(xj) +∑

xj∈S[x,y)

v(xj)

− f(x)

Now, the argument reduces to two cases:


1. if y and x are points of discontinuity, we get

fc(y) − fc(x) = f(y)− u(y)−

∑xj∈S(x,y)

u(xj) +∑

xj∈S(x,y)

v(xj)

− f(x)− v(x)

= f(y) − (f(y)− f(y−))−

∑xj∈S(x,y)

u(xj) +∑

xj∈S(x,y)

v(xj)

− f(x)− (f(x+)− f(x))

≤ f(y−)− f(x+)

≤ f(y)− f(x+) < ε

2. if either x and/ or y are not a point of discontinuity, a similar argument holds

Thus, we see fc is continuous from the right at this x. Now use a similar argument to show continuityfrom the left at x. Together, these arguments show fc is continuous at x.

3.2.1 Worked Out ExampleLet’s define f on [0, 2] by

f(x) =

−2 x = 0x3 0 < x < 19/8 x = 1x4/4 + 1 1 < x < 27 x = 2

1. Find u and v

2. Find Sf

3. Find fc

4. Following the discussion in Section 2.4 explain how to compute the Riemann Integral of f andfind its value (yes, this is in the careful rigorous section and so this problem is a bit out ofplace, but we will be dotting all of our i’s and crossing all of our t’s soon enough!)

Solution First, note f(0−) = −2, f(0) = −2 and f(0+) = 0 and so 0 is a point of discontinuity.Further, f(1−) = 1, f(1) = 9/8 and f(1+) = 5/4 giving another point of discontinuity at 1.Finally, since f(2−) = 5, f(2) = 7 and f(2+) = 7, there is a third point of discontinuity at 2. So,the set of discontinuities of f is S = 0, 1, 2. Thus,

S(0, x] =

∅ 0 < x < 11 1 ≤ x < 21, 2 2 = x

and S[0, x) =

0 0 < x ≤ 10, 1 1 < x ≤ 2

Also,

u(x) =

0 x = 00 0 < x < 19/8− 1 = 1/8 x = 10 1 < x < 27− 5 = 2 2 = x

and v(x) =

0− (−2) = 2 x = 00 0 < x < 15/4− 9/8 = 1/8 x = 10 1 < x < 20 2 = x

3.2. MONOTONE 43

Now, here

Sf (x) =

f(0) = −2, x = 0f(0) +

∑xj∈S(0,x] u(xj) +

∑xj∈S[0,x) v(xj) 0 < x ≤ 2

Thus,

Sf (x) =

−2, x = 0−2 + v(0) = −2 + 2 = 0 0 < x < 1−2 + u(1) + v(0) = −2 + 1/8 + 2 = 1/8 x = 1−2 + u(1) + v(0) + v(1) = −2 + 1/8 + 2 + 1/8 = 1/4 1 < x < 2−2 + u(1) + u(2) + v(0) + v(1) = −2 + 1/8 + 2 + 2 + 1/8 = 9/4 x = 2

So, Sf is the nice step function and fc = f − Sf gives

Sf (x) =

−2, x = 00 0 < x < 11/8 x = 11/4 1 < x < 29/4 x = 2

and fc(x) =

−2− (−2) = 0 x = 0x3 − 0 = x/3 0 < x < 19/8− 1/8 = 1 x = 1x4/4 + 1− 1/4 = x4/4 + 3/4 1 < x < 27− 9/4 = 19/4 x = 2

We see fc is continuous on [0, 2]. Finally, we can compute the Riemann integral of f on [0, 2].Let’s calculate F (t) =

∫ t0f(x) dx. This will have to be done in several parts because of the

way f is defined.

1. On the interval [0, 1], note that f is continuous except at two points, x = 0 and x = 1. Hence,f is Riemann integrable by Theorem 2.3.1. Also, the function x3 is continuous on this intervaland so is also Riemann integrable. Then since f on [0, 1] and x3 match at all but two points on[0, 2], their Riemann integrals must match. Hence, if t is in [−2, 0], we compute F as follows:

F (t) =

∫ t

0

f(x) dx

=

∫ t

0

x3 dx

= x4/4

∣∣∣∣t0

= t4/4

2. On the interval [1, 2], note that f is continuous except at the two points, x = 1 and x = 2.Hence, f is Riemann integrable by Theorem 2.3.1. Also, the function 1 + x4/4 is continuouson this interval and so is also Riemann integrable. Then since f on [1, 2] and 1 + x4/4 matchat all but two points on [1, 2], their Riemann integrals must match. Hence, if t is in [1, 2], wecompute F as follows:

F (t) =

∫ t

0

f(x) dx

=

∫ 1

0

f(x) dx +

∫ t

1

f(s) ds

=

∫ 1

0

x3 dx +

∫ t

1

(1 + x4/4) dx


= x4/4 |10 + (x + x5/5) |t1= 1/4 + (t + t5/5) − (1 + 1/5)

= t5/5 + t − 19/20


F (t) =

t4/4 0 ≤ t ≤ 1t5/5 + t − 19/20 1 ≤ t ≤ 2

Note, we know from the Fundamental Theorem of Calculus, Theorem 2.3.2, that F must be continu-ous. To check this at an interesting point such as t = 1, note F is clearly continuous on either sideof 1 and we note that limt→ 1− F (t) which is F (1−) is 1/4 which is exactly the value of F (1+).Hence, F is indeed continuous at 1!What about the differentiability of F? The Fundamental Theorem of Calculus guarantees that Fhas a derivative at each point where f is continuous and at those points F ′(t) = f(t). Hence, weknow this is true at all t except 0, 1 and 2 because these are points of discontinuity of f . F ′ is nicelydefined at 0 and 1 as a one sided derivative and at all other t save 1 by

F ′(t) =

t3 0 ≤ t < 1t4 + 1 0 < t ≤ 2

However, when we look at the one sided derivatives, we find F ′(0+) = 0 6= f(0) = −2,F ′(2−) = 17 6= f(2) = 7 and F ′(1−) = 1 and F ′(1+) = 2 giving F ′(1) does not even exist.Thus, note F is not the antiderivative of f on [0, 2] because of this mismatch.

3.2.2 Homework

Exercise 3.2.1 Prove fc is monotone.

Exercise 3.2.2 Let’s define f on [0, 2] by

f(x) =

−1 x = 0x2 0 < x < 17/4 x = 1√x+ 3 1 < x < 2

3 x = 2

1. Find u and v

2. Find Sf

3. Find fc

4. Do a nice graph of u, v, f , fc and Sf

5. Following the discussion in Section 2.4 explain how to compute the Riemann Integral of f andfind its value (yes, this is in the careful rigorous section and so this problem is a bit out ofplace, but we will be dotting all of our i’s and crossing all of our t’s soon enough!)

3.3. BOUNDED VARIATION 45

3.3 Functions of Bounded VariationThe next important topic for us is to consider the class of functions of bounded variation. We willdevelop this classically here, but in later chapters, we will define similar concepts using abstractmeasures. We are going to find out that functions of bounded variation can also be represented as thedifference of two increasing functions and that there classical derivative exists everywhere except aset of measure zero (yes, that idea is not defined yet, but we believe in teasers!). Let’s get on with it.

Definition 3.3.1 Functions Of Bounded Variation

Let f : [a, b] → R and let π ∈ Π[a, b] be given by π = x0 = a, x1, . . . , xp = b. Define∆fj = f(xj) − f(xj−1) for 1 ≤ j ≤ p. If there exists a positive real number, M , suchthat ∑

π

|∆fj | ≤M

for all π ∈ Π[a, b], then we say that f is of bounded variation on [a, b]. The set of allfunctions of bounded variation on the interval [a, b] is denoted by the symbol BV [a, b].

Comment 3.3.1

1. Note saying a function f is of bounded variation is equivalent to saying the set ∑π |∆fj | :

π ∈ Π[a, b] is bounded, and, therefore, has a supremum.

2. Also, if f is of bounded variation on [a, b], then, for any x ∈ (a, b), the set a, x, b is apartition of [a, b]. Hence, there existsM > 0 such that | f(x)−f(a) | + | f(b)−f(x) |≤M .But this implies

| f(x) | − | f(a) | ≤ | f(x)− f(a) | + | f(b)− f(x) | ≤ M

This tells us that | f(x) | ≤ | f(a) | + M . Since our choice of x in [a, b] was arbitrary, thisshows that f is bounded, i.e. || f ||∞ < ∞.

We can state the comments above formally as Theorem 3.3.1.

Theorem 3.3.1 Functions Of Bounded Variation Are Bounded

If f is of bounded variation on [a, b], then f is bounded on [a, b].

Theorem 3.3.2 Monotone Functions Are Of Bounded Variation

If f is monotone on [a, b], then f ∈ BV [a, b].

Proof 3.3.1As usual, we assume, for concreteness, that f is monotone increasing. Let π ∈ Π[a, b]. Hence, wecan write π = x0 = a, x1, . . . , xp−1, xp = b. Then∑

π

| ∆fj |=∑π

| f(xj)− f(xj−1) | .

Since f is monotone increasing, the absolute value signs are unnecessary, so that∑π

| ∆fj |=∑π

∆fj =∑π

(f(xj)− f(xj−1)

).


But this is a telescoping sum, so∑π

∆fj = f(xp)− f(x0) = f(b)− f(a).

Since the partition π was arbitrary, it follows that∑π ∆fj ≤ f(b) − f(a) for all π ∈ Π[a, b].

This implies that f ∈ BV [a, b], for if f(b) > f(a), then we can simply let M = f(b) − f(a). Iff(b) = f(a), then f must be constant, and we can let M = f(b) − f(a) + 1 = 1. In either case,f ∈ BV [a, b].

Theorem 3.3.3 Bounded Differentiable Implies Bounded Variation

Suppose f ∈ C[a, b], f is differentiable on (a, b), and || f ′ ||∞<∞. Then f ∈ BV [a, b].

Proof 3.3.2Let π ∈ Π[a, b] so that π = x0 = a, x1, . . . , xp = b. On each subinterval [xj−1, xj ], for1 ≤ j ≤ p, the hypotheses of the Mean Value Theorem are satisfied. Hence, there is a point,yj ∈ (xj−1, xj), with ∆fj = f(xj)− f(xj−1) = f ′(yj)∆xj . So, we have

| ∆fj |=| f ′(yj) | ∆xj ≤ B∆xj ,

where B is the bound on f ′ that we assume exists by hypothesis. Thus, for any π ∈ Π[a, b], we have∑π

| ∆fj |≤ B∑π

∆xj = B(b− a) <∞.

Therefore, f ∈ BV [a, b].

Definition 3.3.2 The Total Variation Of A Function Of Bounded Variation

Let f ∈ BV [a, b]. The real number

V (f ; a, b) = sup

∑π

| ∆fj | : π ∈ Π[a, b]

is called the Total Variation of f on [a, b].

Note that this number always exists if f ∈ BV [a, b].

Comment 3.3.2 For any f ∈ BV [a, b], we clearly have V (f ; a, b) = V (−f ; a, b) and V (f ; a, b) ≥0. Moreover, we also see that V (f ; a, b) = 0 if and only if f is constant on [a, b].

Theorem 3.3.4 Functions Of Bounded Variation Are Closed Under Addition

If f and g are in BV [a, b], then so are f ± g, and V (f ± g; a, b) ≤ V (f ; a, b) +V (g; a, b).

Proof 3.3.3Let π ∈ Π[a, b], so that π = x0 = a, x1, . . . , xp = b. Consider f + g first. We have, for each1 ≤ j ≤ p,

| ∆(f + g)j | = | (f + g)(xj)− (f + g)(xj−1) |≤ | f(xj)− f(xj−1) | + | g(xj)− g(xj−1) |≤ | ∆fj | + | ∆gj | .

3.3. BOUNDED VARIATION 47

This implies that, for any π ∈ Π[a, b],∑π

| ∆(f + g)j |≤∑π

| ∆fj | +∑π

| ∆gj | .

Both quantities on the right-hand side are bounded by V (f ; a, b) and V (g; a, b), respectively. Sinceπ ∈ Π[a, b] was arbitrary, we have

V (f + g; a, b) ≤ V (f ; a, b) + V (g; a, b).

This shows that f+g ∈ BV [a, b] and proves the desired inequality for that case. Since V (−g; a, b) =V (g; a, b), we also have

V (f − g; a, b) ≤ V (f ; a, b) + V (−g; a, b) = V (f ; a, b) + V (g; a, b),

which proves that f − g ∈ BV [a, b].

Theorem 3.3.5 Products Of Functions Of Bounded Variation Are Of Bounded Variation

If f, g ∈ BV [a, b], then fg ∈ BV [a, b] and V (fg; a, b) ≤ || g ||∞ V (f ; a, b)+ || f ||∞V (g; a, b).

Proof 3.3.4By Theorem 3.3.1, we know that f and g are bounded. Hence, the numbers || f ||∞ and || g ||∞ existand are finite. Let h = fg, and let π = x0 = a, x1, . . . , xp = b be any partition. Then

| ∆hj | = | f(xj)g(xj)− f(xj−1)g(xj−1) |= | f(xj)g(xj)− g(xj)f(xj−1) + g(xj)f(xj−1)− f(xj−1)g(xj−1) |≤ | g(xj) || ∆fj | + | f(xj−1) || ∆gj |≤ || g ||∞| ∆fj | + || f ||∞| ∆gj |

Thus, ∑π

| ∆hj | ≤ || g ||∞∑π

| ∆fj | + || f ||∞∑π

| ∆gj |

≤ || g ||∞ V (f ; a, b) + || f ||∞ V (g; a, b)

Since π was arbitrary, we see the right hand side is an upper bound for all the partition sums andhence, the supremum of all these sums must also be less than or equal to the right hand side. Thus,

V (fg; a, b) ≤ || g ||∞ V (f ; a, b) + || f ||∞ V (g; a, b)

Comment 3.3.3 Note that we have verified that BV [a, b] is a commutative algebra (i.e. a ring) offunctions with an identity, since the constant function f = 1 is of bounded variation.

It is natural to ask, then, what the units are in this algebra. That is, what functions have multiplicativeinverses?

Theorem 3.3.6 Inverses Of Functions Of Bounded Variation


Let f be in BV [a, b], and assume that there is a positive m such that | f(x) | ≥ m > 0 forall x ∈ [a, b]. Then 1/f ∈ BV [a, b] and V (1/f ; a, b) ≤ (1/m2)V (f ; a, b).

Proof 3.3.5Let π = x0 = a, x1, . . . , xp be any partition. Then

∣∣∣∣∆( 1

f

)j

∣∣∣∣ =

∣∣∣∣ 1

f(xj)− 1

f(xj−1)

∣∣∣∣=

∣∣∣∣f(xj−1)− f(xj)

f(xj)f(xj−1)

∣∣∣∣=

| ∆fj || f(xj) || f(xj−1) |

≤ ∆fjm2

.

Thus, we have

∑π

| ∆( 1

f

)j| ≤ 1

m2

∑π

| ∆fj |

implying that V (1/f ; a, b) ≤ (1/m2)V (f ; a, b).

Comment 3.3.4

1. Any polynomial, p, is in BV [a, b], and p is a unit if none of its zeros occur in the interval.

2. Any rational function p/q where p and q are of bounded variation on [a, b], is in BV [a, b] aslong as none of the zeros of q occur in the interval.

3. ex ∈ BV [a, b]. In fact, eu(x) ∈ BV [a, b] if u(x) is monotone or has a bounded derivative.

4. sinx and cosx are in BV [a, b] by Theorem 3.3.3.

5. tanx ∈ BV [a, b] if [a, b] does not contain any point of the form (2k + 1)π/2 for k ∈ Z, byTheorem 3.3.6.

6. The function

f(x) =

sin 1

x , 0 < x ≤ 10, x = 0

is not in BV [0, 1]. To see this, choose partition points x0, . . . , xp by x0 = 0, xp = 1, and

xj =2

π(2p− 2j + 1), 1 ≤ j ≤ p− 1.

Then

∆f1 = sin(π(2p− 1)

2

)= ±1,

3.4. TOTAL VARIATION 49

∆f2 = sin(π(2p− 3)

2

)− sin

(π(2p− 1)

2

)= ±2,

and continuing, we find

∆fp−1 = sin(π(2p− 2(p− 1) + 1)

2

)− sin

(π(2p− 2(p− 2) + 1)

2

)= ±2,

∆fp = sin 1− sin(3π/2) = sin 1 + 1.

Thus,

∑π

= | ∆f1 | +

p−1∑j=2

| ∆fj | + sin 1 + 1

= 2(p− 1) + sin 1.

Hence, we can make the value of this sum as large as we desire and so this function is not ofbounded variation.

3.3.1 HomeworkExercise 3.3.1 Prove that if f is of bounded variation on the finite interval [a, b], then α f is also ofbounded variation for any scalar α. Do this proof using the partition approach.

Exercise 3.3.2 Prove that if f and g are of bounded variation on the finite interval [a, b], then αf+βgis also of bounded variation for any scalars α and β. Do this proof using the partition approach.Note, these two exercises essentially show BV [a, b] is a vector space.

Exercise 3.3.3 Prove BV [a, b] is a complete normed linear space with norm || · || defined by

|| f || = | f(a) | + V (f, a, b)

Exercise 3.3.4 Define f on [0, 1] by

f(x) =

x2 cos(x−2) x 6= 0 ∈ [0, 1]0 x = 0

Prove that f is differentiable on [0, 1] but is not of bounded variation. This is a nice example ofsomething we will see later. This f is a function which is continuous but not absolutely continuous.

3.4 The Total Variation FunctionTheorem 3.4.1 The Total Variation Is Additive On Intervals

If f ∈ BV [a, b] and c ∈ [a, b], then f ∈ BV [a, c], f ∈ BV [c, b], and V (f ; a, b) =V (f ; a, c) + V (f ; c, b). That is, the total variation, V , is additive on intervals.

Proof 3.4.1The case c = a or c = b is easy, so we assume c ∈ (a, b). Let π1 ∈ Π[a, c] and π2 ∈ Π[c, b] withπ1 = x0 = a, x1, . . . , xp = c and π2 = y0 = c, y1, . . . , yq = b. Then π1 ∨ π2 is a partition of[a, b] and we know


∑π1∨π2

| ∆fj |=∑π1

| ∆fj | +∑π2

| ∆fj |≤ V (f ; a, b).

Dropping the π2 term, and noting that π1 ∈ Π[a, c] was arbitrary, we see that

supπ1∈Π[a,c]

∑π1

| ∆fj |≤ V (f ; a, b),

which implies that V (f ; a, c) ≤ V (f ; a, b) < ∞. Thus, f ∈ BV [a, c]. A similar argument showsthat V (f ; c, b) ≤ V (f ; a, b), so f ∈ BV [c, b].

Finally, since both π1 and π2 were arbitrary and we know that

∑π1

| ∆fj | +∑π2

| ∆fj |≤ V (f ; a, b),

we see that V (f ; a, c) + V (f ; c, b) ≤ V (f ; a, b).

Now we will establish the reverse inequality. Let π ∈ Π[a, b], so that π = x0 = a, x1, . . . , xp =b. First, assume that c is a partition point of π, so that c = xk0 for some k0. Thus, π =x0, . . . , xk0 ∪ xk0 , . . . , xp. Let π1 = x0, . . . , xk0 ∈ Π[a, c] and let π2 = xk0 , . . . , xp ∈Π[c, b]. From the first part of our proof, we know that f ∈ BV [a, c] and f ∈ BV [c, b], so

∑π

| ∆fj | =∑π1

| ∆fj | +∑π2

| ∆fj |

≤ V (f ; a, c) + V (f ; c, b).

Since π ∈ Π[a, b] was arbitrary, it follows that V (f ; a, b) ≤ V (f ; a, c) + V (f ; c, b). For the othercase, suppose c is not a partition point of π. Then c must lie inside one of the subintervals. That is,c ∈ (xk0−1, xk0) for some k0. Let π′ = x0, . . . , xk0−1, c, xk0 , . . . , xp be a new partition of [a, b].Then π′ refines π. Apply our previous argument to conclude that

∑π′

| ∆fj |≤ V (f ; a, c) + V (f ; c, b).

Finally, we note that ∑π

| ∆fj |≤∑π′

| ∆fj |,

since| f(xk0)− f(xk0−1) |≤| f(xk0)− f(c) | + | f(c)− f(xk0−1) | .

Thus, we have

∑π

| ∆fj |≤ V (f ; a, c) + V (f ; c, b).

Since π was arbitrary, it follows that V (f ; a, b) ≤ V (f ; a, c) + V (f ; c, b). Combining these twoinequalities, we see the result is established.

Definition 3.4.1 The Variation Function Of a Function f Of Bounded Variation

3.4. TOTAL VARIATION 51

Let f ∈ BV [a, b]. The Variation Function of f , or simply the Variation of f , is thefunction Vf : [a, b]→ < defined by

Vf (x) =

0, x = aV (f ; a, x), a < x ≤ b

Theorem 3.4.2 Vf and Vf − f Are Monotone For a Function f of Bounded Variation

If f ∈ BV [a, b], then the functions Vf and Vf − f are monotone increasing on [a, b].

Proof 3.4.2Pick x1, x2 ∈ [a, b] with x1 < x2. By Theorem 3.4.1, f ∈ BV [a, x1] and f ∈ BV [a, x2]. Apply thissame theorem to the interval [a, x1] ∪ [x1, x2] to conclude that f ∈ BV [x1, x2]. Thus

Vf (x2) = V (f ; a, x2) = V (f ; a, x1) + V (f ;x1, x2) = Vf (x1) + V (f ;x1, x2).

It follows that Vf (x2) − Vf (x1) = V (f ;x1, x2) ≥ 0, so Vf is monotone increasing. Now, consider(Vf − f)(x2)− (Vf − f)(x1). We have

(Vf − f)(x2)− (Vf − f)(x1) = Vf (x2)− Vf (x1)− (f(x2)− f(x1))

= V (f ; a, x2)− V (f ; a, x1)− (f(x2)− f(x1))

= V (f ;x1, x2)− (f(x2)− f(x1)).

But x1, x2 is the trivial partition of [x1, x2], so

∑x1,x2

| ∆fj | ≤ supπ∈Π[x1,x2]

∑π

| ∆fj |= V (f ;x1, x2).

Thus, V (f ;x1, x2)− (f(x2)− f(x1)) ≥ 0, implying that Vf − f is monotone increasing.

Theorem 3.4.3 A Function Of Bounded Variation Is The Difference of Two Increasing Func-tions

Every f ∈ BV [a, b] can be written as the difference of two monotone increasing functionson [a, b]. In other words,

BV [a, b] = f : [a, b]→ < | ∃u, v : [a, b]→ <, u, v monotone increasing, f = u− v.

Proof 3.4.3If f = u− v, where u and v are monotone increasing, then u and v are of bounded variation. SinceBV [a, b] is an algebra, it follows that f ∈ BV [a, b].

Conversely, suppose f ∈ BV [a, b], and let u = Vf and v = Vf − f . Then u and v are monotoneincreasing and u− v = f .

Comment 3.4.1 Theorem 3.4.3 tells us if g is of bounded variation on [a, b], then g = u − v whereu and v are monotone increasing. Thus, we can also use the Saltus decomposition of u and v to


conclude

f = (uc + Su) − (vc + Sv)

= (uc − vc) + (Su − Sv)

The first term is the difference of two continuous functions of bounded variation and the second termis the difference of Saltus functions. This is essentially another form of decomposition theorem for afunction of bounded variation.

3.5 Continuous Functions of Bounded VariationTheorem 3.5.1 Functions Of Bounded Variation Always Possess Right and Left Hand Limits

Let f ∈ BV [a, b]. Then the limit f(x+) exists for all x ∈ [a, b) and the limit f(x−) existsfor all x ∈ (a, b].

Proof 3.5.1By Theorem 3.4.2, Vf and Vf −f are monotone increasing. So Vf (x+) and (Vf −f)(x+) both exist.Hence,

f(x+) = limx→x+

f(x)

= limx→x+

[Vf (x)− (Vf − f)(x)]

= limx→x+

Vf (x) + limx→x+

(Vf − f)(x)

= Vf (x+) + (Vf − f)(x+).

So, f(x+) exists. A similar argument shows that f(x−) exists.

Theorem 3.5.2 Functions Of Bounded Variation Have Countable Discontinuity Sets

If f ∈ BV [a, b], then the set of discontinuities of f is countable.

Proof 3.5.2f = u−v where u and v are monotone increasing. By Theorem 3.4.3, S1 = x ∈ [a, b]|uis not continuous atxand S2 = x ∈ [a, b] | vis not continuous atx are countable. The union of these sets is the set of allthe points of possible discontinuity of f , so the set of discontinuities of f is countable.

Theorem 3.5.3 A Function Of Bounded Variation Continuous If and Only If Vf Is Continuous

Let f ∈ BV [a, b]. Then f is continuous at c ∈ [a, b] if and only if Vf is continuous at c.

Proof 3.5.3The case where c = a and c = b are easier, so we will only prove the case where c ∈ (a, b). First,suppose f is continuous at c. We will prove separately that Vf is continuous from the right at c andfrom the left at c.

Let ε > 0 be given. Since f is continuous at c, there is a positive δ such that if x is in (c− δ, c+δ) ⊂ [a, b], then | f(x)− f(c) |< ε/2. Now,

3.5. CONTINUOUS ALSO 53

V (f ; c, b) = supπ∈Π[c,b]

∑π

| ∆fj | .

So, there is a partition π0 such that

V (f ; c, b)− ε

2<

∑π0

| ∆fj |≤ V (f ; c, b). (∗)

If π0′ is any refinement of π0, we see that∑

π0

| ∆fj |≤∑π0′

| ∆fj |,

since adding points to π0 simply increases the sum. Thus,

V (f ; c, b)− ε

2<

∑π0

| ∆fj | ≤∑π0′

| ∆fj | ≤ V (f ; c, b)

for any refinement π0′ of π0. Now, choose a partition, π1 which refines π0 and satisfies || π1 ||< δ.

Then

V (f ; c, b)− ε

2<

∑π1

| ∆fj |≤ V (f ; c, b). (∗∗)

So, if π1 = x0 = c, x1, . . . , xp, then | x1 − x0 |< δ. Thus, we have | x1 − c |< δ. It follows that| f(x1)− f(c) |< ε/2. From Equation ∗∗, we then have

V (f ; c, b)− ε

2<

∑π1

| ∆fj |

= | f(x1)− f(c) | +∑

rest of π1

| ∆fj |

<ε

2+

∑rest of π1

| ∆fj |

<ε

2+ V (f ;x1, b).

So, we see that

V (f ; c, b)− ε

2<ε

2+ V (f ;x1, b),

which implies that

V (f ; c, b)− V (f ;x1, b) <ε

2+ε

2= ε.

But V (f ; c, b)− V (f ;x1, b) = V (f ; c, x1) which is the same as Vf (x1)− Vf (c). Thus, we have

Vf (x1)− Vf (c) < ε.


Now Vf is monotone and hence we have shown that if x ∈ (c, x1),

Vf (x)− Vf (c) ≤ Vf (x1)− Vf (c) < ε.

Since ε > 0 was arbitrary, this verifies the right continuity of Vf at c.The argument for left continuity is similar. We can find a partition π1 of [a, c] with partition points

x0 = a, x1, . . . , xp−1, xp = c such that || π1 ||< δ and

V (f ; a, c)− ε

2< | f(c)− f(xp−1) | +

∑rest of π1

| ∆fj |

≤ | f(c)− f(xp−1) | +V (f ; a, xp−1).

Since || π1 ||< δ, we see as before that | f(c)− f(xp−1) |< ε/2. Thus,

V (f ; a, c)− ε

2<ε

2+ V (f ; a, xp−1),

and it follows that

V (f ; a, c)− V (f ; a, xp−1) < ε,

or

Vf (c)− Vf (xp−1) < ε.

Since Vf is monotone, we then have for any x in (xp−1, c) that

Vf (c)− Vf (x) < Vf (c)− Vf (xp−1) < ε

which shows the left continuity of Vf at c. Hence, Vf is continuous at c.Conversely, suppose Vf is continuous at c ∈ (a, b). Given ε > 0, there is a positive δ such that

(c− δ, c+ δ) ⊂ [a, b] and | Vf (x)− Vf (c) |< ε for all x ∈ (c− δ, c+ δ). Pick any x ∈ (c, c+ δ).Then c, x is a trivial partition of [c, x]. Hence

0 ≤| f(x)− f(c) |≤ V (f ; c, x) = V (f ; a, x)− V (f ; a, c)

or

0 ≤| f(x)− f(c) |≤ Vf (x)− Vf (c) < ε.

Hence, it follows that f is continuous from the right. A similar argument shows that f is continuousfrom the left.

We immediately have this corollary.

Theorem 3.5.4 f Continuous and Bounded Variation If and Only If Vf and Vf − f Are Contin-uous and Increasing

f ∈ C[a, b] ∩ BV [a, b] if and only if Vf and Vf − f are monotone increasing and contin-uous.

Chapter 4

The Theory Of Riemann Integration

We will now develop the theory of the Riemann Integral for a bounded function f on the interval[a, b]. We followed the development of this material in (Fulks (3) 1978) closely at times, althoughFulks does not cover some of the sections very well.

4.1 Defining The Riemann IntegralDefinition 4.1.1 The Riemann Sum

Let f ∈ B[a, b], and let π ∈ Π[a, b] be given by π = x0 = a, x1, . . . , xp = b. Defineσ = s1, . . . , sp, where sj ∈ [xj−1, xj ] for 1 ≤ j ≤ p. We call σ an evaluation set,and we denote this by σ ⊂ π. The Riemann Sum determined by the partition π and theevaluation set σ is defined by

S(f,π,σ) =∑π

f(sj)∆xj

Definition 4.1.2 Riemann Integrability Of a Bounded f

We say f ∈ B[a, b] is Riemann Integrable on [a, b] if there exists a real number, I , suchthat for every ε > 0 there is a partition, π0 ∈ Π[a, b] such that

| S(f,π,σ)− I |< ε

for any refinement, π, of π0 and any evaluation set, σ ⊂ π. We denote this value, I , by

I ≡ RI(f ; a, b)

We denote the set of Riemann integrable functions on [a, b] by RI[a, b]. Also, it is readily seen thatthe numberRI(f ; a, b) in the definition above, when it exists, is unique. So we can speak of RiemannIntegral of a function, f . We also have the following conventions.

1. RI(f ; a, b) = −RI(f ; b, a)

2. RI(f ; a, a) = 0

3. f is called the integrand.

55

56 CHAPTER 4. RIEMANN INTEGRATION

Theorem 4.1.1 RI[a, b] Is A Vector Space and RI(f ; a, b) Is A Linear Mapping

RI[a, b] is a vector space over < and the mapping IR : RI[a, b]→ < defined by

IR(f) = RI(f ; a, b)

is a linear mapping.

Proof 4.1.1Let f1, f2 ∈ RI[a, b], and let α, β ∈ <. For any π ∈ Π[a, b] and σ ⊂ π, we have

S(αf1 + βf2,π,σ) =∑π

(αf1 + βf2)(sj)∆xj

= α∑π

f1(sj)∆xj + β∑π

f2(sj)∆xj

= αS(f1,π,σ) + βS(f2,π,σ).

Since f1 is Riemann integrable, given ε > 0, there is a real number I1 = RI(f1, a, b) and a partitionπ1 ∈ Π[a, b] such that

| S(f1,π,σ)− I1 | <ε

2(| α | +1)(∗)

for all refinements, π, of π1, and all σ ⊂ π.

Likewise, since f2 is Riemann integrable, there is a real number I2 = RE(f2; a, b) and a partitionπ2 ∈ Π[a, b] such that

| S(f2,π,σ)− I2 | <ε

2(| β | +1)(∗∗)

for all refinements, π, of π2, and all σ ⊂ π.

Let π0 = π1 ∨π2. Then π0 is a refinement of both π1 and π2. So, for any refinement, π, of π0, andany σ ⊂ π, we have Equation ∗ and Equation ∗∗ are valid. Hence,

| S(f1,π,σ)− I1 | <ε

2(| α | +1)

| S(f2,π,σ)− I2 | <ε

2(| β | +1).

Thus, for any refinement π of π0 and any σ ⊂ π, it follows that

| S(αf1 + βf2,π,σ)− (αI1 + βI2) | = | αS(f1,π,σ) + βS(f2,π,σ)− αI1 − βI2 |≤ | α || S(f1,π,σ)− I1 | + | β || S(f2,π,σ)− I2 |

< | α | ε

2(| α | +1)+ | β | ε

2(| β | +1)< ε.

4.1. DEFINITION 57

This shows that αf1+βf2 is Riemann integrable and that the value of the integralRI(αf1+βf2; a, b)is given by αRI(f1; a, b) + βRI(f2; a, b). It then follows immediately that IR is a linear mapping.

Theorem 4.1.2 Fundamental Riemann Integral Estimates

Let f ∈ RI[a, b]. Let m = infx f(x) and let M = supx f(x). Then

m(b− a) ≤ RI(f ; a, b) ≤M(b− a).

Proof 4.1.2If π ∈ Π[a, b], then for all σ ⊂ π, we see that∑

π

m∆xj ≤∑π

f(sj)∆xj ≤∑π

M∆xj .

But∑π ∆xj = b− a, so

m(b− a) ≤∑π

f(sj)∆xj ≤M(b− a),

or

m(b− a) ≤ S(f,π,σ) ≤M(b− a),

for any partition π and any σ ⊂ π.Now, let ε > 0 be given. Then there exist π0 ∈ Π[a, b] such that for any refinement, π, of π0 and

any σ ⊂ π,

RI(f ; a, b)− ε < S(f,π,σ) < RI(f ; a, b) + ε.

Hence, for any such refinement, π, and any σ ⊂ π, we have

m(b− a) ≤ S(f,π,σ) < RI(f ; a, b) + ε

and

M(b− a) ≥ S(f,π,σ) > RI(f ; a, b)− ε.

Since ε > 0 is arbitrary, it follows that

m(b− a) ≤ RI(f ; a, b) ≤M(b− a).

Theorem 4.1.3 The Riemann Integral Is Order Preserving

The Riemann integral is order preserving. That is, if f, f1, f2 ∈ RI[a, b], then

(i)f ≥ 0⇒ RI(f ; a, b) ≥ 0;

(ii)f1 ≤ f2 ⇒ RI(f1; a, b) ≤ RI(f2; a, b).


Proof 4.1.3If f ≥ 0 on [a, b], then infx f(x) = m ≥ 0. Hence, by Theorem 4.1.2∫ b

a

f(x)dx ≥ m(b− a) ≥ 0.

This proves the first assertion. To prove (ii), let f = f2 − f1. Then f ≥ 0, and the second resultfollows from the first.

4.2 The Existence of the Riemann Integral

Although we have a definition for what it means for a bounded function to be Riemann integrable,we still do not actually know that RI[a, b] is nonempty! In this section, we will show how we provethat the set of Riemann integrable functions is quite rich and varied.

Definition 4.2.1 Darboux Upper and Lower Sums

Let f ∈ B[a, b]. Let π ∈ Π[a, b] be given by π = x0 = a, x1, . . . , xp = b. Define

mj = infxj−1≤x≤xj

f(x) 1 ≤ j ≤ p,

andMj = sup

xj−1≤x≤xjf(x) 1 ≤ j ≤ p.

We define the Lower Darboux Sum by

L(f,π) =∑π

mj∆xj

and the Upper Darboux Sum by

U(f,π) =∑π

Mj∆xj .

Comment 4.2.1

1. It is straightforward to see that

L(f,π) ≤ S(f,π,σ) ≤ U(f,π)

for all π ∈ Π[a, b].

2. We also have

U(f,π)− L(f,π) =∑π

(Mj −mj)∆xj .

Theorem 4.2.1 π π′ Implies L(f,π) ≤ L(f,π′) and U(f,π) ≥ U(f,π′)

If π π′, that is, if π′ refines π, then L(f,π) ≤ L(f,π′) and U(f,π) ≥ U(f,π′).

4.2. EXISTENCE 59

Proof 4.2.1The general result is established by induction on the number of points added. It is actually quite aninvolved induction. Here are some of the details:

Step 1 We prove the proposition for inserting points z1, . . . , zq into one subinterval of π. Theargument consists of

1. The Basis Step where we prove the proposition for the insertion of a single point into onesubinterval.

2. The Induction Step where we assume the proposition holds for the insertion of q pointsinto one subinterval and then we show the proposition still holds if an additional point isinserted.

3. With the Induction Step verified, the Principle of Mathematical Induction then tells us thatthe proposition is true for any refinement of π which places points into one subintervalof π.

Basis:

Proof Let π ∈ Π[a, b] be given by x0 = a, x1, . . . , xp = b. Suppose we form the refine-ment, π′, by adding a single point x′ to π. into the interior of the subinterval [xk0−1, xk0 ].Let

m′ = inf[xk0−1,x′]

f(x)

m′′ = inf[x′,xk0 ]

f(x).

Note that mk0 = minm′,m′′ and

mk0∆xk0 = mk0(xk0 − xk0−1)

= mk0(xk0 − x′) +mk0(x′ − xk0−1)

≤ m′′(xk0 − x′) +m′(x′ − xk0−1)

≤ m′′∆x′′ +m′∆x′,

where ∆x′′ = xk0 − x′ and ∆x′ = x′ − xk0−1. It follows that

L(f,π′) =∑j 6=k0

mj∆xj +m′∆x′ +m′′∆x′′

≥∑j 6=k0

mj∆xj +mk0∆xk0

≥ L(f,π).

Induction:

Proof We assume that q points z1, . . . , zq have been inserted into the subinterval [xk0−1, xk0 ].Let π′ denote the resulting refinement of π. We assume that

L(f,π) ≤ L(f,π′)


let the additional point added to this subinterval be called x′ and call π′′ the resulting refine-ment of π′. We know that π′ has broken [xk0−1, xk0 ] into q + 1 pieces. For convenience ofnotation, let’s label these q + 1 subintervals as [yj−1, yj ] where y0 is xk0−1 and yq+1 is xk0and the yj values in between are the original zi points for appropriate indices. The new pointx′ is thus added to one of these q+1 pieces, call it [yj0−1, yj0 ] for some index j0. This intervalplays the role of the original subinterval in the proof of the em Basis Step. An argument similarto that in the proof of the Basis Step then shows us that

L(f,π′) ≤ L(f,π′′)

Combining with the first inequality from the Induction hypothesis, we establish the result. Thus,the Induction Step is proved.

Step 2 Next, we allow the insertion of a finite number of points into a finite number of subintervalsof π. The induction is now on the number of subintervals.

1. The Basis Step where we prove the proposition for the insertion of points into one subin-terval.

2. The Induction Step where we assume the proposition holds for the insertion of points intoq subintervals and then we show the proposition still holds if an additional subintervalhas points inserted.

3. With the Induction Step verified, the Principle of Mathematical Induction then tells usthat the proposition is true for any refinement of π which places points into any numberof subintervals of π.

Basis

Proof Step 1 above gives us the Basis Step for this proposition.

Induction

Proof We assume the results holds for p subintervals and show it also holds when one moresubinterval is added. Specifically, let π′ be the refinement that results from adding points to psubintervals of π. Then the Induction hypothesis tells us that

L(f,π) ≤ L(f,π′)

Let π′′ denote the new refinement of π which results from adding more points into one moresubinterval of π. Then π′′ is also a refinement of π′ where all the new points are added to onesubinterval of π′. Thus, Step 1 holds for the pair (π′,π′′). We see

L(f,π′) ≤ L(f,π′′)

and the desired result follows immediately.

A similar argument establishes the result for upper sums.

4.2. EXISTENCE 61

Theorem 4.2.2 L(f,π1) ≤ U(f,π2)

Let π1 and π2 be any two partitions in Π[a, b]. Then L(f,π1) ≤ U(f,π2).

Proof 4.2.6Let π = π1 ∨ π2 be the common refinement of π1 and π2. Then, by the previous result, we have

L(f,π1) ≤ L(f,π) ≤ U(f,π) ≤ U(f,π2).

Theorem 4.2.2 then allows us to define a new type of integrability for the bounded function f .We begin by looking at the infimum of the upper sums and the supremum of the lower sums for agiven bounded function f .

Theorem 4.2.3 The Upper And Lower Darboux Integral Are Finite

Let f ∈ B[a, b]. Let L = L(f,π) |π ∈ Π[a, b] and U = U(f,π) |π ∈ Π[a, b]. De-fine L(f) = sup L , and U(f) = inf U . Then L(f) and U(f) are both finite. Moreover,L(f) ≤ U(f).

Proof 4.2.7By Theorem 4.2.2, the set L is bounded above by any upper sum for f . Hence, it has a finitesupremum and so sup L is finite. Also, again by Theorem 4.2.2, the set U is bounded below by anylower sum for f . Hence, inf U is finite. Finally, since L(f) ≤ U(f,π) and U(f) ≥ L(f,π) for allπ, by definition of the infimum and supremum of a set of numbers, we must have L(f) ≤ U(f).

Definition 4.2.2 Darboux Lower And Upper Integrals

Let f be in B[a, b]. The Lower Darboux Integral of f is defined to be the finite numberL(f) = sup L , and the Upper Darboux Integral of f is the finite number U(f) = inf U .

We can then define what is meant by a bounded function being Darboux Integrable on [a, b].

Definition 4.2.3 Darboux Integrability

Let f be in B[a, b]. We say f is Darboux Integrable on [a, b] if L(f) = U(f). Thecommon value is then called the Darboux Integral of f on [a, b] and is denoted by thesymbol DI(f ; a, b).

Comment 4.2.2 Not all bounded functions are Darboux Integrable. Consider the function f :[0, 1]→ < defined by

f(t) =

1 t ∈ [0, 1] and is rational−1 t ∈ [0, 1] and is irrational

You should be able to see that for any partition of [0, 1], the infimum of f on any subinterval is always−1 as any subinterval contains irrational numbers. Similarly, any subinterval contains rationalnumbers and so the supremum of f on a subinterval is 1. Thus U(f,π) = 1 and L(f,π) = −1for any partition π of [0, 1]. It follows that L(f) = −1 and U(f) = 1. Thus, f is bounded but notDarboux Integrable.

Definition 4.2.4 Riemann’s Criterion for Integrability


Let f ∈ B[a, b]. We say that Riemann’s Criteria holds for f if for every positive ε thereexists a π0 ∈ Π[a, b] such that U(f,π)− L(f,π) < ε for any refinement, π, of π0.

Theorem 4.2.4 The Riemann Integral Equivalence Theorem

Let f ∈ B[a, b]. Then the following are equivalent.

(i) f ∈ RI[a, b].

(ii) f satisfies Riemann’s Criteria.

(iii) f is Darboux Integrable, i.e, L(f) = U(f), and RI(f ; a, b) = DI(f ; a, b).

Proof 4.2.8

(i)⇒ (ii)

Proof Assume f ∈ RI[a, b], and let ε > 0 be given. Let IR be the Riemann integral of f over [a, b].Choose π0 ∈ Π[a, b] such that | S(f,π,σ) − IR |< ε/3 for any refinement, π, of π0 and anyσ ⊂ π. Let π be any such refinement, denoted by π = x0 = a, x1, . . . , xp = b, and let mj ,Mj

be defined as usual. Using the Infimum and Supremum Tolerance Lemmas, we can conclude that, foreach j = 1, . . . , p, there exist sj , tj ∈ [xj−1, xj ] such that

Mj −ε

6(b− a)< f(sj) ≤Mj

mj ≤ f(tj) < mj +ε

6(b− a).

It follows that

f(sj)− f(tj) > Mj −ε

6(b− a)−mj −

ε

6(b− a).

Thus, we have

Mj −mj −ε

3(b− a)< f(sj)− f(tj).

Multiply this inequality by ∆xj to obtain

(Mj −mj)∆xj −ε

3(b− a)∆xj <

(f(sj)− f(tj)

)∆xj .

Now, sum over π to obtain

U(f,π)− L(f,π) =∑π

(Mj −mj)∆xj

<ε

3(b− a)

∑π

∆xj +∑π

(f(sj)− f(tj)

)∆xj .

This simplifies to

∑π

(Mj −mj)∆xj −ε

3<

∑π

(f(sj)− f(tj)

)∆xj . (∗)

4.2. EXISTENCE 63

Now, we have

|∑π

(f(sj)− f(tj)

)∆xj | = |

∑π

f(sj)∆xj −∑π

f(tj)∆xj |

= |∑π

f(sj)∆xj − IR+ IR−∑π

f(tj)∆xj |

≤ |∑π

f(sj)∆xj − IR | + |∑π

f(tj)∆xj − IR |

= | S(f,π,σs)− IR | + | S(f,π,σt)− IR |,

where σs = s1, . . . , sp and σt = t1, . . . , tp are evaluation sets of π. Now, by our choice ofpartition π, we know

| S(f,π,σs)− IR | <ε

3

| S(f,π,σt)− IR | <ε

3.

Thus, we can conclude that

|∑π

(f(sj)− f(tj)

)∆xj |<

2ε

3.

Applying this to the inequality in Equation ∗, we obtain∑π

(Mj −mj)∆xj < ε.

Now, π was an arbitrary refinement of π0, and ε > 0 was also arbitrary. So this shows that fsatisfies Riemann’s condition.

(ii)⇒ (iii)

Proof Now, assume that f satisfies Riemann’s criteria, and let ε > 0 be given. Then there is apartition, π0 ∈ Π[a, b] such that U(f,π) − L(f,π) < ε for any refinement, π, of π0. Thus, by thedefinition of the upper and lower Darboux integrals, we have

U(f) ≤ U(f,π) < L(f,π) + ε ≤ L(f) + ε.

Since ε is arbitrary, this shows that U(f) ≤ L(f). The reverse inequality has already been estab-lished. Thus, we see that U(f) = L(f).

(iii)⇒ (i)

Proof Finally, assume f is Darboux integral which means L(f) = U(f). Let ID denote the valueof the Darboux integral. We will show that f is also Riemann integrable according to the definitionand that the value of the integral is ID.

Let ε > 0 be given. Now, recall that

ID = L(f) = supπL(f,π)

= U(f) = infπU(f,π)


Hence, by the Supremum Tolerance Lemma, there exists π1 ∈ Π[a, b] such that

ID − ε = L(f)− ε < L(f,π1) ≤ L(f) = ID

and by the Infimum Tolerance Lemma, there exists π2 ∈ Π[a, b] such that

ID = U(f) ≤ U(f,π2) < U(f) + ε = ID + ε.

Let π0 = π1 ∨π2 be the common refinement of π1 and π2. Now, let π be any refinement of π0, andlet σ ⊂ π be any evaluation set. Then we have

ID − ε < L(f,π1) ≤ L(f,π0) ≤ L(f,π) ≤ S(f,π,σ) ≤ U(f,π) ≤ U(f,π0) ≤ U(f,π2) < ID + ε.

Thus, it follows that

ID − ε < S(f,π,σ) < ID + ε.

Since the refinement, π, of π0 was arbitrary, as were the evaluation set, σ, and the tolerance ε, itfollows that for any refinement, π, of π0 and any ε > 0, we have

| S(f,π,σ)− ID |< ε.

This shows that f is Riemann Integrable and the value of the integral is ID.

Comment 4.2.3 By Theorem 4.2.4, we now know that the Darboux and Riemann integral are equiv-alent. Hence, it is now longer necessary to use a different notation for these two different approachesto what we call integration. From now on, we will use this notation

RI(f ; a, b) ≡ DI(f ; a, b) ≡∫

f(t) dt

where the (t) in the new integration symbol refers to the name we wish to use for the independentvariable and dt is a mnemonic to remind us that the || π || is approaching zero as we chooseprogressively finer partitions of [a, b]. This is, of course, not very rigorous notation. A better notationwould be

RI(f ; a, b) ≡ DI(f ; a, b) ≡ I(f ; a, b)

where the symbol I denotes that we are interested in computing the integral of f using the equivalentapproach of Riemann or Darboux. Indeed, the notation I(f ; a, b) does not require the uncomfortablelack of rigor that the symbol dt implies. However, for historical reasons, the symbol

∫f(t) dt will

be used.Also, the use of the

∫f(t) dt allows us to very efficiently apply the integration techniques of

substitution and so forth as we have shown in Chapter 2.

4.3 Properties Of The Riemann Integral

We can now prove a series of properties of the Riemann integral. Let’s start with a lemma aboutinfinimums and supremums.

Lemma 4.3.1 Fundamental infimum and supremum equalities

4.3. PROPERTIES 65

If f is a bounded function on the finite interval [a, b], then

1. supa≤x≤b f(x) = − infa≤x≤b(−f(x)) and − supa≤x≤b(−f(x)) =infa≤x≤b(f(x))

2.

supx,y∈[a,b]

(f(x)− f(y)) = supy∈[a,b]

supy∈[a,b]

(f(x)− f(y))

= supx∈[a,b]

supy∈[a,b]

(f(x)− f(y)) = M −m

where M = supa≤x≤b f(x) and m = infa≤x≤b f(x).

3. supx,y∈[a,b] | f(x)− f(y) |= M −m

Proof 4.3.1

First let Q = supa≤x≤b(−f) and q = infa≤x≤b(−f).(1):Let (f(xn) be a sequence which converges to M . Then since −f(xn) ≥ q for all n, letting n→∞,we find −M ≥ q.

Now let (−f(zn)) be a sequence which converges to q. Then, we have f(zn) ≤ M for all n andletting n→]infty, we have −q ≤M or q ≥ −M .

Combining, we see −q = M which is the first part of the statement; i.e. supa≤x≤b f(x) =− infa≤x≤b(−f(x)). Now just replace all the f ’s by −f ’s in this to get supa≤x≤b(−f(x)) =− infa≤x≤b(−− f(x)) or − supa≤x≤b(−f(x)) = infa≤x≤b(f(x)) which is the other identity.

(2):We know

f(x)− f(y) ≤ supa≤x≤b

f(x)− f(y) ≤ supx∈[a,b]

− inf(a≤y≤b

f(y) = M −m

f(x)− f(y) ≤ f(x)− infa≤y≤b

f(y) ≤ supx∈[a,b]

− infa≤y≤b

f(y) = M −m

Thus,

supx,y∈[a,b]

(f(x)− f(y)) ≤M −m

So one side of the inequality is clear. Now let f(xn) be a sequence converging to M and f(yn) be asequence converging to m. Then, we have

f(xn)− f(yn) ≤ supx,y∈[a,b]

(f(x)− f(y))

Letting n→∞, we see

M −m ≤ supx,y∈[a,b]

(f(x)− f(y))


This is the other side of the inequality. We have thus shown that the equality is valid.(3):Note

| f(x)− f(y) | =

f(x)− f(y), f(x) ≥ f(y)f(y)− f(x), f(x) < f(y)

In either case, we have | f(x) − f(y) |≤ Mj − mj for all x, y using Part (2) implying thatsupx,y | f(x)− f(y) |≤Mj −mj .

To see the reverse inequality holds, we first note that if Mj = mj , we see the reverse inequality holdstrivially as supx,y | f(x) − f(y) | ≥ 0 = Mj − mj . Hence, we may assume without loss ofgenerality that the gap Mj −mj is positive.

Then, given 0 < ε < (1/2(Mj −m− j), there exist, sj , tj ∈ [xj−1, xj ] such that Mj − ε/2 < f(sj)and mj + ε/2 > f(tj), so that f(sj) − f(tj) > Mj −mj − ε. by our choice of ε, these terms arepositive and so we also have | f(sj)− f(tj) |> Mj −mj − ε.

It follows that

supx,y∈[xj−1,xj ]

| f(x)− f(y) | ≥ | f(sj)− f(tj) |> Mj −mj − ε | .

Since we can make ε arbitrarily small, this implies that


| f(x)− f(y) | ≥ Mj −mj .

This establishes the reverse inequality and proves the claim.

Theorem 4.3.2 Properties Of The Riemann Integral

4.3. PROPERTIES 67

Let f, g ∈ RI[a, b]. Then

(i) | f |∈ RI[a, b];

(ii) ∣∣∣∣∣∫ b

a

f(x)dx

∣∣∣∣∣ ≤∫ b

a

| f | dx;

(iii) f+ = maxf, 0 ∈ RI[a, b];

(iv) f− = max−f, 0 ∈ RI[a, b];

(v) ∫ b

a

f(x)dx =

∫ b

a

[f+(x)− f−(x)]dx =

∫ b

a

f+(x)dx−∫ b

a

f−(x)dx∫ b

a

| f(x) | dx =

∫ b

a

[f+(x) + f−(x)]dx =

∫ b

a

f+(x)dx+

∫ b

a

f−(x)dx;

(vi) f2 ∈ RI[a, b];

(vii) fg ∈ RI[a, b];

(viii) If there exists m,M such that 0 < m ≤| f |≤M , then 1/f ∈ RI[a, b].

Proof 4.3.2

(i):

Proof Note given a partition π = x0 = a, x1, . . . , xp = b, for each j = 1, . . . , p, from Lemma4.3.1, we know


(f(x)− f(y)) = Mj −mj

Now, let m′j and M ′j be defined by

m′j = inf[xj−1,xj ]

| f(x) |, M ′j = sup[xj−1,xj ]

| f(x) | .

Then, applying Lemma 4.3.1 to |f |, we have

M ′j −m′j = supx,y∈[xj−1,xj ]

| f(x) | − | f(y) | .

Thus, for each j = 1, . . . , p, we have

Mj −mj = supx,y∈[xj−1,xj ]

| f(x)− f(y) | .

So, since | f(x) | − | f(y) |≤| f(x) − f(y) | for all x, y, it follows that M ′j −m′j ≤ Mj −mj ,implying that

∑π(M ′j − m′j)∆xj ≤

∑π(Mj − mj)∆xj . This means U(|f |,π) − L(|f |,π) ≤

U(f,π) − L(f,π) for the chosen π. Since f is integrable by hypothesis, by Theorem 4.2.4, we


know the Riemann criterion must also hold for f . Thus, given ε > 0, there is a partition π0 so thatU(f,π)−L(f,π) < ε for any refinement π of π0. Therefore |f | also satisfies the Riemann Criterionand so | f | is Riemann integrable.

The other results now follow easily.(ii):

Proof We have f ≤| f | and f ≥ − | f |, so that∫ b

a

f(x)dx ≤∫ b

a

| f(x) | dx∫ b

a

f(x)dx ≥ −∫ b

a

| f(x) | dx,

from which it follows that

−∫ b

a

| f(x) | dx ≤∫ b

a

f(x)dx ≤∫ b

a

| f(x) | dx

and so ∣∣∣∣∣∫ b

a

f

∣∣∣∣∣ ≤∫ b

a

| f |,

(iii) and (iv):

Proof This follows from the facts that f+ = 12 (| f | +f) and f− = 1

2 (| f | −f) and the Riemannintegral is a linear mapping.

(v):

Proof This follows from the facts that f = f+ − f− and | f |= f+ + f− and the linearity of theintegral.

(vi) :

Proof Note that, since f is bounded, there exists K > 0 such that | f(x) |≤ K for all x ∈ [a, b].Then, applying Lemma 4.3.1 to f2, we have


(f2(x)− f2(y)) = Mj(f2)−mj(f

2)

where [xj−1, xj ] is a subinterval of a given partition π and Mj(f2) = supx∈[xj−1,xj ] f

2(x) andmj(f

2) = infx∈[xj−1,xj ] f2(x). Thus, for this partition, we have

U(f2,π)− L(f2,π) =∑π

(Mj(f2)−mj(f

2)) ∆xj

But we also know


(f2(x)− f2(y)) = supx,y∈[xj−1,xj ]

(f(x) + f(y))(f(x)− f(y))

≤ 2K supx,y∈[xj−1,xj ]

((f(x)− f(y)) = Mj −mj .

4.3. PROPERTIES 69

Thus,

U(f2,π)− L(f2,π) =∑π

(Mj(f2)−mj(f

2)) ∆xj

≤ 2K∑π

(Mj −mj) ∆xj = U(f,π)− L(fπ).

Now since f is Riemann Integrable, it satisfies the Riemann Criterion and so given ε > 0, there is apartition π0 so that U(f,π)− L(fπ) < ε/(2K) for any refinement π of π0. Thus, f2 satisfies theRiemann Criterion too and so it is integrable.

(vii):

Proof To prove that fg is integrable when f and g are, simply note that

fg = (1/2)

((f + g)2 − f2 − g2

).

Property (vi) and the linearity of the integral then imply fg is integrable.

(viii):

Proof Suppose f ∈ RI[a, b] and there exist M,m > 0 such that m ≤| f(x) |≤M for all x ∈ [a, b].Note that

1

f(x)− 1

f(y)=f(y)− f(x)

f(x)f(y).

Let π = x0 = a, x1, . . . , xp = b be a partition of [a, b], and define

M ′j = sup[xj−1,xj ]

1

f(x)

m′j = inf[xj−1,xj ]

1

f(x).

Then we have

M ′j −m′j = supx,y∈[xj−1,xj ]

f(y)− f(x)

f(x)f(y)

≤ supx,y∈[xj−1,xj ]

| f(y)− f(x) || f(x) || f(y) |

≤ 1

m2sup

x,y∈[xj−1,xj ]

| f(y)− f(x) |

≤ Mj −mj

m2.

Since f ∈ RI[a, b], given ε > 0 there is a partition π0 such that U(f,π)− L(f,π) < m2ε for anyrefinement, pi, of π0. Hence, the previous inequality implies that, for any such refinement, we have

U( 1

f,π)− L

( 1

f,π)

=∑π

(M ′j −m′j)∆xj


≤ 1

m2

∑π

(Mj −mj)∆xj

≤ 1

m2

(U(f,π)− L(f,π)

)<

m2ε

m2= ε.

Thus 1/f satisfies the Riemann Criterion and hence it is integrable.

4.4 What Functions Are Riemann Integrable?Now we need to show that the set RI[a, b] is nonempty. We begin by showing that all continuousfunctions on [a, b] will be Riemann Integrable.

Theorem 4.4.1 Continuous Implies Riemann Integrable

If f ∈ C[a, b], then f ∈ RI[a, b].

Proof 4.4.1Since f is continuous on a compact set, it is uniformly continuous. Hence, given ε > 0, there is aδ > 0 such that x, y ∈ [a, b], | x − y |< δ ⇒| f(x) − f(y) |< ε/(b − a). Let π0 be a partitionsuch that || π0 ||< δ, and let π = x0 = a, x1, . . . , xp = b be any refinement of π0. Then π alsosatisfies || π ||< δ. Since f is continuous on each subinterval [xj−1, xj ], f attains its supremum,Mj , and infimum, mj , at points sj and tj , respectively. That is, f(sj) = Mj and f(tj) = mj foreach j = 1, . . . , p. Thus, the uniform continuity of f on each subinterval implies that, for each j,

Mj −mj =| f(sj)− f(tj) |<ε

b− a.

Thus, we have

U(f,π)− L(f,π) =∑π

(Mj −mj)∆xj <ε

b− a∑π

∆xj = ε.

Since π was an arbitrary refinement of π0, it follows that f satisfies Riemann’s criterion. Hence,f ∈ RI[a, b].

Theorem 4.4.2 Constant Functions Are Riemann Integrable

If f : [a, b] → < is a constant function, f(t) = c for all t in [a, b], then f is RiemannIntegrable on [a, b] and

∫ baf(t)dt = c(b− a).

Proof 4.4.2For any partition π of [a, b], since f is a constant, all the individual mj’s and Mj’s associated withπ take on the value c. Hence, U(f,π)−U(f,π) = 0 always. It follows immediately that f satisfiesthe Riemann Criterion and hence is Riemann Integrable. Finally, since f is integrable, by Theorem4.1.2, we have

c(b− a) ≤ RI(f ; a, b) ≤ c(b− a).

Thus,∫ baf(t)dt = c(b− a).

4.5. MORE PROPERTIES 71

Theorem 4.4.3 Monotone Implies Riemann Integrable

If f is monotone on [a, b], then f ∈ RI[a, b].

Proof 4.4.3As usual, for concreteness, we assume that f is monotone increasing. We also assume f(b) > f(a),for if not, then f is constant and must be integrable by Theorem 4.4.2. Let ε > 0 be given, and let π0

be a partition of [a, b] such that || π0 ||< ε/(f(b) − f(a)). Let π = x0 = a, x1, . . . , xp = b beany refinement of π0. Then π also satisfies || π ||< ε/(f(b) − f(a)). Thus, for each j = 1, . . . , p,we have

∆xj <ε

f(b)− f(a).

Since f is increasing, we also know that Mj = f(xj) and mj = f(xj−1) for each j. Hence,

U(f,π)− L(f,π) =∑π

(Mj −mj)∆xj

=∑π

[f(xj)− f(xj−1)]∆xj

<ε

f(b)− f(a)

∑π

[f(xj)− f(xj−1)].

But this last sum is telescoping and sums to f(b)− f(a). So, we have

U(f,π)− L(f,π) <ε

f(b)− f(a)(f(b)− f(a)) = ε.

Thus, f satisfies Riemann’s criterion.

Theorem 4.4.4 Bounded Variation Implies Riemann Integrable

If f ∈ BV [a, b], then f ∈ RI[a, b].

Proof 4.4.4Since f is of bounded variation, there are functions u and v, defined on [a, b] and both monotoneincreasing, such that f = u − v. Hence, by the linearity of the integral and the previous theorem,f ∈ RI[a, b].

4.5 Further Properties of the Riemann Integral

We first want to establish the familiar summation property of the Riemann integral over an interval[a, b] = [a, c] ∪ [c, b]. Most of the technical work for this result is done in the following Lemma.

Lemma 4.5.1 The Upper And Lower Darboux Integral Is Additive On Intervals


Let f ∈ B[a, b] and let c ∈ (a, b). Let∫ b

a

f(x) dx = L(f) and∫ b

a

f(x) dx = U(f)

denote the lower and upper Darboux integrals of f on [a, b], respectively. Then we have∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx

∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

Proof 4.5.1We prove the result for the upper integrals as the lower integral case is similar. Let π be given by π =x0 = a, x1, . . . , xp = b. We first assume that c is a partition point of π. Thus, there is some index1 ≤ k0 ≤ p−1 such that xk0 = c. For any interval [α, β], let Uβα (f,π) denote the upper sum of f forthe partition π over [α, β]. Now, we can rewrite π as π = x0, x1, . . . , xk0∪xk0 , xk0+1, . . . , xp.Let π1 = x0, . . . , xk0 and π2 = xk0 , . . . , xp. Then π1 ∈ Π[a, c], π2 ∈ Π[c, b], and

U ba(f,π) = U ca(f,π1) + U bc (f,π2)

≥∫ c

a

f(x)dx+

∫ b

c

f(x)dx,

by the definition of the upper sum. Now, if c is not in π, then we can refine π by adding c, obtainingthe partition π′ = x0, x1, . . . , xk0 , c, xk0+1, . . . , xp. Splitting up π′ at c as we did before intoπ1 and π2, we see that π′ = π1 ∨ π2 where π1 = x0, . . . , xk0 , c and π2 = c, xk0+1, . . . , xp.Thus, by our properties of upper sums, we see that

U ba(f,π) ≥ U ba(f,π′) = U ca(f,π1) + U bc (f,π2) ≥∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

Combining both cases, we can conclude that for any partition π ∈ Π[a, b], we have

U ba(f,π) ≥∫ c

a

f(x)dx+

∫ b

c

f(x)dx,

which implies that ∫ b

a

f(x)dx ≥∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

Now we want to show the reverse inequality. Let ε > 0 be given. By the definition of the upperintegral, there exists π1 ∈ Π[a, c] and π2 ∈ Π[c, b] such that

U ca(f,π1) <

∫ c

a

f(x)dx+ε

2

U bc (f,π2) <

∫ b

c

f(x)dx+ε

2.

4.5. MORE PROPERTIES 73

Let π = π1 ∪ π2 ∈ Π[a, b]. It follows that

U ba(f,π) = U ca(f,π1) + U bc (f,π2) <

∫ c

a

f(x)dx+

∫ b

c

f(x)dx+ ε.

But, by definition, we have

∫ b

a

f(x)dx ≤ U ba(f,π)

for all π. Hence, we see that

∫ b

a

f(x)dx <

∫ c

a

f(x)dx+

∫ b

c

f(x)dx+ ε.

Since ε was arbitrary, this proves the reverse inequality we wanted. We can conclude, then, that

∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

Theorem 4.5.2 The Riemann Integral Exists On Subintervals

If f ∈ RI[a, b] and c ∈ (a, b), then f ∈ RI[a, c] and f ∈ RI[c, b].

Proof 4.5.2Let ε > 0 be given. Then there is a partition π0 ∈ Π[a, b] such that U ba(f,π) − Lba(f,π) < εfor any refinement, π, of π0. Let π0 be given by π0 = x0 = a, x1, . . . , xp = b. Define π′0 =π0 ∪ c, so there is some index k0 such that xk0 ≤ c ≤ xk0+1. Let π1 = x0, . . . , xk0 , c andπ2 = c, xk0+1, . . . , xp. Then π1 ∈ Π[a, c] and π2 ∈ Π[c, b]. Let π′1 be a refinement of π1. Thenπ′1 ∪ π2 is a refinement of π0, and it follows that

U ca(f,π′1)− Lca(f,π′1) =∑π′1

(Mj −mj)∆xj

≤∑π′1∪π2

(Mj −mj)∆xj

≤ U ba(f,π′1 ∪ π2)− Lba(f,π′1 ∪ π2).

But, since π′1 ∪ π2 refines π0, we have

U ba(f,π′1 ∪ π2)− Lba(f,π′1 ∪ π2) < ε,

implying that

U ca(f,π′1)− Lca(f,π′1) < ε

for all refinements, π′1, of π1. Thus, f satisfies Riemann’s criterion on [a, c], and f ∈ RI[a, c]. Theproof on [c, b] is done in exactly the same way.

Theorem 4.5.3 The Riemann Integral Is Additive On Subintervals


If f ∈ RI[a, b] and c ∈ (a, b), then∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

Proof 4.5.3Since f ∈ RI[a, b], we know that ∫ b

a

f(x)dx =

∫ b

a

f(x)dx.

Further, we also know that f ∈ RI[a, c] and f ∈ RI[c, b] for any c ∈ (a, b). Thus,

∫ c

a

f(x)dx =

∫ c

a

f(x)dx

∫ b

c

f(x)dx =

∫ b

c

f(x)dx.

So, applying Lemma 4.5.1, we conclude that, for any c ∈ (a, b),

∫ b

a

f(x)dx =

∫ b

a

f(x) dx =

∫ c

a

f(x) dx+

∫ b

c

f(x) dx =

∫ c

a

f(x) dx+

∫ b

c

f(x) dx.

4.6 The Fundamental Theorem Of CalculusThe next result is the well-known Fundamental of Theorem of Calculus.

Theorem 4.6.1 The Fundamental Theorem Of Calculus

Let f ∈ RI[a, b]. Define F : [a, b]→ < by

F (x) =

∫ x

a

f(t)dt.

Then

(i) F ∈ BV [a, b];

(ii) F ∈ C[a, b];

(iii) if f is continuous at c ∈ [a, b], then F is differentiable at c and F ′(c) = f(c).

Proof 4.6.1First, note that f ∈ RI[a, b] ⇒ f ∈ R[a, x] for all x ∈ [a, b], by our previous results. Hence, F iswell-defined. We will prove the results in order.(i) F ∈ BV [a, b]:

4.6. FUNDAMENTAL THEOREM 75

Proof Let π ∈ Π[a, b] be given by π = x0 = a, x1, . . . , xp = b. Then the fact that f ∈ R[a, xj ]implies that f ∈ R[xj−1, xj ] for each j = 1, . . . , p. Thus, we have

mj∆xj ≤∫ xj

xj−1

f(t)dt ≤Mj∆xj .

This implies that, for each j, we have∣∣∣∣∣∫ xj

xj−1

f(t)dt

∣∣∣∣∣ ≤|| f ||∞ ∆xj .

Thus,

| ∆Fj | = | F (xj)− F (xj−1) |

=

∣∣∣∣∣∫ xj

a

f(t)dt−∫ xj−1

a

f(t)dt

∣∣∣∣∣=

∣∣∣∣∣∫ xj

xj−1

f(t)dt

∣∣∣∣∣≤ || f ||∞ ∆xj .

Summing over π, we obtain∑π

| ∆Fj |≤|| f ||∞∑π

∆xj = (b− a) || f ||∞<∞.

Since the partition π was arbitrary, we conclude that F ∈ BV [a, b].

(ii) F ∈ C[a, b]:

Proof Now, let x, y ∈ [a, b] be such that x < y. Then

inf[x,y]

f(t) (y − x) ≤∫ y

x

f(t)dt ≤ sup[x,y]

f(t) (y − x),

which implies that

| F (y)− F (x) |=

∣∣∣∣∣∫ y

x

f(t)dt

∣∣∣∣∣ ≤|| f ||∞ (y − x).

A similar argument shows that if y, x ∈ [a, b] satisfy y < x, then

| F (y)− F (x) |=

∣∣∣∣∣∫ y

x

f(t)dt

∣∣∣∣∣ ≤|| f ||∞ (x− y).

Let ε > 0 be given. Then if

| x− y |< ε

|| f ||∞ +1,

we have

| F (y)− F (x) |≤|| f ||∞| y − x |<|| f ||∞|| f ||∞ +1

ε < ε.


Thus, F is continuous at x and, consequently, on [a, b].

(iii) (if) f is continuous at c ∈ [a, b], then F is differentiable at c and F ′(c) = f(c):

Proof Finally, assume f is continuous at c ∈ [a, b], and let ε > 0 be given. Then there exists δ > 0such that x ∈ (c− δ, c+ δ) ∩ [a, b] implies | f(x)− f(c) |< ε/2. Pick h ∈ < such that 0 <| h |< δand c+ h ∈ [a, b]. Let’s assume, for concreteness, that h > 0. Define

m = inf[c,c+h]

f(t) and M = sup[c,c+h]

f(t).

If c < x < c+ h, then we have x ∈ (c− δ, c+ δ) ∩ [a, b] and −ε/2 < f(x)− f(c) < ε/2. That is,

f(c)− ε

2< f(x) < f(c) +

ε

2∀x ∈ [c, c+ h].

Hence, m ≥ f(c)− ε/2 and M ≤ f(c) + ε/2. Now, we also know that

mh ≤∫ c+h

c

f(t)dt ≤Mh.

Thus, we have

F (c+ h)− F (c)

h=

∫ c+ha

f(t)dt−∫ caf(t)dt

h=

∫ c+hc

f(t)dt

h.

Combining inequalities, we find

f(c)− ε

2≤ m ≤ F (c+ h)− F (c)

h≤M ≤ f(c) +

ε

2

yielding ∣∣∣∣∣F (c+ h)− F (c)

h− f(c)

∣∣∣∣∣ ≤ ε

2< ε

if x ∈ [c, c+ h].The case where h < 0 is handled in exactly the same way. Thus, since ε was arbitrary, this showsthat F is differentiable at c and F ′(c) = f(c). Note that if c = a or c = b, we need only consider thedefinition of the derivative from one side.

Comment 4.6.1 We call F (x) the indefinite integral of f . F is always better behaved than f , sinceintegration is a smoothing operation. We can see that f need not be continuous, but, as long as it isintegrable, F is always continuous.

The next result is one of the many mean value theorems in the theory of integration. It is a moregeneral form of the standard mean value theorem given in beginning calculus classes.

Theorem 4.6.2 The Mean Value Theorem For Riemann Integrals

Let f ∈ C[a, b], and let g ≥ 0 be integrable on [a, b]. Then there is a point, c ∈ [a, b], suchthat ∫ b

a

f(x)g(x)dx = f(c)

∫ b

a

g(x)dx.


Proof 4.6.5Since f is continuous, it is also integrable. Hence, fg is integrable. Let m and M denote the lowerand upper bounds of f on [a, b], respectively. Then mg(x) ≤ f(x)g(x) ≤ Mg(x) for all x ∈ [a, b].Since the integral preserves order, we have

m

∫ b

a

g(x)dx ≤∫ b

a

f(x)g(x)dx ≤M∫ b

a

g(x)dx.

If the integral of g on [a, b] is 0, then this shows that the integral of fg will also be 0. Hence, in thiscase, we can choose any c ∈ [a, b] and the desired result will follow. If the integral of g is not 0, thenit must be positive, since g ≥ 0. Hence, we have, in this case,

m ≤∫ baf(x)g(x)dx∫ bag(x)dx

≤M.

Now, f is continuous, so it attains the values M and m at some points. Hence, by the intermediatevalue theorem, there must be some c ∈ [a, b] such that

f(c) =

∫ baf(x)g(x)dx∫ bag(x)dx

.

This implies the desired result.

The next result is another standard mean value theorem from basic calculus. It is a direct con-sequence of the previous theorem, by simply letting g(x) = 1 for all x ∈ [a, b]. This result can beinterpreted as stating that integration is an averaging process.

Theorem 4.6.3 Average Value For Riemann Integrals

If f ∈ C[a, b], then there is a point c ∈ [a, b] such that

1

b− a

∫ b

a

f(x)dx = f(c).

The next result is the standard means for calculating definite integrals in basic calculus. We startwith a definition.

Definition 4.6.1 The Antiderivative of f

Let f : [a, b] → < be a bounded function. Let G : [a, b] → < be such that G′ exists on[a, b] and G′(x) = f(x) for all x ∈ [a, b]. Such a function is called an antiderivative or aprimitive of f .

Comment 4.6.2 The idea of an antiderivative is intellectually distinct from the Riemann integral ofa bounded function f . Consider the following function f defined on [−1, 1].

f(x) =

x2 sin(1/x2), x 6= 0, x ∈ [−1, 1]0, x = 0

It is easy to see that this function has a removable discontinuity at 0. Moreover, f is even differen-tiable on [−1, 1] with derivative

f ′(x) =

2x sin(1/x2)− (2/x) cos(1/x2), x 6= 0, x ∈ [−1, 1]0, x = 0


Note f ′ is not bounded on [−1, 1] and hence it can not be Riemann Integrable. Now to connect thisto the idea of antiderivatives, just relabel the functions. Let g be defined by

g(x) =

2x sin(1/x2)− (2/x) cos(1/x2), x 6= 0, x ∈ [−1, 1]0, x = 0

then define G by

G(x) =

x2 sin(1/x2), x 6= 0, x ∈ [−1, 1]0, x = 0

We see that G is the antiderivative of g even though g itself does not have a Riemann integral. Again,the point is that the idea of the antiderivative of a function is intellectually distinct from that of beingRiemann integrable.

Theorem 4.6.4 Cauchy’s Fundamental Theorem

Let f : [a, b]→ < be integrable. Let G : [a, b]→ < be any antiderivative of f . Then∫ b

a

f(t)dt = G(t)∣∣∣ba

= G(b)−G(a).

Proof 4.6.6Since G′ exists on [a, b], G must be continuous on [a, b]. Let ε > 0 be given. Since f is integrable,there is a partition π0 ∈ Π[a, b] such that for any refinement, π, of π0 and any σ ⊂ π, we have∣∣∣∣∣S(f,π,σ)−

∫ b

a

f(x)dx

∣∣∣∣∣ < ε.

Let π be any refinement of π0, given by π = x0 = a, x1, . . . , xp = b. The Mean Value Theoremfor differentiable functions then tells us that there is an sj ∈ (xj−1, xj) such thatG(xj)−G(xj−1) =G′(sj)∆xj . Since G′ = f , we have G(xj) −G(xj−1) = f(sj)∆xj for each j = 1, . . . , p. The setof points, s1, . . . , sp, is thus an evaluation set associated with π. Hence,

∑π

[G(xj)−G(xj−1)] =∑π

G′(sj)∆xj =∑π

f(sj)∆xj

The first sum on the left is a collapsing sum, hence we have

⇒ G(b)−G(a) = S(f,π, s1, . . . , sp).

We conclude ∣∣∣∣∣G(b)−G(a)−∫ b

a

f(x)dx

∣∣∣∣∣ < ε.

Since ε was arbitrary, this implies the desired result.

Comment 4.6.3 Not all functions (in fact, most functions) will have closed form, or analyticallyobtainable, antiderivatives. So, the previous theorem will not work in such cases.

Theorem 4.6.5 The Recapture Theorem


If f is differentiable on [a, b], and if f ′ ∈ RI[a, b], then∫ x

a

f ′(t)dt = f(x)− f(a).

Proof 4.6.7f is an antiderivative of f . Now apply Cauchy’s Fundamental Theorem 4.6.4.

Another way to evaluate Riemann integrals is to directly approximate them using an appropri-ate sequence of partitions. Theorem 4.6.6 is a fundamental tool that tells us when and why suchapproximations will work.

Theorem 4.6.6 Approximation Of The Riemann Integral

If f ∈ RI[a, b], then given any sequence of partitions πn with any associated sequenceof evaluation sets σn that satisfies || πn ||→ 0, we have

limn→∞

S(f,πn,σn) =

∫ b

a

f(x) dx

Proof 4.6.8Since f is integrable, given a positive ε, there is a partition π0 so that

| S(f,π,σ)−∫ b

a

f(x)dx | < ε/2, π0 π, σ ⊆ π. (∗)

Let the partition π0 be x0, x1, . . . , xP and let ξ be defined to be the smallest ∆xj from π0. Thensince the norm of the partitions πn goes to zero, there is a positive integer N so that

|| πn || < min (ξ, ε/(4P || f ||)∞)) (∗)

Now pick any n > N and label the points of πn as y0, y1, . . . , yQ. We see that the points in πnare close enough together so that at most one point of π0 lies in any subinterval yj−1, yj ] from πn.This follows from our choice of ξ. So the intervals of πn split into two pieces: those containing apoint of π0 and those that do not have a π0 inside. Let A be the first collection of intervals andB, the second. Note there are P points in π0 and so there are P subintervals in B. Now considerthe common refinement πn ∨ π0. The points in the common refinement match πn except on thesubintervals from B. Let [yj−1, yj ] be such a subinterval and let γj denote the point from π0 whichis in this subinterval. Let’s define an evaluation set σ for this refinement πn ∨ π0 as follows.

1. if we are in the subintervals labeled A , we choose as our evaluation point, the evaluationpoint sj that is already in this subinterval since σn ⊆ πn. Here, the length of the subintervalwill be denoted by δj(A ) which equals yj − yj−1 for appropriate indices.

2. if we are in the the subintervals labeled B, we have two intervals to consider as [yj−1, yj ] =[yj−1, γj ] ∪ [γj , yj ]. Choose the evaluation point γj for both [yj−1, γ] and [γ, yj ]. Here, thelength of the subintervals will be denoted by δj(B). Note that δj(B) = γj − yj−1 or yj − γj .

Then we have

S(f,πn ∨ π0,σ) =∑A

f(sj)δj(A ) +∑B

f(γj)δj(A )


=∑A

f(sj)(yj − yj−1) +∑B

(f(γj)(yj − γj) + f(γj)(γj − yj−1))

=∑A

f(sj)(yj − yj−1) +∑B

(f(γj)(yj − yj−1))

Thus, since the Riemann sums over πn and πn ∨ π0 with these choices of evaluation sets match onA , we have using Equation ∗ that

| S(f,πn,σn)− S(f,πn ∨ π0,σ) | = |∑A

(f(sj)− f(γj)(yj − yj−1) |

≤∑A

(| f(sj) | + | f(γj) |) (yj − yj−1) |

≤ P 2 || f ||∞ || πn ||

< P 2 || f ||∞ε

4P || f ||∞= ε/2

We conclude that for our special evaluation set σ for the refinement πn ∨ π0 that

|| S(f,πn,σn)−∫ b

a

f(x)dx | = | S(f,πn,σn)− S(f,πn ∨ π0,σ) + S(f,πn ∨ π0,σ)−∫ b

a

f(x)dx |

≤ | S(f,πn,σn)− S(f,πn ∨ π0,σ) | + | S(f,πn ∨ π0,σ)−∫ b

a

f(x)dx |

< ε/2 + ε/2 = ε

using Equation ∗ as πn ∨ π0 refines π0. Since we can do this analysis for any n > N , we see wehave shown the desired result.

4.6.1 HomeworkExercise 4.6.1 Let f(x) = x2 on the interval [−1, 3]. Use Theorem 4.6.6 to prove that

∫ 3

−1f(x)dx =

28/3.

Hint 4.6.1 We know f is Riemann integrable because it is continuous and so this theorem can beapplied. Use the uniform approximations xi = −1 + 4i/n for i = 0 to i = n to define partitions πn.Then using left or right hand endpoints on each subinterval to define the evaluation set σn, you canprove directly that

∫ 3

−1x2dx = lim S(f,πn,σn) = 28/3. Make sure you tell me all the reasoning

involved.

Exercise 4.6.2 If f is continuous, evaluate

limx→a

x

x− a

∫ x

a

f(t)dt

Exercise 4.6.3 Prove if f is continuous on [a, b] and∫ baf(x)g(x)dx = 0 for all choices of integrable

g, then f is identically 0.

4.7 Substitution Type ResultsUsing the Fundamental Theorem of Calculus, we can derive many useful tools.

4.7. SUBSTITUTION 81

Theorem 4.7.1 Integration By Parts

Assume u : [a, b] → < and v : [a, b] → < are differentiable on [a, b] and u′ and v′ areintegrable. Then ∫ x

a

u(t)v′(t) dt = u(t)v(t)

∣∣∣∣∣x

a

−∫ x

a

v(t)u′(t) dt

Proof 4.7.1Since u and v are differentiable on [a, b], they are also continuous and hence, integrable. Now applythe product rule for differentiation to obtain

(u(t)v(t))′

= u′(t)v(t) + u(t)v′(t)

By Theorem 4.3.2, we know products of integrable functions are integrable. Also, the integral islinear. Hence, the integral of both sides of the equation above is defined. We obtain

∫ x

a

(u(t)v(t))′dt =

∫ x

a

u′(t)v(t) dt +

∫ x

a

u(t)v′(t) dt

Since (uv)′ is integrable, we can apply the Recapture Theorem to see

u(t)v(t)

∣∣∣∣∣x

a

=

∫ x

a

u′(t)v(t) dt +

∫ x

a

u(t)v′(t) dt

This is the desired result.

Theorem 4.7.2 Substitution In Riemann Integration

Let f be continuous on [c, d] and u be continuously differentiable on [a, b] with u(a) = cand u(b) = d. Then ∫ d

c

f(u) du =

∫ b

a

f(u(t)) u′(t) dt

Proof 4.7.2Let F be defined on [c, d] by F (u) =

∫ ucf(t)dt. Then since f is continuous, F is continuous and

differentiable on [c, d] by the Fundamental Theorem of Calculus. We know F ′(u) = f(u) and so

F ′(u(t)) = f(u(t)), a ≤ t ≤ b

implyingF ′(u(t)) u′(t) = f(u(t))u′(t) , a ≤ t ≤ b

By the Chain Rule for differentiation, we also know

(F u)′(t) = F (u(t))u′(t) , a ≤ t ≤ b.


and hence (F u)′(t) = f(u(t))u′(t) on [a, b].Now define g on [a, b] by

g(t) = (f u)(t) u′(t) = f(u(t)) u′(t)

= (F u)′(t).

Since g is continuous, g is integrable on [a, b]. Now define G on [a, b] by G(t) = (F u)(t). ThenG′(t) = f(u(t))u′(t) = g(t) on [a, b] and G′ is integrable. Now, apply the Cauchy FundamentalTheorem of Calculus to G to find ∫ b

a

g(t) dt = G(b) − G(a)

or ∫ b

a

f(u(t)) u′(t) dt = F (u(b)) − F (u(a))

=

∫ u(b)=d

c

f(t)dt −∫ u(a)=c

c

f(t)dt

=

∫ d

c

f(t)dt.

Theorem 4.7.3 Leibnitz’s Rule

Let f be continuous on [a, b], u : [c, d] → [a, b] be differentiable on [c, d] and v : [c, d] →[a, b] be differentiable on [c, d]. Then(∫ v(x)

u(x)

f(t) dt

)′= f(v(x))v′(x) − f(u(x)u′(x)

Proof 4.7.3Let F be defined on [a, b] by F (y) =

∫ yaf(t)dt. Since f is continuous, F is also continuous and

moreover, F is differentiable with F ′(y)) = f(y). Since v is differentiable on [c, d], we can use theChain Rule to find

(F v)′(x) = F ′(v(x)) v′(x)

= f(v(x)) v′(x)

This says (∫ v(x)

a

f(t) dt

)′= f(v(x))v′(x)

Next, define G on [a, b] by G(y) =∫ byf(t)dt =

∫ baf(t) −

∫ yaf(t)dt. Apply the Fundamental

Theorem of Calculus to conclude

G′(y) = −(∫ y

a

f(t)dt

)= −f(y)

4.8. SAME INTEGRAL? 83

Again, apply the Chain Rule to see

(G u)′(x) = G′(u(x)) u′(x)

= −f(u(x)) u′(x).

We conclude (∫ b

u(x)

f(t) dt

)′= −f(u(x))u′(x)

Now combine these results as follows:∫ b

a

f(t)dt =

∫ v(x)

a

f(t)dt +

∫ u(x)

v(x)

f(t)dt +

∫ b

u(x)

f(t)dt

or

(F v)(x) + (G u)(x)−∫ b

a

f(t)dt = −∫ u(x)

v(x)

f(t)dt

=

∫ v(x)

u(x)

f(t)dt

Then, differentiate both sides to obtain

(F v)′(x) + (G u)′(x) = f(v(x))v′(x) − f(u(x))u′(x)

=

(∫ v(x)

u(x)

f(t)dt

)′which is the desired result.

4.8 When Do Two Functions Have The Same Integral?

The last results in this chapter seek to find conditions under which the integrals of two functions, fand g, are equal.

Lemma 4.8.1 f Zero On (a, b) Implies Zero Riemann Integral

Let f ∈ B[a, b], with f(x) = 0 on (a, b). Then f is integrable on [a, b] and∫ b

a

f(x)dx = 0.

Proof 4.8.1If f is identically 0, then the result is follows easily. Now, assume f(a) 6= 0 and f(x) = 0 on (a, b].Let ε > 0 be given, and let δ > 0 satisfy

δ <ε

| f(a) |.


Let π0 ∈ Π[a, b] be any partition such that || π0 ||< δ. Let π = x0 = a, x1, . . . , xp be anyrefinement of π0. Then U(f,π) = max(f(a), 0)∆x1 and L(f,π) = min(f(a), 0)∆x1. Hence, wehave

U(f,π)− L(f,π) = [max(f(a), 0)−min(f(a), 0)]∆x1 =| f(a) | ∆x1.

But

| f(a) | ∆x1 <| f(a) | δ <| f(a) | ε

| f(a) |= ε.

Hence, if π is any refinement of π0, we have U(f,π)− L(f,π) < ε. This shows that f ∈ RI[a, b].Further, we have

U(f,π) = max(f(a), 0)∆x1 ⇒ U(f) = infπU(f,π) = 0,

since we can make ∆x1 as small as we wish. Likewise, we also see that L(f) = supπ L(f,π) = 0,implying that

U(f) = L(f) =

∫ b

a

f(x)dx = 0.

The case where f(b) 6= 0 and f(x) = 0 on [a, b) is handled in the same way. So, assume thatf(a), f(b) 6= 0 and f(x) = 0 for x ∈ (a, b). Let ε > 0 be given, and choose δ > 0 such that

δ <ε

2 max| f(a) |, | f(b) |.

Let π0 be a partition of [a, b] such that | π0 |< δ, and let π be any refinement of π0. Then

U(f,π) = max(f(a), 0)∆x1 + max(f(b), 0)∆xp

L(f,π) = min(f(a), 0)∆x1 + min(f(b), 0)∆xp.

It follows that

U(f,π)− L(f, pi) = [max(f(a), 0)−min(f(a), 0)]∆x1 + [max(f(b), 0)−min(f(a), 0)]∆xp

= | f(a) | ∆x1+ | f(b) | ∆xp< | f(a) | δ+ | f(b) | δ< ε.

Since we can make ∆x1 and ∆xp as small as we wish, we see

∫ b

a

f(x)dx = 0.

Lemma 4.8.2 f = g on (a, b) Implies Riemann Integrals Match

4.8. SAME INTEGRAL? 85

Let f, g ∈ RI[a, b] with f(x) = g(x) on (a, b). Then∫ b

a

f(x)dx =

∫ b

a

g(x)dx.

Proof 4.8.2Let h = f − g, and apply the previous lemma.

Theorem 4.8.3 Two Riemann Integrable Functions Match At All But Finitely Many PointsImplies Integrals Match

Let f, g ∈ RI[a, b], and assume that f = g except at finitely many points c1, . . . , ck. Then∫ b

a

f(x)dx =

∫ b

a

g(x)dx.

Proof 4.8.3We may re-index the points c1, . . . , ck, if necessary, so that c1 < c2 < · · · < ck. Then applyLemma 4.8.2 on the intervals (cj−1, cj) for all allowable j. This shows∫ cj

cj−1

f(t)dt =

∫ cj

cj−1

g(t)dt.

Then, since ∫ b

a

f(t)dt =

k∑j=1

∫ cj

cj−1

f(t)dt

the results follows.

Theorem 4.8.4 f Bounded and Continuous At All But One Point Implies f is Riemann Inte-grable

If f is bounded on [a, b] and continuous except at one point c in [a, b], then f is Riemannintegrable.

Proof 4.8.4For convenience, we will assume that c is an interior point, i.e. c is in (a, b). We will show thatf satisfies the Riemann Criterion and so it is Riemann integrable. Let ε > 0 be given. Since f isbounded on [a, b], there is a real number M so that f(x) < M for all x in [a, b]. We know f iscontinuous on [a, c − ε/(6M)] and f is continuous on [c + ε/(6M), b]. Thus, f is integrable onboth of these intervals and f satisfies the Riemann Criterion on both intervals. For this ε there is apartition π0 of [a, c− ε/(6M)] so that

U(f,P )− L(f,P ) < ε/3, if π0 P

and there is a partition π1 of [c+ ε/(6M), b] so that

U(f,Q)− L(f,Q) < ε/3, if π0 Q.


Let π2 be the partition we get by combining π0 with the points c− ε/(6M), c+ ε/(6M) and π1.Then, we see

U(f,π2)− L(f,π2) = U(f,π0)− L(f,π0) +

(sup

x∈[c−ε/(6M),c+ε/(6M)]

f(x)

)ε/3 + U(f,π1)− L(f,π1)

< ε/3 +Mε/(3M) + ε/3 = ε

Then if π2 π on [a, b], we have

U(f,π)− L(f,π) < ε

This shows f satisfies the Riemann criterion and hence is integrable if the discontinuity c is interiorto [a, b]. The argument at c = a and c = b is similar but a bit simpler as it only needs to be donefrom one side. Hence, we conclude f is integrable on [a, b] in all cases..

It is then easy to extend this result to a function f which is bounded and continuous on [a, b]except at a finite number of points x1, x2, . . . , xk for some positive integer k. We state this asTheorem 4.8.5.

Theorem 4.8.5 f Bounded and Continuous At All But Finitely Many Points Implies f is Rie-mann Integrable

if f is bounded on [a, b] and continuous except at finitely many points x1, x2, . . . , xk in[a, b], then f is Riemann integrable.

Proof 4.8.5We may assume without loss of generality that the points of discontinuity are ordered as a < x1 <x2 < . . . < xk < b. Then f is continuous except at x1 on [a, x1] and hence by Theorem 4.8.4 f isintegrable on [a, x1]. Now apply this argument on each of the subintervals xk−1, xk] in turn.

Chapter 5

Further Riemann Integration Results

In this chapter, we will explore certain aspects of Riemann Integration that are more subtle. We beginwith a limit interchange theorem. A good reference for this is (Fulks (3) 1978).

5.1 The Limit Interchange Theorem for Riemann Integration

Suppose you knew that the sequence of functions xn contained in RI[a, b] converged uniformlyto the function x on [a, b]. Is it true that

∫ bax(t)dt = limn→∞ xn(t)dt? The answer to this question

is Yes! and it is our Theorem 5.1.1.

Theorem 5.1.1 The Riemann Integral Limit Interchange Theorem

Let xn be a sequence of Riemann Integrable functions on [a, b] which converge uniformlyto the function x on [a, b]. Then x is also Riemann Integrable on [a, b] and∫ b

a

x(t)dt = limn→∞

xn(t)dt

Proof 5.1.1First, we show that x is Riemann integrable on [a, b]. Let ε be given. Then since xn convergesuniformly to x on [a, b],

∃ δ > 0 3 | xn(t)− x(t) | <ε

5(b− a)∀ n > N, t ∈ [a, b] (α)

Fix any n1 > N . Then since xn1 is integrable,

∃ π0 ∈ Π[a, b] 3 U(xn1,π) − L((xn1

,π) < fracε5 ∀ π0 π (β)

Since xn converges uniformly to x on [a, b], you should be able to show that x is bounded on [a, b].Hence, we can define

Mj = sup[xj−1,xj ]

x(t), M1j = sup

[xj−1,xj ]

xn1(t)

mj = inf[xj−1,xj ]

x(t), m1j = inf

[xj−1,xj ]xn1

(t)

87

88 CHAPTER 5. FURTHER RIEMANN RESULTS

Using the Infimum and Supremum Tolerance Lemma, there are points sj and tj in [xj−1, xj ] so that

Mj −ε

5(b− a)< x(sj) ≤ Mj (γ)

and

mj ≤ x(tj) < mj +ε

5(b− a)(ξ)

Thus,

U(x,π) − L(x,π) =∑π

(Mj −mj)∆xj

The term on the right hand side can be rewritten using the standard add and subtract trick as

∑π

(Mj − x(sj) + x(sj)− xn1

(sj) + xn1(sj)− xn1

(tj) + xn1(tj)− x(tj) + x(tj)−mj

)∆xj

We can then overestimate this term using the triangle inequality to find

U(x,π) − L(x,π) ≤∑π

(Mj − x(sj))∆xj +∑π

(x(sj)− xn1(sj))∆xj +

∑π

(xn1(sj)− xn1

(tj))∆xj

+∑π

(xn1(tj)− x(tj))∆xj +

∑π

(x(tj)−mj)∆xj

The first term can be estimated by Equation γ and the fifth term by Equation ξ to give

U(x,π) − L(x,π) <ε

5(b− a)

∑π

∆xj +∑π

(x(sj)− xn1(sj))∆xj +∑π

(xn1(sj)− xn1(tj))∆xj

+∑π

(xn1(tj)− x(tj))∆xj +

ε

5(b− a)

∑π

∆xj

Thus,

U(x,π) − L(x,π) < 2ε

5+∑π

(x(sj)− xn1(sj))∆xj

+∑π

(xn1(sj)− xn1

(tj))∆xj +∑π

(xn1(tj)− x(tj))∆xj

Now apply the estimate from Equationα to the first and third terms of the equation above to conclude

U(x,π) − L(x,π) < 4ε

5+∑π

(xn1(sj)− xn1

(tj))∆xj

Finally, note

| xn1(sj)− xn1

(tj) | ≤ M1j − m1

j

and so ∑π

(xn1(sj)− xn1

(tj))∆xj ≤∑π

(M1j − m1

j )∆xj

5.1. LIMIT INTERCHANGE 89

< ε/5

by Equation β. Thus, U(x,π) − L(x,π) < ε. Since the partition π refining π0 was arbitrary, wesee x satisfies the Riemann Criterion and hence, is Riemann integrable on [a, b].

It remains to show the limit interchange portion of the theorem. Since xn converges uniformly tox, given a positive ε, there is an integer N so that

supa≤t≤b

| xn(t)− x(t) | < ε/(b− a), if n > N. (ζ)

Now for any n > N , we have

|∫ b

a

x(t)dt −∫ b

a

xn(t)dt | = |∫ b

a

(x(t)− xn(t)

)dt |

≤∫ b

a

∣∣∣∣∣x(t)− xn(t)

∣∣∣∣∣dt≤

∫ b

a

supa≤t≤b

| xn(t)− x(t) | dt

<

∫ b

a

ε/(b− a)dt

= ε

using Equation ζ. This says lim∫ baxn(t)dt =

∫ bax(t)dt.

The next result is indispensable in modern analysis. Fundamentally, it states that a continuousreal-valued function defined on a compact set can be uniformly approximated by a smooth function.This is used throughout analysis to prove results about various functions. We can often verify aproperty of a continuous function, f , by proving an analogous property of a smooth function thatis uniformly close to f . We will only prove the result for a closed finite interval in <. The generalresult for a compact subset of a more general set called a Topological Space is a modification of thisproof which is actually not that more difficult, but that is another story. We follow the developmentof (Simmons (10) 1963) for this proof.

Theorem 5.1.2 Weierstrass Approximation Theorem

Let f be a continuous real-valued function defined on [0, 1]. For any ε > 0, there is apolynomial, p, such that |f(t)− p(t)| < ε for all t ∈ [0, 1], that is || p− f ||∞< ε

Proof 5.1.2We first derive some equalities. We will denote the interval [0, 1] by I . By the binomial theorem, forany x ∈ I , we have

n∑k=0

(n

k

)xk(1− x)n−k = (x+ 1− x)n = 1. (α)

Differentiating both sides of Equation α, we get

0 =

n∑k=0

(n

k

)(kxk−1(1− x)n−k − xk(n− k)(1− x)n−k−1

)


=

n∑k=0

(n

k

)xk−1(1− x)n−k−1

(k(1− x) − x(n− k)

)

=

n∑k=0

(n

k

)xk−1(1− x)n−k−1

(k − nx)

)

Now, multiply through by x(1− x), to find

0 =

n∑k=0

(n

k

)xk(1− x)n−k(k − nx).

Differentiating again, we obtain

0 =

n∑k=0

(n

k

)d

dx

(xk(1− x)n−k(k − nx)

).

This leads to a series of simplifications. It is pretty messy and many texts do not show the details, butwe think it is instructive.

0 =

n∑k=0

(n

k

)[−nxk(1− x)n−k + (k − nx)

((k − n)xk(1− x)n−k−1 + kxk−1(1− x)n−k

)]=

n∑k=0

(n

k

)[−nxk(1− x)n−k + (k − nx)(1− x)n−k−1xk−1

((k − n)x+ k(1− x)

)]=

n∑k=0

(n

k

)(− nxk(1− x)n−k + (k − nx)2(1− x)n−k−1xk−1

)= −n

n∑k=0

(n

k

)xk(1− x)n−k +

n∑k=0

(n

k

)(k − nx)2xk−1(1− x)n−k−1

Thus, since the first sum is 1, we have

n =

n∑k=0

(n

k

)(k − nx)2xk−1(1− x)n−k−1

and multiplying through by x(1− x), we have

nx(1− x) =

n∑k=0

(n

k

)(k − nx)2xk(1− x)n−k

x(1− x)

n=

n∑k=0

(n

k

)(k − nxn

)2

xk(1− x)n−k

This last equality then leads to the

n∑k=0

(n

k

)(x− k

n

)2

xk(1− x)n−k =x(1− x)

n(β)

5.1. LIMIT INTERCHANGE 91

We now define the nth order Bernstein Polynomial associated with f by

Bn(x) =

n∑k=0

(n

k

)xk(1− x)n−kf

(kn

).

Note that

f(x)−Bn(x) =

n∑k=0

(n

k

)xk(1− x)n−k

[f(x)− f

(kn

)].

Also note that f(0) − Bn(0) = f(1) − Bn(1) = 0, so f and Bn match at the endpoints. It followsthat

| f(x)−Bn(x) | ≤n∑k=0

(n

k

)xk(1− x)n−k

∣∣∣f(x)− f(kn

)∣∣∣. (γ)

Now, f is uniformly continuous on I since it is continuous. So, given ε > 0, there is a δ > 0 suchthat |x− k

n | < δ ⇒ |f(x)− f( kn )| < ε2 . Consider x to be fixed in [0, 1]. The sum in Equation γ has

only n+ 1 terms, so we can split this sum up as follows. Let K1,K2 be a partition of the index set0, 1, ..., n such that k ∈ K1 ⇒ |x− k

n | < δ and k ∈ K2 ⇒ |x− kn | ≥ δ. Then

| f(x)−Bn(x) |≤∑k∈K1

(n

k

)xk(1−x)n−k

∣∣∣f(x)−f(kn

)∣∣∣+ ∑k∈K2

(n

k

)xk(1−x)n−k

∣∣∣f(x)−f(kn

)∣∣∣.which implies

|f(x)−Bn(x)| ≤ ε

2

∑k∈K1

(n

k

)xk(1− x)n−k +

∑k∈K2

(n

k

)xk(1− x)n−k

∣∣∣f(x)− f(kn

)∣∣∣=

ε

2+∑k∈K2

(n

k

)xk(1− x)n−k

∣∣∣f(x)− f(kn

)∣∣∣.Now, f is bounded on I , so there is a real numberM > 0 such that |f(x)| ≤M for all x ∈ I . Hence∑

k∈K2

(n

k

)xk(1− x)n−k

∣∣∣f(x)− f(kn

)∣∣∣ ≤ 2M∑k∈K2

(n

k

)xk(1− x)n−k.

Since k ∈ K2 ⇒ |x− kn | ≥ δ, using Equation β, we have

δ2∑k∈K2

(n

k

)xk(1− x)n−k ≤

∑k∈K2

(n

k

)(x− k

n

)2

xk(1− x)n−k ≤ x(1− x)

n.

This implies that ∑k∈K2

(n

k

)xk(1− x)n−k ≤ x(1− x)

δ2n.

and so combining inequalities

2M∑k∈K2

(n

k

)xk(1− x)n−k ≤ 2Mx(1− x)

δ2n

We conclude then that


∑k∈K2

(n

k

)xk(1− x)n−k

∣∣∣f(x)− f(kn

)∣∣∣ ≤ 2Mx(1− x)

δ2n.

Now, the maximum value of x(1− x) on I is 14 , so

∑k∈K2

(n

k

)xk(1− x)n−k

∣∣∣f(x)− f(kn

)∣∣∣ ≤ M

2δ2n.

Finally, choose n so that n > Mδ2ε . Then M

nδ2 < ε implies M2nδ2 <

ε2 . So, Equation γ becomes

| f(x)−Bn(x) |≤ ε

2+ε

2= ε.

Note that the polynomial Bn does not depend on x ∈ I , since n only depends on M , δ, and ε, all ofwhich, in turn, are independent of x ∈ I . So, Bn is the desired polynomial, as it is uniformly withinε of f .

Comment 5.1.1 A change of variable translates this result to any closed interval [a, b].

5.2 Showing Functions Are Riemann IntegrableWe already know that continuous functions, monotone functions and functions of bounded variationare classes of functions which are Riemann Integrable on the interval [a, b]. A good reference forsome of the material in this section is (Douglas (2) 1996) although it is mostly in problems and notin the text! Hence, since f(x) =

√x is continuous on [0,M ] for any positive M , we know f is

Riemann integrable on this interval. What about the composition√g where g is just known to be non

negative and Riemann integrable on [a, b]? If g were continuous, since compositions of continuousfunctions are also continuous, we would have immediately that

√g is Riemann Integrable. However,

it is not so easy to handle this case. Let’s try this approach. Using Theorem 5.1.2, we know givena finite interval [c, d], there is a sequence of polynomials pn(x) which converge uniformly to

√x

on [c, d]. Of course, the polynomials in this sequence will change if we change the interval [c, d],but you get the idea. To apply this here, note that since g is Riemann Integrable on [a, b], g mustbe bounded. Since we assume g is non negative, we know that there is a positive number M so thatg(x) is in [0,M ] for all x in [a, b]. Thus, there is a sequence of polynomials pn which convergeuniformly to

√· on [0,M ].

Next, using Theorem 4.3.2, we know a polynomial in g is also Riemann integrable on [a, b](f2 = f · f so it is integrable and so on). Hence, pn(f) is Riemann integrable on [a, b]. Then givenε > 0, we know there is a positive N so that

| pn(u)−√u | < ε, if n > N and u ∈ [0,M ].

Thus, in particular, since g(x) ∈ [0,M ], we have

| pn(g(x))−√g(x) | < ε, if n > N and x ∈ [a, b].

We have therefore proved that pn g converges uniformly to√g on [0,M ]. Then by Theorem 5.1.1,

we see√g is Riemann integrable on [0,M ].

If you think about it a bit, you should be able to see that this type of argument would work forany f which is continuous and g that is Riemann integrable. We state this as Theorem 5.2.1.

Theorem 5.2.1 f Continuous and g Riemann Integrable Implies f g is Riemann Integrable

5.3. CONTENT ZERO 93

If f is continuous on g([a, b]) where g is Riemann Integrable on [a, b], then f g is RiemannIntegrable on [a, b].

Proof 5.2.1

Exercise 5.2.1 This proof is for you.

In general, the composition of Riemann Integrable functions is not Riemann integrable. Here isthe standard counterexample. This great example comes from (Douglas (2) 1996). Define f on [0, 1]by

f(y) =

1 if y = 00 if 0 < y ≤ 1

and g on [0, 1] by

g(x) =

1 if x = 01/p if x = p/q, (p, q) = 1, x ∈ (0, 1] and x is rational0 if x ∈ (0, 1] and x is irrational

We see immediately that f is integrable on [0, 1] by Theorem 4.8.4. We can show that g is alsoRiemann integrable on [0, 1], but we will leave this as an exercise.

Exercise 5.2.2

1. Show g is continuous at each irrational points in [01, ] and discontinuous at all rational pointsin [0, 1].

2. Show g is Riemann integrable on [0, 1] with value∫ 1

0g(x)dx = 0.

Now f g becomes

f(g(x)) =

f(1) if x = 0f(1/p) if x = p/q, (p, q) = 1, x ∈ (0, 1] and x rationalf(0) if 0 < x ≤ 1 and x irrational

=

1 if x = 00 if if x rational ∈ (0, 1]1 if if x irrational ∈ (0, 1]

The function f g above is not Riemann integrable as U(f g) = 1 and L(f g) = 0. Thus, wehave found two Riemann integrable functions whose composition is not Riemann integrable!

5.3 Sets Of Content ZeroWe already know the length of the finite interval [a, b] is b − a and we exploit this to develop theRiemann integral when we compute lower, upper and Riemann sums for a given partition. We alsoknow that the set of discontinuities of a monotone function is countable. We have seen that contin-uous functions with a finite number of discontinuities are integrable and in the last section, we sawa function which was discontinuous on a countably infinite set and still was integrable! Hence, weknow that a function is integrable should imply something about its discontinuity set. However, theconcept of length doesn’t seem to apply as there are no intervals in these discontinuity sets. With thatin mind, let’s introduce a new notion: the content of a set. We will follow the development of a setof content zero as it is done in (Sagan (9) 1974).


Definition 5.3.1 Sets Of Content Zero

A subset S of < is said to have content zero if and only if given any positive ε we can find asequence of bounded open intervals Jεn = (an, bn) either finite in number or infinite sothat

S ⊆ ∪ Jn,

with the total length ∑(bn − an) < ε

If the sequence only has a finite number of intervals, the union and sum are written from 1to N where N is the number of intervals and if there are infinitely many intervals, the sumand union are written from 1 to∞.

Comment 5.3.1

1. A single point c in < has content zero because c ∈ (c− ε/2, c+ ε/2) for all positive ε.

2. A finite number of points S = c1, . . . , ck in < has content zero because Bi = ci ∈ (ci −ε/(2k), ci+ε/(2k)) for all positive ε. Thus, S ⊆ ∪ki=1Bi and the total length of these intervalsis smaller than ε.

3. The rational numbers have content zero also. Let ci be any enumeration of the rationals.Let Bi = (ci− ε/(2i), ci + ε/(2i)) for any positive ε. The Q is contained in the union of theseintervals and the length is smaller than ε

∑∞i=1 1/2i = ε.

4. Finite unions of sets of content zero also have content zero.

5. Subsets of sets of content zero also have content zero.

Hence, the function g above is continuous on [0, 1] except on a set of content zero. We make thismore formal with a definition.

Definition 5.3.2 Continuous Almost Everywhere

The function f defined on the interval [a, b] is said to be continuous almost everywhere ifthe set of discontinuities of f has content zero. We abbreviate the phrase almost everywhereby writing a.e.

We are now ready to prove an important theorem which is known as the Riemann - LebesgueLemma. This is also called Lebesgue’s Criterion For the Riemann Integrability of BoundedFunctions . We follow the proof given in (Sagan (9) 1974).

Theorem 5.3.1 Riemann - Lebesgue Lemma

(i) f ∈ B[a, b] and continuous a.e. implies f ∈ RI[a, b].

(ii) f ∈ RI[a, b] implies f is continuous a.e.

Proof 5.3.1The proof of this result is fairly complicated. So grab a cup of coffee, a pencil and prepare for a longbattle!(i):


Proof We will prove this by showing that for any positive ε, we can find a partition π0 so that theRiemann Criterion is satisfied. First, since f is bounded, there is are numbers m and M so thatm ≤ f(x) ≤ M for all x in [a, b]. If m and M we the same, then f would be constant and it wouldtherefore be continuous. If this case, we know f is integrable. So we can assume without loss ofgenerality that M −m > 0. Let D denote the set of points in [a, b] where f is not continuous. Byassumption, the content of D is zero. Hence, given a positive ε there is a sequence of bounded openintervals Jn = (an, bn) (we will assume without loss of generality that there are infinitely many suchintervals) so that

D ⊆ ∪Jn,∑

(bn − an) < ε/(2(M −m)).

Now if x is from [a, b], x is either in D or in the complement of D, DC . Of course, if x ∈ DC , thenf is continuous at x. The set

E = [a, b] ∩

(∪Jn

)Cis compact and so f must be uniformly continuous on E. Hence, for the ε chosen, there is a δ > 0 sothat

| f(y)− f(x) |< ε/(8(b− a)), (∗)

if y ∈ (x− δ, x+ δ) ∩ E. Next, note that

O = Jn, Bδ/2(x) | x ∈ E

is an open cover of [a, b] and hence must have a finite sub cover. Call this finite sub cover O′ andlabel its members as follows:

O′ = Jn1 , . . . , Jnr , Bδ/2(x1), . . . , Bδ/2(xs)

Then it is also true that we know that

[a, b] ⊆ O′′ = Jn1 , . . . , Jnr , Bδ/2(x1) ∩ E, . . . , Bδ/2(xs) ∩ E

All of the intervals in O′′ have endpoints. Throw out any duplicates and arrange these endpoints inincreasing order in [a, b] and label them as y1, . . . , yp−1. Then, let

π0 = y0 = a, y1, y2, . . . , yp−1, yp = b

be the partition formed by these points. Recall where the points yj come from. The endpoints of theBδ/2(xi) ∩E sets are not in any of the intervals Jnk . So suppose two successive points yj−1 and yjsatisfied yj−1 is in an interval Jnk and the next point yj was an endpoint of aBδ/2(xi)∩E set whichis also inside Jnk . By our construction, this can not happen as all of the Bδ/2(xi) ∩ E are disjointfrom the Jnk sets. Hence, the next point yj either must be in the set Jnk also or it must be outside. Ifyj−1 is inside and yj is outside, this is also a contradiction as this would give us a third point, call itz temporarily, so that

yj−1 < z < yj

with z a new distinct endpoint of the finite cover O′′. Since we have already ordered these points,this third point is not a possibility. Thus, we see (yj−1, yj) is in some Jnk or neither of the points isin any Jnk . Hence, we have shown that given the way the points yj were chosen, either (yj−1, yj)is inside some interval Jnq or it’s closure [yj−1, yj ] lies in none of the Jnq for any 1 ≤ q ≤ r. Butthat means (yj−1, yj) lies in some Bδ/2(xi). Note this set uses the radius δ/2 and so we can say the


closed interval [yj−1, yj ] must be contained in some Bδ(xi).

Now we separate the index set 1, 2, . . . , p into two disjoint sets. We define A1 to be the set ofall indices j so that (yj−1, yj) is contained in some Jnk . Then we set A2 to be the complement ofA1 in the entire index set, i.e. A2 = 1, 2, . . . , p − A1. Note, by our earlier remarks, if j is in A2,[yj−1, yj ] is contained in some Bδ(xi) ∩ E. Thus,

U(f,π0)− L(f,π0) =

n∑j=1

(Mj −mj

)∆yj

=∑j∈A1

(Mj −mj

)∆yj +

∑j∈A2

(Mj −mj

)∆yj

Let’s work with the first sum: we have

∑j∈A1

(Mj −mj

)∆yj ≤

(M −m

) ∑j∈A1

∆yj

< (M −m) ε/(2(M −m)) = ε/2

Now if j is in A2, then [yj−1, yj ] is contained in some Bδ(xi) ∩ E. So any two points u and v in[yj−1, yj ] satisfy | u − xi |< δ and | v − xi |< δ. Since these points are this close, the uniformcontinuity condition, Equation ∗, holds. Therefore

| f(u)− f(v) | ≤ | f(u)− f(xi) | + | f(v)− f(xi) |< ε/(4(b− a)).

This holds for any u and v in [yj−1, yj ]. In particular, we can use the Supremum and InfimumTolerance Lemma to choose uj and vj so that

Mj − ε/(8(b− a)) < f(uj), mj + ε/(8(b− a)) > f(vj).

It then follows thatMj −mj < f(uj)− f(vj) + ε/(4(b− a)).

Now, we can finally estimate the second summation term. We have

∑j∈A2

(Mj −mj

)∆yj <

∑j∈A2

(| f(uj)− f(vj) | +ε/(4(b− a))

)∆yj

<∑j∈A2

(| f(uj)− f(vj) |

)∆yj + ε/(4(b− a))

∑j∈A2

∆yj

< ε/(4(b− a))∑j∈A2

∆yj + ε/(4(b− a))∑j∈A2

∆yj

< ε/2

Combining our estimates, we have

U(f,π0)− L(f,π0) =∑j∈A1

(Mj −mj

)∆yj +

∑j∈A2

(Mj −mj

)∆yj

< ε/2 + ε/2 = ε.


Any partition π that refines π0 will also satisfy U(f,π) − L(f,π) < ε. Hence, f satisfies theRiemann Criterion and so f is integrable.

(ii):

Proof We begin by noting that if f is discontinuous at a point x in [a, b], if and only if there is apositive integer m so that

∀δ > 0, ∃y ∈ (x− δ, x+ δ) ∩ [a, b] 3 | f(y)− f(x) |≥ 1/m.

This allows us to define some interesting sets. Define the set Em by

Em = x ∈ [a, b] | ∀δ > 0 ∃y ∈ (x− δ, x+ δ) ∩ [a, b] 3 | f(y)− f(x) |≥ 1/m,

Then, the set of discontinuities of f ,D can be expressed asD = ∪∞j=1Em.

Now let π = x0, x1, . . . , xn be any partition of [a, b]. Then, given any positive integer m, theopen subinterval [xk−1, xk] either intersects Em or it does not. Define

A1 =

k ∈ 1, . . . , n | (xk−1, xk) ∩ Em 6= ∅

,

A2 =

k ∈ 1, . . . , n | (xk−1, xk) ∩ Em = ∅

By construction, we have A1 ∩A2 = ∅ and A1 ∪A2 = 1, . . . , n.

We assume f is integrable on [a, b]. So, by the Riemann Criterion, given ε > 0, and a positive integerm, there is a partition π0 such that

U(f,π)− L(f,π) < ε/(2m), ∀π0 π. (∗∗)

It follows that if π0 = y0, y1, . . . , yn, then

U(f,π0)− L(f,π0) =

n∑k=1

(Mk −mk)∆yk

=∑k∈A1

(Mk −mk)∆yk +∑k∈A2

(Mk −mk)∆yk

If k is in A1, then by definition, there is a point uk in Em and a point vk in (yk−1, yk) so that| f(uk)− f(vk) |≥ 1/m. Also, since uk and vk are both in (yk−1, yk),

Mk −mk ≥| f(uk)− f(vk) | .

Thus, ∑k∈A1

(Mk −mk)∆yk ≥∑k∈A1

| f(uk)− f(vk) | ∆yk ≥ (1/m)∑k∈A1

∆yk.

Also, the second term,∑k∈A2

(Mk −mk)∆yk is non-negative and so using Equation ∗∗, we find

ε/(2m) > U(f,π0 − L(f,π0 ≥ (1/m)∑k∈A1

∆yk.


which implies∑k∈A1

∆yk < ε/2.The partition π0 divides [a, b] as follows:

[a, b] =

(∪k∈A1

(yk−1, yk)

)∪

(∪k∈A2

(yk−1, yk)

)∪(y0, . . . , yn

)= C1 ∪ C2 ∪ π0

By the way we constructed the sets Em, we know Em does not intersect C2. Hence, we can say

Em =

(C1 ∩ Em

)∪

(Em ∩ π0

)

Therefore, we have C1 ∩Em ⊆ ∪k∈A1 (yk−1, yk) with∑k∈A1

∆yk < ε/2. Since ε is arbitrary, wesee C1 ∩ Em has content zero. The other set Em ∩ π0 consists of finitely many points and so it alsohas content zero by the comments at the end of Definition 5.3.1. This shows that Em has content zerosince it is the union of two sets of content zero. We finish by noting D = ∪Em also has content zero.The proof of this we leave as an exercise.

Exercise 5.3.1 Prove that if Fn ⊆ [a, b] has content zero for all n, then F = ∪Fn also has contentzero.

Part III

Riemann - Stieljes Integrals

99

Chapter 6

The Riemann-Stieltjes Integral

In classical analysis, the Riemann-Stieltjes integral was the first attempt to generalize the idea of thesize, or measure, of a subset of the real numbers. Instead of simply using the length of an interval asa measure, we can use any function that satisfies the same properties as the length function.

Let f and g be any bounded functions on the finite interval [a, b]. If π is any partition of [a, b] andσ is any evaluation set, we can extend the notion of the Riemann sum S(f,π,σ to the more generalRiemann - Stieljes sum as follows:

Definition 6.0.3 The Riemann - Stieljes Sum

Let f, g ∈ B[a, b], π ∈ Π[a, b] and σ ⊆ π. Let the partition points in π bex0, x1, . . . , xp and the evaluation points be s1, s2, . . . , sp as usual. Define

∆gj = g(xj)− g(xj − i), 1 ≤ j ≤ p.

and the Riemann - Stieljes sum for integrand f and integrator g for partition π and eval-uation set π by

S(f, g,π,σ) =∑j∈π

f(sj) ∆gj

This is also called the Riemann - Stieljes sum for the function f with respect to thefunction g for partition π and evaluation set σ.

Of course, you should compare this definition to Definition 4.1.1 to see the differences! We canthen define the Riemann - Stieljes integral of f with respect to g using language very similar to thatof Definition 4.1.2.

Definition 6.0.4 The Riemann - Stieljes Integral

101

102 CHAPTER 6. RIEMANN-STIELJES

Let f, g ∈ B[a, b]. If there is a real number I so that for all positive ε, there is a partitionπ0 ∈ Π[a, b] so that ∣∣∣∣∣S(f, g,π,σ)− I

∣∣∣∣∣ < ε

for all partitions π that refine π0 and evaluation sets σ from π, then we say f is Riemann- Stieljes integrable with respect to g on [a, b]. We call the value I the Riemann - Stieljesintegral of f with respect to g on [a, b]. We use the symbol

I = RS(f, g; a, b)

to denote this value. We call f the integrand and g the integrator.

As usual, there is the question of what pairs of functions (f, g) will turn out to have a finiteRiemann - Stieljes integral. The collection of the functions f fromB[a, b] that are Riemann - Stieljesintegrable with respect to a given integrator g from B[a, b] is denoted by RS[g, a, b].

Comment 6.0.2 If g(x) = x on [a, b], then RS[g, a, b] = RI[a, b] and RS(f, g; a, b) =∫ baf(x)dx.

Comment 6.0.3 We will use the standard conventions: RS(f, g; a, b) = −RS(f, g; b, a) andRS(f, g; a; a) =0.

6.1 Standard Properties Of The Riemann - Stieljes Integral

We can easily prove the usual properties that we expect an integration type mapping to have.

Theorem 6.1.1 The Linearity of the Riemann - Stieljes Integral

If f1 and f2 are in RS[g, a, b], then

(i)c1f1 + c2f2 ∈ RS[g, a, b], ∀c1, c2 ∈ <

(ii)RS(c1f1 + c2f2, g; a, b) = c1RS(f1, g; a, b) + c2RS(f2, g; a, b)

If f ∈ RS[g1, a, b] and f ∈ RS[g2, a, b] then

(i)f ∈ RS[c1g1 + c2g2, a, b], ∀c1, c2 ∈ <

(ii)RS(f, c1g1 + c2g2; a, b) = c1RS(f, g1; a, b) + c2RS(f, g2; a, b)

Proof 6.1.1

Exercise 6.1.1 We leave these proofs to you as an exercise.

The proof of these statements is quite similar in spirit to those of Theorem 4.1.1. You shouldcompare the techniques!

6.1. PROPERTIES 103

To give you a feel for the kind of partition arguments we use for Riemann - Stieljes proofs (youwill no doubt enjoy working out these details for yourselves in various exercises), we will go throughthe proof of the standard Integration By Parts formula in this context.

Theorem 6.1.2 Riemann Stieljes Integration By Parts

If f ∈ RS[g, a, b], then g ∈ RS[f, a, b] and

RS(g, f ; a, b) = f(x)g(s)

∣∣∣∣∣b

a

−RS(f, g; a, b)

Proof 6.1.2Since f ∈ RS[g, a, b], there is a number If = RS(f, g; a, b) so that given a positive ε, there is apartition π0 such that ∣∣∣∣∣S(f, g,π,σ − If

∣∣∣∣∣ < ε, π0 π, σ ⊆ π. (α)

For such a partition π and evaluation set σ ⊆ π, we have

π = x0, x1, . . . , xp,σ = s1, . . . , sp

and

S(g, f,π,σ) =∑π

g(sj)∆fj .

We can rewrite this as

S(g, f,π,σ =∑π

g(sj)f(xj) −∑π

g(sj)f(xj−1) (β)

Also, we have the identity (it is a collapsing sum)

∑π

(f(xj)g(xj)− f(xj−1)g(xj−1)

)= f(b)g(b)− f(a)g(a). (γ)

Thus, using Equation β and Equation γ, we have

f(b)g(b)− f(a)g(a) − S(g, f,π,σ) =∑π

f(xj)

(g(xj)− g(sj)

)(ξ)

+∑π

f(xj−1)

(g(sj)− g(xj−1)

)

Since σ ⊆ π, we have the ordering

a = x0 ≤ s1 ≤ x1 ≤ s2 ≤ x2 ≤ . . . ≤ xp−1 ≤ sp ≤ xp = b.


Hence, the points above are a refinement of π we will call π′. Relabel the points of π′ as

π′ = y0, y1, . . . , yq

and note that the original points of π now form an evaluation set σ′ of π′. We can therefore rewriteEquation ξ as

f(b)g(b)− f(a)g(a) − S(g, f,π,σ) =∑π

f(yj)∆gj = S(f, g,π′,σ′)

Let Ig = f(b)g(b)−f(a)g(a)− If . Then since π0 π π′, we can apply Equationα to conclude

ε >

∣∣∣∣∣S(f, g,π′,σ′ − If

∣∣∣∣∣=

∣∣∣∣∣f(b)g(b)− f(a)g(a) − S(g, f,π,σ) − If

∣∣∣∣∣=

∣∣∣∣∣S(g, f,π,σ) − Ig

∣∣∣∣∣Since our choice of refinement π of π0 and evaluation set σ was arbitrary, we have shown thatg ∈ RS[f, a, b] with value

RS(g, f, a, b) = f(x)g(x)

∣∣∣∣∣b

a

−RS(f, g, a, b).

6.2 Step Function IntegratorsWe now turn our attention to the question of what pairs of functions might have a Riemann - Stieljesintegral. All we know so far is that if g(x) = x on [a, b] is labeled as g = id, then RS[f, id, a, b] =RI[f, a, b].

First, we need to define what we mean by a Step Function.

Definition 6.2.1 Step Function

We say g ∈ B[a, b] is a Step Function if g only has finitely many jump discontinuities on[a, b] and g is constant on the intervals between the jump discontinuities. Thus, we mayassume there is a non negative integer p so that the jump discontinuities are ordered andlabeled as

c0 < c1 < c2 < . . . < cp

and g is constant on each subinterval (ck−1, ck) for 1 ≤ k ≤ p.

Comment 6.2.1 We can see g(c−k ) and g(c+k ) both exist and are finite with g(c−k ) the value g hason (ck−1, ck) and g(c+k ) the value g has on (ck, ck+1). At the endpoints, g(a+) and g(b−) are alsodefined. The actual finite values g takes on at the points cj are completely arbitrary.

We can prove a variety of results about Riemann Stieljes integrals with step function integrations.

Lemma 6.2.1 One Jump Step Functions As Integrators One

6.2. STEP INTEGRATORS 105

Let g ∈ B[a, b] be a step function having only one jump at some c in [a, b]. Let f ∈ B[a, b].Then f ∈ C[a, b] implies f ∈ RS[g, a, b] and

• If c ∈ (a, b), then RS(f, g; a, b) = f(c)[g(c+)− g(c−)].

• If c = a, then RS(f, g; a, b) = f(a)[g(a+)− g(a)].

• If c = b, then RS(f, g; a, b) = f(b)[g(b)− g(b−)].

Proof 6.2.1Let π be any partition of [a, b]. We will assume that c is a partition point of π because if not, we canuse the argument we have used before to construct an appropriate refinement as done, for example,in the proof of Lemma 4.5.1. Letting the partition points be

π = x0, x1, . . . , xp,

we see there is a partition point xk0 = c with k0 6= 0 or p. Hence, on [xk0−1, xk0 ] = [xk0−1, c],∆gk0 = g(c) − g(xk0−1). However, since g has a single jump at c, we see that the value g(xk0−1)must be g(c−). Thus, ∆gk0 = g(c)− g(c−). A similar argument shows that ∆gk0 = g(c+)− g(c).Further, since g has only one jump, all the other terms ∆gk are zero. Hence, for any evaluation setσ in π, we have σ = s1, . . . , sp and

S(f, g,π,σ) = f(sk0)∆gk0 + f(sk0+1∆gk0+1

= f(sk0)

(g(c)− g(c−)

)+ f(sk0+1

(g(c+)− g(c)

)

=

(f(sk0)− f(c) + f(c)

)(g(c)− g(c−)

)

+

(f(sk0+1 − f(c) + f(c)

)(g(c+)− g(c)

)

Thus, we obtain

S(f, g,π,σ) =

(f(sk0)− f(c)

)(g(c)− g(c−)

)

+

(f(sk0+1 − f(c)

)(g(c+)− g(c)

)(α)

+ f(c)

(g(c+)− g(c−)

)

We know f is continuous at c. Let A = max

(| g(c) − g(c−) |, | g(c+) − g(c) |

). Then A > 0

because g has a jump at c. Since f is continuous at c, given ε > 0, there is a δ > 0, so that

| f(x)− f(c) | < ε/(2A), x ∈ (c− δ, c+ δ) ∩ [a, b]. (β)

In fact, since c is an interior point of [a, b], we can choose δ so small that (c − δ, c + δ) ⊆ [a, b].Now, if π0 is any partition with || π0 ||< δ containing c as a partition point, we can argue as we did


in the prefatory remarks to this proof. Thus, there is an index k0 so that

[xk0−1, xk0 = c] ⊆ (c− δ, c], [c = xk0 , xk0+1] ⊆ [c, c+ δ).

This implies that[xk0−1, xk0+1] ⊆ (c− δ, c+ δ)

and so the evaluation points, labeled as usual, sk0 and sk0+1 are also in (c − δ, c + δ). ApplyingEquation β, we have

| f(sk0)− f(c)| < ε/(2A), | f(sk0+1)− f(c)| < ε/(2A.

From Equation α, we then have∣∣∣∣∣S(f, g,π,σ)− f(c)

(g(c+)− g(c−)

)∣∣∣∣∣ ≤

∣∣∣∣∣(f(sk0)− f(c)

) (g(c)− g(c−)

)∣∣∣∣∣+

∣∣∣∣∣(f(sk0+1 − f(c)

) (g(c+)− g(c)

)∣∣∣∣∣< ε/(2A)

∣∣∣∣g(c)− g(c−)

∣∣∣∣ + ε/(2A)

∣∣∣∣g(c+)− g(c)

∣∣∣∣< ε

Finally, if π0 π, then || π ||< δ also and the same argument shows that for any evaluation setσ ⊆ π, we have ∣∣∣∣∣S(f, g,π,σ)− f(c)

(g(c+)− g(c−)

)∣∣∣∣∣ < ε

This proves that f ∈ RS[g, a, b] andRS(f, g; a, b) = f(c)

(g(c+)−g(c−)

). Now, if c = a or c = b,

the arguments are quite similar, except one sided and we findRS(f, g; a, b) = f(a)

(g(a+)−g(a)

)

or RS(f, g; a, b) = f(b)

(g(b)− g(b−)

).

Lemma 6.2.2 One Jump Step Functions As Integrators Two

Let g ∈ B[a, b] be a step function having only one jump at some c in [a, b]. Let f ∈ B[a, b].If c ∈ (a, b), f(c−) = f(c) and g(c+) = g(c), then f ∈ RS[g, a, b]. We can rephrase thisas: if c is an interior point, f is continuous from the left at c and g is continuous from theright at c, then f ∈ RS[g, a, b] and

• If c ∈ (a, b), then RS(f, g; a, b) = f(c)[g(c)− g(c−)].

• If c = a, then RS(f, g; a, b) = f(a)[g(a)− g(a)] = 0.

• If c = b, then RS(f, g; a, b) = f(b)[g(b)− g(b−)].

Proof 6.2.2

6.2. STEP INTEGRATORS 107

To prove this result, we first use the initial arguments of Lemma 6.2.1 and then we note in this case,f is continuous from the left at c so f(c−) = f(c). Further, g is continuous from the right; hence,g(c) = g(c+). Thus, Equation α reduces to

S(f, g,π,σ) =

(f(sk0)− f(c)

)(g(c)− g(c−)

)+ f(c)

(g(c)− g(c−)

)(α′)

Let L =| g(c)− g(c−) |. Then, given ε > 0, since f is continuous from the left, there is a δ > 0 sothat

| f(x)− f(c) |< ε/L, x ∈ (c− δ, c] ⊆ [a, b].

As usual, we can restrict our attention to partitions that contain the point c. We continue to usexi’s and sj’s to represent points in these partitions and associated evaluation sets. Let π be such apartition with xk0 = c and || π ||< δ. Let σ be any evaluation set of π. Then, we have

[xk0−1, xk0 ] ⊆ (c− δ, c]

and thus| f(sk0)− f(c) |< ε/L.

Hence, ∣∣∣∣∣S(f, g,π,σ)− f(c)

(g(c+ − g(c)

)∣∣∣∣∣ =

∣∣∣∣f(sk0 − f(c)

∣∣∣∣ ∣∣∣∣g(c)− g(c−)

∣∣∣∣< ε.

Finally, just as in the previous proof, if π0 π, then || π ||< δ also and the same argument showsthat for any evaluation set σ ⊆ π, we have∣∣∣∣∣S(f, g,π,σ)− f(c)

(g(c)− g(c−)

)∣∣∣∣∣ < ε

This proves that f ∈ RS[g, a, b] and RS(f, g; a, b) = f(c)

(g(c) − g(c−)

). Now, if c = a or

c = b, the arguments are again similar, except one sided and we find RS(f, g; a, b) = f(a)

(g(a)−

g(a)

)= 0 or RS(f, g; a, b) = f(b)

(g(b)− g(b−)

).

Lemma 6.2.3 One Jump Step Functions As Integrators Three

Let g ∈ B[a, b] be a step function having only one jump at some c in [a, b]. Let f ∈ B[a, b].If c ∈ (a, b), f(c+) = f(c) and g(c−) = g(c), then f ∈ RS[g, a, b]. We can rephrase thisas: if c is an interior point, f is continuous from the right at c and g is continuous from theleft at c, then f ∈ RS[g, a, b] and

• If c ∈ (a, b), then RS(f, g; a, b) = f(c)[g(c+)− g(c)].

• If c = a, then RS(f, g; a, b) = f(a)[g(a+)− g(a)].

• If c = b, then RS(f, g; a, b) = f(b)[g(b)− g(b)] = 0.


Proof 6.2.3This is quite similar to the argument presented for Lemma 6.2.2. We find f ∈ RS[g, a, b] and

RS(f, g; a, b) = f(c)

(g(c+) − g(c)

). Now, if c = a or c = b, the arguments are again similar,

except one sided and we find RS(f, g; a, b) = f(a)

(g(a+) − g(a)

)= 0 or RS(f, g; a, b) =

f(b)

(g(b)− g(b)

)= 0.

We can then generalize to a finite number of jumps.

Lemma 6.2.4 Finite Jump Step Functions As Integrators

Let g be a step function on [a, b] with jump discontinuities at

a ≤ c0, c1, . . . , ck−1, ck ≤ b.

Assume f ∈ B[a, b]. Then, if

(i) f is continuous at cj , or

(ii) f is left continuous at cj and g is right continuous at cj , or

(iii) f is right continuous at cj and g is left continuous at cj ,

then, f ∈ RS[f, g, a, b] and

RS(f, g, a, b) = f(a)

(g(a+)−g(a)

)+

k∑j=0

f(cj)

(g(c+j )−g(c−j )

)+f(b)

(g(b)−g(b−)

).

Proof 6.2.4Use Lemma 6.2.1, Lemma 6.2.2 and Lemma 6.2.3 repeatedly.

6.3 Monotone Integrators

The next step is to learn how to deal with integrators that are monotone functions. To do this, weextend the notion of Darboux Upper and Lower Sums in the obvious way.

Definition 6.3.1 Upper and Lower Riemann - Stieljes Darboux Sums

6.3. MONOTONE INTEGRATORS 109

Let f ∈ B[a, b] and g ∈ B[a, b] be monotone increasing. Let π be any partition of [a, b]with partition points

π = x0, x1, . . . , xp

as usual. Define

Mj = supx∈[xj−1,xj ]

f(x), mj = infx∈[xj−1,xj ]

f(x).

The Lower Riemann - Stieljes Darboux Sum for f with respect to g on [a, b] for the partitionπ is

L(f, g,π) =∑π

mj∆gj

and the Upper Riemann - Stieljes Darboux Sum for f with respect to g on [a, b] for thepartition π is

U(f, g,π) =∑π

Mj∆gj

Comment 6.3.1 It is clear that for any partition π and associated evaluation set σ, that we have theusual inequality chain:

L(f, g,π) ≤ S(f, g,π,σ ≤ U(f, g,π)

The following theorems have proofs very similar to the ones we did for Theorem 4.2.1 and The-orem 4.2.2.

Theorem 6.3.1 π π′ Implies L(f, g,π) ≤ L(f, g,π′) and U(f, g,π) ≥ U(f, g,π′)

Assume g is a bounded monotone increasing function on [a, b] and f ∈ B[a, b]. Then ifπ π′, then L(f, g,π) ≤ L(f, g,π′) and U(f, g,π) ≥ U(f, g,π′).

Theorem 6.3.2 L(f, g,π1) ≤ U(f, g,π2)

Let π1 and π2 be any two partitions in Π[a, b]. Then L(f, g,π1) ≤ U(f, g,π2).

These two theorems allow us to prove the following

Theorem 6.3.3 The Upper And Lower Riemann - Stieljes Darboux Integral Are Finite

Let f ∈ B[a, b] and let g be a bounded monotone increasing function on [a, b]. Let U =L(f, g,π) |π ∈ Π[a, b] and V = U(f, g,π) |π ∈ Π[a, b]. Define L(f, g) = sup U ,and U(f, g) = inf V . Then L(f, g) and U(f, g) are both finite. Moreover, L(f, g) ≤U(f, g).

We can then define upper and lower Riemann - Stieljes integrals analogous to the way we definedthe upper and lower Riemann integrals.

Definition 6.3.2 Upper and Lower Riemann - Stieljes Integrals

Let f ∈ B[a, b] and g be a bounded, monotone increasing function on [a, b]. The Upperand Lower Riemann - Stieljes integrals of f with respect to g are U(f, g) and L(f, g),respectively.

Thus, we can define the Riemann - Stieljes Darboux integral of f ∈ B[a, b] with respect to thebounded monotone increasing integrator g.


Definition 6.3.3 The Riemann - Stieljes Darboux Integral

Let f ∈ B[a, b] and g be a bounded, monotone increasing function on [a, b]. We say f isRiemann - Stieljes Darboux integrable with respect to the integrator g ifU(f, g) = L(f, g).We denote this common value by RSD(f, g, a, b).

6.4 The Riemann - Stieljes Equivalence TheoremThe connection between the Riemann - Stieljes and Riemann - Stieljes Darboux integrals is obtainedusing an analog of the familiar Riemann Condition we have seen before in Definition 4.2.4.

Definition 6.4.1 The Riemann - Stieljes Criterion For Integrability

Let f ∈ B[a, b] and g be a bounded monotone increasing function on [a, b]. We say theRiemann Condition or Criterion holds for f with respect to g if there is a partition of [a, b],π0 so that

U(f, g,π)− L(f, g,π) < ε, π0 π.

We can then prove an equivalence theorem for Riemann - Stieljes and Riemann - Stieljes Darbouxintegrability.

Theorem 6.4.1 The Riemann Stieljes Integral Equivalence Theorem

Let f ∈ B[a, b] and g be a bounded monotone increasing function on [a, b]. Then thefollowing are equivalent.

(i) f ∈ RS[g, a, b].

(ii) Riemann’s Criterion holds for f with respect to g.

(iii) f is Riemann - Stieljes Darboux Integrable, i.e, L(f, g) = U(f, g), andRS(f, g; a, b) = RSD(f, g; a, b).

Proof 6.4.1The arguments are essentially the same as presented in the proof of Theorem 4.2.4 and hence, youwill be asked to go through the original proof and replace occurrences of ∆xj with ∆gj and b − awith g(b)− g(a).

Comment 6.4.1 We have been very careful to distinguish between Riemann - Stieljes and Riemann -Stieljes Darboux integrability. Since we now know they are equivalent, we can begin to use a commonnotation. Recall, the common notation for the Riemann integral is

∫ baf(x)dx. We will now begin

using the notation∫ baf(x)dg(x) to denote the common value RS(f, g; a, b) = RSD(f, g; a, b).

We thus know intbaf(x)dx is equivalent to the Riemann - Stieljes integral of f with respect to theintegrator g(x) = x. Hence, in this case, we could write g(x) = id(x) = x, where id is the identityfunction. We could then use the notation

∫ baf(x)dx =

∫ baf(x)did. However, that is cumbersome.

We can easily remember that the identity mapping is simply x itself. So replace did by dx to obtain∫ baf(x)dx. The use of the (x) in these notations has always been helpful to allow us to handle

substitution type rules, but it is certainly somewhat awkward. A reasonable change of notation wouldbe to go to using boldface for the f and g in these integrals and write

∫ bafdg giving

∫ bafdx for the

simpler Riemann integral.

6.5. FURTHER PROPERTIES 111

You can see no matter what we do the symbolism becomes awkward. For example, supposef(x) = sin(x2) on [0, π] and g(x) = x2. Then, how do we write

∫ π0fdg? We will usually abuse

our integral notation and write∫ π

0sin(x2)d(x2).

6.5 Further Properties Of The Riemann-Stieljes Integral

We can prove the following useful collection of facts about Riemann - Stieljes integrals.

Theorem 6.5.1 Properties Of The Riemann Stieljes Integral

Let the integrator g be bounded and monotone increasing on [a, b]. Assume f1, f2 and f3

are in RS[f, g, a, b]. Then

(i) | f |∈ RS[g, a, b];

(ii) ∣∣∣∣∣∫ b

a

f(x)dg(x)

∣∣∣∣∣ ≤∫ b

a

| f | dg(x);

(iii) f+ = maxf, 0 ∈ RS[g, a, b];

(iv) f− = max−f, 0 ∈ RS[g, a, b];

(v) ∫ b

a

f(x)dg(x) =

∫ b

a

[f+(x)− f−(x)]dg(x)

=

∫ b

a

f+(x)dg(x)−∫ b

a

f−(x)dg(x)∫ b

a

| f(x) | dg(x) =

∫ b

a

[f+(x) + f−(x)]dg(x)

=

∫ b

a

f+(x)dg(x) +

∫ b

a

f−(x)dg(x);

(vi) f2 ∈ RS[g, a, b];

(vii) f1f2 ∈ RS[g, a, b];

(viii) If there exists m such that 0 < m ≤ f(x) for all x in [a, b], then 1/f ∈ RS[g, a, b].

Proof 6.5.1The arguments are straightforward modifications of the proof of Theorem 4.3.2 using b−a = g(b)−g(a) and ∆xj = ∆gj .

We can also easily prove the following fundamental estimate.

Theorem 6.5.2 Fundamental Riemann Stieljes Integral Estimates


Let g be bounded and monotone increasing on [a, b] and let f ∈ RS[g, a, b]. Let m =infx f(x) and let M = supx f(x). Then

m(g(b)− g(a)) ≤∫ b

a

f(x)dg(x) ≤M(g(b)− g()a).

In addition, Riemann - Stieljes integrals are also order preserving as we can modify the proof ofTheorem 4.1.3 quite easily.

Theorem 6.5.3 The Riemann Stieljes Integral Is Order Preserving

Let g be bounded and monotone increasing on [a, b] and f, f1, f2 ∈ RS[g, a, b] with f1 ≤f2 on [a, b]. Then the Riemann Stieljes integral is order preserving in the sense that

(i)

f ≥ 0⇒∫ b

a

f(x)dg(x) ≥ 0;

(ii)

f1 ≤ f2 ⇒∫ b

a

f1(x)dg(x) ≤∫ b

a

f2(x)dg(x).

We also want to establish the familiar summation property of the Riemann Stieljes integral over aninterval [a, b] = [a, c] ∪ [c, b]. We can modify the proof of the corresponding result in Lemma 4.5.1

as usual to obtain Lemma 6.5.4.

Lemma 6.5.4 The Upper And Lower Riemann - Stieljes Darboux Integral Is Additive On In-tervals

Let g be bounded and monotone increasing on [a, b] and f ∈ B[a, b]. Let c ∈ (a, b). Define∫ b

a

f(x) dg(x) = L(f, g) and∫ b

a

f(x) dg(x) = U(f, g)

denote the lower and upper Riemann - Stieljes Darboux integrals of f on with respect to gon [a, b], respectively. Then we have∫ b

a

f(x)dg(x) =

∫ c

a

f(x)dg(x) +

∫ b

c

f(x)dg(x)

∫ b

a

f(x)dg(x) =

∫ c

a

f(x)dg(x) +

∫ b

c

f(x)dg(x).

Lemma 6.5.4 allows us to prove existence of the Riemann - Stieljes on [a, b] implies it alsoexists on subintervals of [a, b] and the Riemann - Stieljes value is additive. The proofs are obviousmodifications of the proofs of Theorem 4.5.2 and Theorem 4.5.3, respectively.

Theorem 6.5.5 The Riemann Stieljes Integral Exists On Subintervals

Let g be bounded and monotone increasing on [a, b]. If f ∈ RS[g, a, b] and c ∈ (a, b),then f ∈ RS[g, a, c] and f ∈ RS[g, c, b].

6.6. BOUNDED VARIATION INTEGRATORS 113

Theorem 6.5.6 The Riemann Integral Is Additive On Subintervals

If f ∈ RS[g, a, b] and c ∈ (a, b), then∫ b

a

f(x)dg(x) =

∫ c

a

f(x)dg(x) +

∫ b

c

f(x)dg(x).

6.6 Bounded Variation Integrators

We now turn our attention to integrators which are of bounded variation. By Theorem 3.4.3, we knowthat if g ∈ BV [a, b], then we can write g = u− v where u and v are monotone increasing on [a, b].Note if h is any other monotone increasing function on [a, b], we could also use the decomposition

g = (u+ h)− (v + h)

as well, so this representation is certainly not unique. We must be very careful when we extend theRiemann - Stieljes integral to bounded variation integrators. For example, even if f ∈ RS[g, a, b]it does not always follow that f ∈ RS[u, a, b] and /or f ∈ RS[v, a, b]! However, we can provethat this statement is true if we use a particular decomposition of f . Let u(x) = Vg(x) and v(x) =Vg(x) − g(x) be our decomposition of g. Then, we will be able to show f ∈ RS[g, a, b] impliesf ∈ RS[Vg, a, b] and f ∈ RS[Vg − g, a, b].

Theorem 6.6.1 f Riemann Stieljes Integrable With Respect To g Of Bounded Variation ImpliesIntegrable With Respect To Vg and Vg − g.

Let g ∈ BV [a, b] and f ∈ RS[g, a, b]. Then f ∈ RS[Vg, a, b] and f ∈ RS[Vg − g, a, b].

Proof 6.6.1For convenience of notation, let u = Vg and v = Vg − g. First, we show that f ∈ RS[u, a, b] byshowing the Riemann - Stieljes Criterion holds for f with respect to u on [a, b]. Fix a positive ε. Thenthere is a partition π0 so that

| S(f, g,π,σ)−∫ b

a

f(x)dg(x) |< ε

for all refinements π of π0 and evaluation sets σ of π. Thus, given two such evaluation sets σ1 andσ2 of a refinement π, we have

| S(f, g,π,σ1)− S(f, g,π,σ2) | ≤ | S(f, g,π,σ1)−∫ b

a

f(x)dg(x) |

+ | S(f, g,π,σ2)−∫ b

a

f(x)dg(x) |

< 2ε.

Hence, we know for σ1 = s1, . . . , sp and σ2 = s′1, . . . , s′p, that

| S(f, g,π,σ1 − S(f, g,π,σ2 | < 2ε (α)


Now, u(b) = Vg(b) = supπ∑π | ∆gj |. Thus, by the Supremum Tolerance Lemma, there is a

partition π1 so thatu(b)− ε <

∑π1

| ∆gj |≤ u(b).

Then if π refines π1, we have

u(b)− ε <∑π1

| ∆gj |≤∑π

| ∆gj |≤ u(b).

and so for all π1 π,

u(b)− ε <∑π

| ∆gj |≤ u(b). (β)

Now let π2 = π0 ∨ π1 and choose any partition π that refines π2. Then,∑π

(Mj −mj

)| ∆uj | − | ∆gj | ≤

∑π

(Mj +mj

)∆uj − | ∆gj |

≤ 2M∑π

| ∆uj | − | ∆gj |

where M =|| f ||∞. But the term∑π ∆uj is a collapsing sum which becomes u(b)− u(a) = u(b)

as u(a) = 0. We conclude∑π

(Mj −mj

)| ∆uj | − | ∆gj | ≤ 2M

(u(b)−

∑π

∣∣∣∣∆gj∣∣∣∣Now by Equation α, for all refinements of π2, we have

u(b)−∑π

| ∆gj |< ε.

Hence, ∑π

(Mj −mj

)| ∆uj | − | ∆gj | ≤ 2M ε. (γ)

Next, for any refinement of π of π2, let the partition points be x0, . . . , xn as usual and define

J+(π) = j ∈ π|∆gj ≥ 0, J−(π) = j ∈ π|∆gj < 0.

By the Infimum and Supremum Tolerance Lemma, if j ∈ J+(π),

∃s′j ∈ [xj−1, xj ] 3 mj ≤ f(s′j) < mj + ε/2, ∃sj ∈ [xj−1, xj ] 3 Mj − ε/2 < f(sj) ≤Mj .

It follows

f(sj)− f(s′j) > Mj −mj − ε, j ∈ J+(π). (ξ)

On the other hand, if j ∈ J−(π), we can find sj and s′j in [xj−1, xj ] so that

∃s′j ∈ [xj−1, xj ] 3 mj ≤ f(sj) < mj + ε/2, ∃sj ∈ [xj−1, xj ] 3 Mj − ε/2 < f(s′j) ≤Mj .

6.6. BOUNDED VARIATION INTEGRATORS 115

This leads to

f(s′j)− f(sj) > Mj −mj − ε, j ∈ J−(π). (ζ)

Thus,∑π

(Mj −mj

)| ∆gj | =

∑j∈J+(π)

(Mj −mj

)∆gj +

∑j∈J−(π)

(Mj −mj

)(−∆gj

)

<∑

j∈J+(π)

(f(sj)− f(s′j)

)∆gj + ε

∑j∈J+(π)

∆gj

+∑

j∈J−(π)

(f(s′j)− f(sj)

)(−∆gj

)+ ε

∑j∈J+(π)

(−∆gj

)

=∑j∈π

(f(sj)− f(s′j)

)∆gj + ε

∑j∈π| ∆gj | .

Also, by the definition of the variation function of g, we have∑j∈π| ∆gj |≤ u(b) = Vg(b).

Since the points s1, . . . , sn and s′1, . . . , s′n are evaluation sets of π, we can apply Equation αto conclude

| S(f, g,π,σ1 − S(f, g,π,σ2 | =∑j∈π

(f(sj)− f(s′j)

)∆gj

< 2ε.

Hence, ∑π

(Mj −mj

)| ∆gj | < 2ε+ εu(b) = (2 + u(b))ε. (θ)

Then, using Equation γ and Equation θ, we find∑π

(Mj −mj

)∆uj =

∑π

(Mj −mj

) (∆uj− | ∆gj |

)+∑π

(Mj −mj

) (| ∆gj |

)< 2Mε + (2 + u(b))ε = (2M + 2 + u(b))ε.

Letting A = 2M + 2 + u(b), and recalling that u = Vg , we have

U(f, Vg,π)− L(f, Vg,π) < Aε

for any refinement π of π2. Hence, f satisfies the Riemann - Stieljes Criterion with respect to Vg on[a, b]. We conclude f ∈ RS[Vg, a, b].

Thus, f ∈ RS[g, a, b] and f ∈ RS[Vg, a, b] and by Theorem 6.1.1, we have f ∈ RS[Vg − g, a, b]also.

Theorem 6.6.2 Products And Reciprocals Of Functions Riemann Stieljes Integrable With Re-spect To g Of Bounded Variation Are Also Integrable


Let g ∈ BV [a, b] and f, f1, f2 ∈ RS[g, a, b]. Then

(i) f2 ∈ RS[g, a, b]

(ii) f1f2 ∈ RS[g, a, b]

(iii) If there is a positive constant m, so that |f(x)| > m for all x in [a, b], then 1/f ∈Rs[g, a, b].

Proof 6.6.2(i)

Proof Since f ∈ RS[g, a, b], f ∈ RS[Vg, a, b] and f ∈ RS[Vg − g, a, b] by Theorem 6.6.1. Hence,by Theorem 6.5.1, f2 ∈ RS[Vg, a, b] and f2 ∈ RS[Vg−g, a, b]. Then, by the linearity of the RiemannStieljes integral for monotone integrators, Theorem 6.1.1, we have f2 ∈ RS[Vg−(Vg−g) = g, a, b].

(ii)

Proof f1, f2 ∈ RS[g, a, b] implies f1, f2 ∈ RS[Vg, a, b] and f1, f2 ∈ RS[Vg − g, a, b]. Thus, usingreasoning just like that in Part (i), we have f1f2 ∈ RS[g, a, b].

(iii)

Proof By our assumptions, we know 1/f ∈ RS[Vg, a, b] and 1/f ∈ RS[Vg − g, a, b]. Thus, by thelinearity of the Riemann Stieljes integral with respect to monotone integrators, 1/f ∈ RS[g, a, b].

Theorem 6.6.3 The Riemann Stieljes Integral Is Additive On Subintervals

Let g ∈ BV [a, b] and f ∈ RS[g, a, b]. Then, if a ≤ c ≤ b,∫ b

a

f(x)dg(x) =

∫ c

a

f(x)dg(x) +

∫ b

c

f(x)dg(x).

Proof 6.6.6From Theorem 6.5.6, we know∫ b

a

f(x)dVg(x) =

∫ c

a

f(x)dVg(x) +

∫ b

c

f(x)dVg(x)∫ b

a

f(x)d(Vg − g)(x) =

∫ c

a

f(x)d(Vg − g)(x) +

∫ b

c

f(x)d(Vg − g)(x).

Also, we know ∫ b

a

f(x)dg(x) =

∫ b

a

f(x)dVg(x)−∫ b

c

f(x)d(Vg − g)(x)

and so the result follows.

Chapter 7

Further Riemann - Stieljes Results

We know quite a bit about the Riemann Stieljes integral in theory. However, we do not know how tocompute a Riemann Stieljes integral and we only know that Riemann Stieljes integrals exist for a fewtype of integrators: those that are bounded with a finite number of jumps and the identity integratorg(x) = x. It is time to learn more.

7.1 The Riemann - Stieljes Fundamental Theorem Of Calculus

As you might expect, we can prove a Riemann - Stieljes variant of the Fundamental Theorem OfCalculus.

Theorem 7.1.1 Riemann Stieljes Fundamental Theorem Of Calculus

Let g ∈ BV [a, b]m f ∈ RS[g, a, b]. Define F : [a, b]→ < by

F (x) =

∫ x

a

f(t)dg(t).

Then

(i) F ∈ BV [a, b],

(ii) If g is continuous at c in [a, b], then F is continuous at c.

(iii) If g is monotone and if at c is in [a, b], g′(c) exists and f is continuous at c, thenF ′(c) exists with

F ′(c) = f(c) g′(c).

Proof 7.1.1First, assume g is monotone increasing and g(a) < g(b). Let π be a partition of [a, b]. Then, weimmediately have the fundamental estimates

m(g(b)− g(a)) ≤ L(f, g) ≤ U(f, g) ≤M(g(b)− g(a)),

where m and M are the infimum and supremum of f on [a, b] respectively. Since f ∈ RS[g, a, b], wethen have

m(g(b)− g(a)) ≤∫ b

a

fdg ≤M(g(b)− g(a)).

117

118 CHAPTER 7. FURTHER RIEMANN-STIELJES

or

m ≤∫ bafdg

g(b)− g(a)≤M.

LetK(a, b) =∫ bafdg/(g(b)−g(a)). Then,m ≤ K(a, b) ≤M and

∫ bafdg = K(a, b)(g(b)−g(a)).

Now assume x < y in [a, b]. Since f ∈ RS[g, a, b], by Theorem 6.5.5, f ∈ RS[g, x, y]. By theargument just presented, we can show there is a number K(x, y) so that

K(x, y) =

∫ y

x

fdg/(g(y)− g(x)),

m ≤ inft∈[x,y]

f(t) ≤ K(x, y) ≤ supt∈[x,y]

f(t) ≤M (α)∫ y

x

fdg = K(x, y)(g(y)− g(x))

(i)

Proof We show f ∈ BV [a, b]. Let π be a partition of [a, b]. Then, labeling the partition points inthe usual way,

∑π

| ∆Fj | =∑π

| ∆F (xj)− F (xj−1 |

=∑π

|∫ xj

xj−1

fdg |

=∑π

| K(xj−1, xj) || g(xj)− g(xj−1) |=∑π

| K(xj−1, xj) || ∆gj |

using Equation α on each subinterval [xj−1, xj ]. However, we know each m ≤ K(xj−1, xj) ≤ Mand so ∑

π

| ∆Fj | ≤ ‖ f ‖∞∑π

| ∆gj |

= ‖ f ‖∞ (g(b)− g(a)),

as g is monotone increasing. Since this inequality holds for all partitions of [a, b], we see

V (F ; a, b) ≤‖ f ‖∞ (g(b)− g(a))

implying F ∈ BV [a, b].

(ii)

Proof Let g be continuous at c. Then given a positive ε, there is a δ > 0, so that

| g(c)− g(y) | < ε/(1+ ‖ f ‖∞), | y − c |< δ, y ∈ [a, b].

For any such y, apply Equation α to the interval [c, y] or [y, c] depending on whether y > c or vice -versa. For concreteness, let’s look at the case y > c. Then, there is a K(c, y) so that m ≤ K(c, y) ≤M and

∫ ycf(t)dg(t) = K(c, y)(g(y)− g(c)). Thus, since y is within δ of c, we have∣∣∣∣∫ y

c

f(t)dg(t)

∣∣∣∣ =| K(c, y) | | g(y)− g(c) |≤‖ f ‖∞ ε/(1+ ‖ f ‖∞) < ε.


We conclude that if y ∈ [c, c+δ), then∣∣∣∣∫ yc f(t)dg(t)

∣∣∣∣ < ε. A similar argument holds for y ∈ (c−δ, c].

Combining, we see y ∈ (c− δ, c+ δ) and in [a, b] implies∣∣∣∣F (y)− F (c)

∣∣∣∣ =

∣∣∣∣∫ y

c

f(t)dg(t)

∣∣∣∣ < ε.

So F is continuous at c.

(iii)

Proof If c ∈ [a, b], g′(c) exists and f is continuous at c, we must show that F ′(c) = f(c)g′(c). Leta positive ε be given. Then,

∃δ1 3

∣∣∣∣∣g(y)− g(c)

y − c− g′(c)

∣∣∣∣∣ < ε, 0 <| y − c |< δ1, y ∈ [a, b]. (β)

and

∃δ2 3∣∣∣∣f(y)− f(c) | < ε, | y − c |< δ2, y ∈ [a, b]. (γ)

Choose any δ < min(δ1, δ2). Let y be in(

(c− δ, c)∪ (c, c+ δ)

)∩ [a, b]. We are interested in the

interval I with endpoints c and y which is either of the form [c, y] or vice - versa. Apply Equation αto this interval. We find there is a K(I) that satisfies

inft∈I

f(t) ≤ K(I) ≤ supt∈I

f(t)

and ∫ y

c

f(t)dg(t) = K([c, y])(g(y)− g(c)), y > c

or ∫ c

y

f(t)dg(t) = K([y, c])(g(c)− g(y)), y < c

or

−∫ y

c

f(t)dg(t) = K([y, c])(g(c)− g(y)), y < c

which gives ∫ y

c

f(t)dg(t) = K([y, c])(g(y)− g(c)), y < c.

So we conclude we can write ∫ y

c

f(t)dg(t) = K(I)(g(y)− g(c)).


where K(I) denotes K([c, y]) or K([y, c]) depending on where y is relative to c. Next, since δ <min(δ1, δ2), both Equation α and Equation β holds. Thus,

f(c)− ε < f(t) < f(c) + ε, y ∈(

(c− δ, c) ∪ (c, c+ δ)

)∩ [a, b].

This tells us that supt∈I f(t) ≤ f(c) + ε and inft∈I f(t) ≥ f(c)− ε. Thus,

f(c)− ε ≤ K([c, y]),K([y, c]) ≤ f(c) + ε

or | K([c, y])− f(c) |< ε and | K([y, c])− f(c) |< ε. Finally, consider∣∣∣∣∣F (y)− F (c)

y − c− f(c)g′(c)

∣∣∣∣∣ =

∣∣∣∣∣K(I)(g(y)− g(c)

y − c− f(c)g′(c)

∣∣∣∣∣=

∣∣∣∣∣K(I)(g(y)− g(c)

y − c− f(c)g′(c) +K(I)g′(c)−K(I)g′(c)

∣∣∣∣∣≤ | K(I) |

∣∣∣∣∣ (g(y)− g(c)

y − c− g′(c)

∣∣∣∣∣+

∣∣∣∣K(I)− f(c)

∣∣∣∣ | g′(c) |< ‖ f ‖∞ ε+ | g′(c) | ε.

Since ε is arbitrary, this shows F is differentiable at c with value f(c)g′(c).

This proves the proposition for the case that g is monotone. To finish the proof, we note if g ∈BV [a, b], then g = Vg − (Vg − g) is the standard decomposition of g into the difference of twomonotone increasing functions. Let F1(x) =

∫ xaf(t)d(Vg)(t) and F2(x) =

∫ xaf(t)d(Vg − g)(t).

From Part (i), we see F = F1−F2 is of bounded variation. Next, if g is continuous at c, so is Vg andVg−g by Theorem 3.5.3. So by Part (ii), F1 and F2 are continuous at c. This implies F is continuousat c.

7.2 Existence Results

We begin by looking at continuous integrands.

Theorem 7.2.1 Integrand Continuous and Integrator Of Bounded Variation Implies Riemann- Stieljes Integral Exists

If f ∈ C[a, b] and g ∈ BV [a, b], then f ∈ RS[g, a, b].

Proof 7.2.1Let’s begin by assuming g is monotone increasing. We may assume without loss of generality thatg(a) < g(b). Let K = g(b) − g(a) > 0. Since f is continuous on [a, b], f is uniformly continuouson [a, b]. Hence, given a positive ε, there is a positive δ so that

| f(s)− f(t) |< ε/K, | t− s |< δ, t, s ∈ [a, b].

Now, repeat the proof of Theorem 4.4.1 which shows that if f is continuous on [a, b], then f ∈RI[a, b], but replace all the ∆xj by ∆gj . This shows that f satisfies the Riemann - Stieljes Criterionfor integrability. Thus, by the equivalence theorem, f ∈ RS[g, a, b].

7.2. EXISTENCE 121

Next, let g ∈ BV [a, b]. Then g = Vg − (Vg − g) as usual. Since Vg and Vg − g are monotoneincreasing, we can apply our first argument to conclude f ∈ RS[Vg, a, b] and f ∈ RS[Vg − g, a, b].Then, by the linearity of the Riemann - Stieljes integral with respect to the integrator, Theorem 6.1.1,we have f ∈ RS[g, a, b] with ∫ b

a

fdg =

∫ b

a

fdvg −∫ b

a

fd(Vg − g).

Next, we let the integrand be of bounded variation.

Theorem 7.2.2 Integrand Bounded Variation and Integrator Continuous Implies Riemann -Stieljes Integral Exists

If f ∈ BV [a, b] and g ∈ C[a, b], then f ∈ RS[g, a, b].

Proof 7.2.2If f ∈ BV [a, b] and g ∈ C[a, b]], then by the previous theorem, Theorem 7.2.1, g ∈ RS[f, a, b]. Nowapply integration by parts, Theorem 6.1.2, to conclude f ∈ RS[g, a, b].

What if the integrator is differentiable?

Theorem 7.2.3 Integrand Continuous and Integrator Continuously Differentiable Implies Rie-mann - Stieljes Integrable

Let f ∈ C[a, b] and g ∈ C1[a, b]. Then f ∈ RS[g, a, b], fg′ ∈ RI[a, b] and∫ b

a

f(x)dg(x) =

∫ b

a

f(x)g′(x)dx

where the integral on the left side is a traditional Riemann integral.

Proof 7.2.3Pick an arbitrary positive ε. Since g′ is continuous on [a, b], g′ is uniformly continuous on [a, b].Thus, there is a positive δ so that

| g′(s)− g′(t) | < ε, | s− t |< δ, s, t ∈ [a, b]. (α)

Since g′ is continuous on [a, b], there is a number M so that | g(x) |≤ M for all x in [a, b].We conclude that g ∈ BV [a, b] by Theorem 3.3.3. Now apply Theorem 7.2.1, to conclude f ∈RS[g, a, b]. Thus, there is a partition π0 of [a, b], so that∣∣∣∣S(f, g,π,σ)−

∫ b

a

fdg

∣∣∣∣ < ε, π0 π, σ ⊆ π. (β)

Further, since fg′ is continuous on [a, b], fg′ ∈ RI[a, b] and so∫ bafg′ exists also.

Now let π1 be a refinement of π0 with || π1 ||< δ. Then we can apply Equation β to conclude∣∣∣∣S(f, g,π,σ)−∫ b

a

fdg

∣∣∣∣ < ε, π1 π, σ ⊆ π. (γ)


Next, apply the Mean Value Theorem to g on the subintervals [xj−1, xj ] from partition π for whichEquation γ holds. Then, ∆gj = g′(tj)(xj − xj−1) for some tj in (xj−1, xj). Hence,

S(f, g,π,σ) =∑π

f(sj)∆gj =∑π

f(sj)g′(tj)∆xj .

Also, we seeS(fg′,π,σ) =

∑π

f(sj)g′(sj)∆xj .

Thus, we can compute∣∣∣∣S(f, g,π,σ)− S(fg′,π,σ)

∣∣∣∣ =

∣∣∣∣∑π

f(sj)

(g′(tj)− g′(sj)∆xj

)≤ || f ||∞

∑π

∣∣∣∣g′(tj)− g′(sj)∆xj∣∣∣∣.By Equation α, since || π ||< δ, |tj − sj | < δ and so |g′(tj)− g′(sj)| < ε. We conclude∣∣∣∣S(f, g,π,σ)− S(fg′,π,σ)

∣∣∣∣ < ε || f ||∞∑π

∆xj = ε || f ||∞ (b− a). (ξ)

Thus,∣∣∣∣S(fg′,π,σ)−∫ b

a

fdg

∣∣∣∣ ≤ ∣∣∣∣S(f, g,π,σ)− S(fg′,π,σ)

∣∣∣∣ +

∣∣∣∣S(f, g,π,σ)−∫ b

a

fdg

∣∣∣∣< ε || f ||∞ (b− a) + ε

by Equation γ and Equation ξ. This proves the desired result.

It should be easy to see that the assumptions of Theorem 7.2.3 can be relaxed. Consider

Theorem 7.2.4 Integrand Riemann Integrable and Integrator Continuously Differentiable Im-plies Riemann - Stieljes Integrable

Let f ∈ C[a, b] and g ∈ C1[a, b]. Then f ∈ RS[g, a, b], fg′ ∈ RI[a, b] and∫ b

a

f(x)dg(x) =

∫ b

a

f(x)g′(x)dx

where the integral on the left side is a traditional Riemann integral.

Proof 7.2.4We never use the continuity of f in the proof given for Theorem 7.2.3. All we use is the fact that f isRiemann integrable. Hence, we can use the proof of Theorem 7.2.3 without change to find∣∣∣∣S(fg′,π,σ)−

∫ b

a

fdg

∣∣∣∣ < ε || f ||∞ (b− a) + ε.

This tells us that fg′ is Riemann integrable on [a, b] with value∫ bafdg.

7.3. COMPUTATIONS 123

7.3 Worked Out Examples Of Riemann Stieljes ComputationsHow do we compute a Riemann Stieljes integral? Let’s look at some example.

Example 7.3.1 Let f and g be defined on [0, 2] by

f(x) =

x, x ∈ Q ∩ [0, 2]2− x, x ∈ Ir ∩ [0, 2],

g(x) =

1, 0 ≤ x < 13, 1 ≤ x ≤ 2.

Does∫fdg exist?

Solution We can answer this two ways so far. Method 1: We note f is continuous at 1 (you should beable to do a traditional ε− δ proof of this fact!) and since g has a jump at 1, we can look at Lemma6.2.1 - Lemma 6.2.3 to see that f is indeed Riemann - Stieljes with respect to g. The value is given by∫ 2

0

fdg = f(1)(g(1+)− g(1−) = 1(3− 1) = 2.

Method 2: We can compute the integral using a partition approach. Let π be a partition of [0, 2].We may assume without loss of generality that 1 ∈ π (recall all of our earlier arguments that allowus to make this statement!). Hence, there is an index k0 such that xk0 = 1. We have

L(f, g,π) =

(inf

x∈[xk0−1,1]f(x)

)(g(1)− g(xk0−1)

)+

(inf

x∈[1,xk0+1]f(x)

)(g(xk0+1)− g(1))

).

Now use how g is defined to see,

L(f, g,π) =

(inf

x∈[xk0−1,1]f(x)

)(3− 1)

)+

(inf

x∈[1,xk0+1]f(x)

)(3− 3)

).

Hence,

L(f, g,π) = 2

(inf

x∈[xk0−1,1]f(x)

).

If you graphed x and 2 − x simultaneously on [0, 2], you would see that they cross at 1 and x isbelow 2− x before 1. This graph works well for f even though we can only use the graph of x whenx is rational and the graph of 2 − x when x is irrational. We can see in our mind how to do thevisualization. For this mental picture, you should be able to see that the infimum of f on [xk0−1, 1]will be the value xk0−1. We have thus found that L(f, g,π) = 2xk0−1. A similar argument will showthat U(f, g,π) = 2(2− xk0−1). This immediately implies that L(f, g) = U(f, g) = 2.

Example 7.3.2 Let f be any bounded function which is discontinuous from the left at 1 on [0, 2].Again, let g be defined on [0, 2] by

g(x) =

1, 0 ≤ x < 13, 1 ≤ x ≤ 2.

Does∫fdg exist?

Solution First, since we know f is not continuous from the left at 1 and g is continuous from the rightat 1, the conditions of Lemma 6.2.2 do not hold. So it is possible this integral does not exist. We willin fact show this using arguments that are similar to the previous example. Again, π is a partitionwhich has xk0 = 1. We find

L(f, g,π) =

(inf

x∈[xk0−1,1]f(x)

)(3− 1)

), U(f, g,π) =

(sup

x∈[xk0−1,1]

f(x)

)(3− 1)

).


Since we can choose xk0 − 1 as close to 1 as we wish, we see

infx∈[xk0−1,1]

f(x)→ min(f(1−), f(1))

supx∈[xk0−1,1]

f(x)→ max(f(1−), f(1))

But f is discontinuous from the left at 1 and so f(1−) 6= f(1). For concreteness, let’s assumef(1−) < f(1) (the argument the other way is very similar). We see L(f, g) = 2f(1−) and U(f, g) =2f(1). Since these values are not the same, f is not Riemann Stieljes integrable with respect to g bythe Riemann - Stieljes equivalence theorem, Theorem 6.4.1.

Example 7.3.3 Let f be any bounded function which is continuous from the left at 1 on [0, 2]. Again,let g be defined on [0, 2] by

g(x) =

1, 0 ≤ x < 13, 1 ≤ x ≤ 2.

Does∫fdg exist?

Solution First, since we know f is continuous from the left at 1 and g is continuous from the right at1, the conditions of Lemma 6.2.2 do hold. So this integral does exist. Using Lemma 6.2.2, we see∫ 2

0

fdg = f(1)(g(1+)− g(1−)) = 2f(1).

We can also show this using partition arguments as we have done before. Again, π is a partitionwhich has xk0 = 1. Again, we have

L(f, g,π) =

(inf

x∈[xk0−1,1]f(x)

)(3− 1)

), U(f, g,π) =

(sup

x∈[xk0−1,1]

f(x)

)(3− 1)

).

Since we can choose xk0 − 1 as close to 1 as we wish, we see

infx∈[xk0−1,1]

f(x)→ min(f(1−), f(1))

supx∈[xk0−1,1]

f(x)→ max(f(1−), f(1))

But f is continuous from the left at 1, f(1−) = f(1). We see L(f, g) = 2f(1) and U(f, g) = 2f(1).Since these values are the same, f is Riemann Stieljes integrable with respect to g by the Riemann -Stieljes equivalence theorem, Theorem 6.4.1.

Example 7.3.4 Define a step function g on [0, 12] by

g(x) =

0, 0 ≤ x < 2∑bxcj=2 (j − 1)/36, 2 ≤ x < 8

21/36 +∑bxcj=8 (13− j)/36, 8 ≤ x ≤ 12

where bxc is the greatest integer which is less than or equal to x. The function g is everywherecontinuous from the right and represents the probability of rolling a number j ≤ x. It is called thecumulative probability distribution function of a fair pair of dice. The Riemann - Stieljes integralµ =

∫ 12

0xdg(x) is called the mean of this distribution. The variance of this distribution is denoted

by σ2 (unfortunate choice, isn’t it as that is the letter we use to denote evaluation sets of partitions!)

7.3. COMPUTATIONS 125

and defined to be

σ2 =

∫ 12

0

(x− µ)2dg(x).

Compute µ and σ2.

Solution Since f(x) = x is continuous on [0, 12], Lemma 6.2.4 applies and we have∫ 12

0

xdg(x) =

12∑j=2

j

(g(j+)− g(j−)

).

The evaluations are a bit messy.

36(g(2+)− g(2−)) = 36(g(2)− g(2−)) = 1− 0 = 1

36(g(3+)− g(3−)) = 36(g(3)− g(3−)) = 3− 1 = 2

36(g(4+)− g(4−)) = 36(g(4)− g(4−)) = 6− 3 = 3

36(g(5+)− g(5−)) = 36(g(5)− g(5−)) = 10− 6 = 4

36(g(6+)− g(6−)) = 36(g(6)− g(6−)) = 15− 10 = 5

36(g(7+)− g(7−)) = 36(g(7)− g(7−)) = 21− 15 = 6

36(g(8+)− g(8−)) = 36(g(8)− g(8−)) = 26− 21 = 5

36(g(9+)− g(9−)) = 36(g(9)− g(9−)) = 30− 26 = 4

36(g(10+)− g(10−)) = 36(g(10)− g(10−)) = 33− 30 = 3

36(g(11+)− g(11−)) = 36(g(11)− g(11−)) = 35− 33 = 2

36(g(12+)− g(12−)) = 36(g(12)− g(12−)) = 36− 35 = 1

Thus, ∫ 12

0

xdg(x) =

(2(1) + 3(2) + 4(3) + 5(4) + 6(5) + 7(6)

+8(5) + 9(4) + 10(3) + 11(2) + 12(1)

)/36

=

(2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12

)/36

= 252/36 = 7.

So, the mean or expected value of a single roll of a fair pair of dice is 7. To find the variance, wecalculate

σ2 =

∫ 12

0

(x− 7)dg(x)

=

12∑j=2

(j − 7)2

(g(j+)− g(j−)

)

=

(25(1) + 16(2) + 9(3) + 4(4) + 1(5) + 0(6) + 1(5) + 4(4) + 9(3) + 16(2) + 25(1)

)/36


=

(25 + 32 + 27 + 16 + 5 + 5 + 16 + 27 + 32 + 25

)/36

= 210/36 = 35/6.

Example 7.3.5 Let f(x) = ex and let g be defined on [0, 2] by

g(x) =

x2, 0 ≤ x ≤ 1x2 + 1, 1 < x ≤ 2.

Show∫fdg exists and evaluate it.

Solution Since g is monotone,∫ 2

0fdg exists. We can thus decompose g into its continuous and saltus

part. We find

gc(x) = x2, sg(x) =

0, 0 ≤ x ≤ 11, 1 < x ≤ 2.

The saltus integral is evaluated using Lemma 6.2.1. The integrand is continuous and the jump is at1, so we have ∫ 2

0

f dsg =

∫ 2

0

ex dsg(x)

= e1(sg(1+)− sg(1−)) = e(1− 0) = e.

and for the continuous part, we can use the fact the integrator is continuously differentiable on [0, 2]to apply Theorem 7.2.3 to obtain∫ 2

0

f dgc =

∫ 2

0

ex d(x2) =

∫ 2

0

ex 2xdx = 2(e2 + 1)

Thus, ∫ 2

0

fdg =

∫ 2

0

f dgc +

∫ 2

0

f dsg

= 2(e2 + 1) + e.

We can also do this by integration by parts, Theorem 6.1.2. Since f ∈ RS[g, 0, 2], it follows thatg ∈ RS[f, 0, 2] and ∫ 2

0

f(x)dg(x) = ex g(x)

∣∣∣∣20

−∫ 2

0

g(x)df(x)

= e2g(2)− g(0) −∫ 2

0

g(x)d(ex)

= e2 −∫ 2

0

g(x)exdx.

Example 7.3.6 Let f(x) = ex and let g be defined on [0, 2] by

g(x) =

x2, 0 ≤ x < 1sin(x), 1 ≤ x ≤ 2.


7.4. HOMEWORK 127

Solution We know that g is of bounded variation on [1, 2] because it is continuously differentiablewith bounded derivative there. But what about on [0, 1]? We know that the function h(x) = x2 on[0, 1] is of bounded variation on [0, 1] because it is also continuously differentiable with a boundedderivative. If π is any partition of [0, 1] then we must have, using standard notation for the partitionpoints of π, that

∑π

| ∆gj | =

p−1∑j=0

| ∆gj | + | g(1)− g(xp−1 |

≤ V (h, 0, 1) + 2 ‖ g ‖∞ .

Since the choice of partition on [0, 1] is arbitrary, we see g ∈ BV [0, 1]. Thus, combining, we havethat g ∈ BV [0, 2]. It then follows that f ∈ RS[g, 0, 2]. Now note that on [0, 1], we can writeg(x) = h(x) + u(x) where

u(x) =

0, 0 ≤ x < 1sin(1)− 1, x = 1.

Then, to evaluate∫ 2

0fdg we write∫ 2

0

fdg =

∫ 1

0

fdg +

∫ 2

1

fdg

=

∫ 1

0

fd(h+ u) +

∫ 2

1

fd(sin(x))

=

∫ 1

0

fd(h) +

∫ 1

0

fd(u) +

∫ 2

1

f cos(x)dx

=

∫ 1

0

ex2xdx+ f(1)(u(1)− u(1−)) +

∫ 2

1

ex cos(x)dx

=

∫ 1

0

ex2xdx+ e(sin(1)− 1) +

∫ 2

1

ex cos(x)dx

and these integrals are standard Riemann integrals that can be evaluated by parts.

7.4 HomeworkExercise 7.4.1 Define g on [0, 2] by

g(x) =

−2 x = 0x3 0 < x < 19/8 x = 1x4/4 + 1 1 < x < 27 x = 2

This function is from a previous exercise.

1. Show that if f(x) = x4 on [0, 2], then f ∈ RS[g, 0, 2].

2. Compute∫ 2

0fdg.

3. Explain why g ∈ RS[f, 0, 2].

4. Compute∫ 2

0gdf .


Exercise 7.4.2 Define g on [0, 2] by

g(x) =

−1 x = 0x2 0 < x < 17/4 x = 1√x+ 3 1 < x < 2

3 x = 2

This function is also from a previous exercise.

1. Show that if f(x) = x2 + 5 on [0, 2], then f ∈ RS[g, 0, 2].

2. Compute∫ 2

0fdg.

3. Explain why g ∈ RS[f, 0, 2].

4. Compute∫ 2

0gdf .

Exercise 7.4.3 Let f and g be defined on [0, 4] by

f(x) =

x, x ∈ Q ∩ [0, 4]2x, x ∈ Ir ∩ [0, 4],

g(x) =

1, 0 ≤ x < 12, 1 ≤ x < 23, 2 ≤ x < 34, 3 ≤ x ≤ 4.

Does∫fdg exist and if so what is its value?

Exercise 7.4.4 Let f(x) = x3 and let g be defined on [0, 3] by

g(x) =

x2, 0 ≤ x ≤ 2x2 + 4, 2 < x ≤ 3.


Exercise 7.4.5 Let f(x) = x2 + 3x+ 10 and let g be defined on [−1, 5] by

g(x) =

x3, −1 ≤ x ≤ 2−10x2, 2 < x ≤ 5.


Exercise 7.4.6 The following are definitions of integrands f1, f2 and f3 and integrators g1, g2 andg3 on [0, 2]. For each pair of indices i, j determine if

∫ 2

0fidgj exists. If the integral exists, compute

the value and if the integral does not exist, provide a proof of its failure to exist.

f1(x) =

1, 0 ≤ x < 1x− 1, 1 ≤ x ≤ 2,

f2(x) =

1, x = 0x, 0 < x ≤ 2

f3(x) =

2, x = 01, 0 < x < 1x− 1, 1 ≤ x ≤ 2

g1(x) =

x, 0 ≤ x < 1x+ 1, 1 ≤ x ≤ 2,

g2(x) =

x, 0 ≤ x ≤ 1x+ 1, 1 < x < 24, x = 2,

g3(x) =

−1, x = 0x, 0 < x ≤ 1x+ 1, 1 < x < 24, x = 2.

Exercise 7.4.7 Prove

7.4. HOMEWORK 129

Theorem 7.4.1 Limit Interchange Theorem For Riemann - Stieljes Integrals

Assume g ∈ BV [a, b] and fn ⊆ RS[g, a, b] converges uniformly to f0 on [a, b]. Then

(i) f0 ∈ RS[g, a, b],

(ii) If Fn(x) =∫ xafn(t)dg(t) and F0(x) =

∫ xaf0(t)dg(t), then Fn converges uniformly

to F0 on [a, b].

(iii)

limn

∫ b

a

fn(t)dg(t) =

∫ b

a

f0(t)dg(t).

Exercise 7.4.8 Let g be strictly monotone on [a, b]. For f1, f2 inC[a, b], define ω : C[a, b]×C[a, b]→< by ω(f1, f2) =

∫ baf1(t)f2(t)dg(t).

(i) Prove that ω is an inner product on C[a, b].

(ii) Prove if ω(f, h) = 0 for all h ∈ RS[g, a, b], then f = 0.

Part IV

Abstract Measure Theory One

131

Chapter 8

Measurable Functions and Spaces

If you have been looking closely at how we prove the properties of Riemann and Riemann Stieljesintegration, you will have noted that these proofs are intimately tied to the way we use partitions todivide the function domain into small pieces. We are now going to explore a new way to associate agiven bounded function with a real number which can be interpreted as the integral.

Let X be a nonempty set. In mathematics, we study sets such as X when various properties andstructures have been added. For example, we might wantX to have a metric d to allow us to measurean abstract version of distance between points in X . We could study sets X which have a linear orvector space structure and if this resulting vector space possessed a norm ‖ · ‖, we could determinean abstract version of the magnitude of objects in X . Here, we want to look at collections of subsetsof the set X and impose some conditions on the structure of these collections.

Definition 8.0.1 Sigma Algebras

Let X be a nonempty set. A family of subsets S is called a σ - algebra if

(i) ∅, X ∈ S.

(ii) If A ∈ S, so is AC . We say S is closed under complementation or complements.

(iii) If An∞n=1 ∈ S, then ∪∞n=1 An ∈ S. We say S is closed under countable unions.

The pair (X,S) will be called a measurable space and if A ∈ S , we will call A an Smeasurable set. If the underlying σ - algebra is understood, we usually just say, A is ameasurable subset of X .

A common tool we use in working with countable collections of sets are De Morgan’s Laws.

Lemma 8.0.2 De Morgan’s Laws

133

134 CHAPTER 8. MEASURABILITY

Let X be a nonempty set and Aα|α ∈ Λ be any collection of subsets of X . Hence, theindex set Λ may be finite, countably infinite or arbitrary cardinality. Then

(i) (∪α Aα

)C= ∩α ACα

(ii) (∩α Aα

)C= ∪α ACα

Proof 8.0.1This is a standard proof and is left to you as an exercise.

8.1 ExamplesLet’s work through a series of examples of σ algebras.

Example 8.1.1 Let X be any not empty set and let S = A|A ⊆ X. This is the collection of allsubsets and is sometimes called the power set of X . It is often denoted by the symbol P(X). Thiscollection clearly is a σ algebra. Hence, (P(X), X) is a measurable space and all subsets of X areP(X) measurable.

Example 8.1.2 Let X be any set and S = ∅, X. Then this collection is also a σ algebra, albeitnot a very interesting one! With this σ algebra, X is a measurable space with only two measurablesets.

Example 8.1.3 Let X be the set of counting numbers and let S = ∅,O,E, X where O is the oddcounting numbers and E, the odd. It is easy to see ( X , S) is a measurable space.

Example 8.1.4 LetX be any uncountable set and let S = A ⊆ X|A is countable or AC is countable.It is easy to see ∅ and X itself are in S. If A ∈ S, then there are two cases: A is countable and/or AC is countable. In both cases, it is straightforward to reason that AC is also in S. It remainsto show that S is closed under countable unions. To do this, assume we have a sequence of sets Anfrom S. Consider A = ∪n An. There are several cases to consider.

1. If all the An are countable, then so is the countable union implying A ∈ S.

2. If all the An are not countable, then each ACn is countable. Thus, ∩n ACn = (∪n An)C iscountable. Again, this tells us A ∈ S.

3. If a countable number of An and a countable number of ACn are uncountable, then we have,since X is uncountable,(

∪nAn)C

=

(∩nACn

)=

(∩(An countable) A

Cn

)∩(∩(An uncountable) A

Cn

)

8.2. BOREL SIGMA ALGEBRA 135

=

(∩(ACn uncountable) A

Cn

)∩(∩(ACn countable) A

Cn

)

Now, for any index n, we must have ∩n ACn ⊆ ACn . Thus, since some ACn are countable, wemust have ∩n ACn is countable. By De Morgan’s Laws, it follows that (∪n An)C is countable.This implies A ∈ S.

We conclude (X,S) is a measurable space.

Example 8.1.5 Let X be any nonempty set and let S1 and S2 be two sigma - algebras of X . Let

S3 = A ⊆ X|A ∈ S1 and A ∈ S2≡ S1 ∩ S2.

It is straightforward to see that (X,S3) is a measurable space.

Example 8.1.6 Let X be any nonempty set. Let A be any nonempty collection of subsets of X . Notethat P(X), the collection of all subsets of X , is a sigma - algebra of X and hence, (X,P(X)) is ameasurable space that contains A. By Example 8.1.5, we know if S1 and S2 are two other sigma -algebras that contain A, then S1 ∩ S2 is a new sigma - algebra that also contains A. This suggestswe search for the smallest sigma - algebra that contains A.

Definition 8.1.1 The Sigma - Algebra Generated By Collection A

The sigma - algebra generated by a collection of subsetsA in a nonempty setX , is denotedby σ(A) and is defined by

σ(A) = ∩S | A ⊆ S .

Since any sigma - algebra S that contains A by definition satisfies σ(A) ⊆ S, it is easy tosee why we interpret this generated sigma - algebra as the smallest sigma - algebra thatcontains the collection A.

8.2 The Borel Sigma - Algebra of <We now discuss a very important sigma algebra of subsets of the real line called the Borel sigma -algebra which is denoted by B. Define four collections of subsets of < as follows:

1. A is the collection of finite open intervals of the form (a, b),

2. B is the collection of finite half open intervals of the form (a, b],

3. C is the collection of finite half open intervals of the form [a, b) and

4. D is the collection of finite closed intervals of the form [a, b].

It is possible to show that

σ(A) = σ(B) = σ(C) = σ(A).

This common sigma - algebra is what we will call the Borel sigma - algebra of<. It should be evidentto you that a set can be very complicated and still be in B. Some of these equalities will be left to


you as homework exercises, but we will prove that σ(A) = σ(D). Let S be any sigma - algebrathat containsA. We know that

[a, b] = (−∞, b] ∩ [a,∞)

= (b,∞)C ∩ (−∞, a)C

=

((−∞, a) ∪ (b,∞)

)C.

In the representation of [a, b] above, note we can write

(−∞, a) =

n=bac⋃−∞

(n, a)

(b,∞) =

∞⋃n=dbe

(b, n).

Since, S is a sigma - algebra containingA, the unions on the right hand sides in the equations abovemust be in S. This immediately tells us that [a, b] is also in S. Hence, since [a, b] is arbitrary, weconclude D is contained in S also. Further, this is true for any sigma - algebra that contains A andso we have thatD ⊆ σ(A). Thus, by definition, we can say σ(D) ⊆ σ(A).

To show the reverse containment is quite similar. Let S be any sigma - algebra that contains D.We know that

(a, b) = (−∞, b) ∩ (a,∞)

= [b,∞)C ∩ (−∞, a]C

=

((−∞, a] ∪ [b,∞)

)C.

In the representation of (a, b) above, note we can write

(−∞, a] =

∞⋃bac

[−n, a]

and

[b,∞) =

∞⋃dbe

[b, n].

Since, S is a sigma - algebra containingD, the unions on the right hand sides in the equations abovemust be in S. This immediately tells us that (a, b) is also in S. Hence, since (a, b) is arbitrary, weconclude A is contained in S also. Again, since this is true for any sigma - algebra that contains A,we have that A ⊆ σ(D). Thus, by definition, we can say σ(A) ⊆ σ(D). Combining, we have theequality we seek.

8.2.1 Homework

Exercise 8.2.1 Prove σ(A) = σ(B).

Exercise 8.2.2 Prove σ(B) = σ(C).

8.3. EXTENDED BOREL SIGMA ALGEBRA 137

8.3 The Extended Borel Sigma Algebra

It is often very convenient to deal with a number system that explicitly adjoins the symbols∞ and−∞ to the standard real line <. This is actually called the two - point compactification of <, but thatis another story!

Definition 8.3.1 The Extended Real Number System

The extended real number systems is denoted by < and is defined as the real numbers withtwo additional elements:

< = < ∪ +∞ ∪ −∞.

We want arithmetic involving the new symbols ±∞ to reflect our everyday experience withlimits of sequences of numbers which either grow without bound positively or negatively.Hence, we use the conventions for all real numbers x:

(±∞) + (±∞) = x + (±∞) = (±∞) + x = ±∞,(±∞) · (±∞) = ∞,(±∞) · (∓∞) = = (∓∞) · (±∞) = −∞,

x · (±∞) = = (±∞) · x = ±∞ if x > 0,

x · (±∞) = = (±∞) · x = 0 if x = 0,

x · (±∞) = = (±∞) · x = ∓∞ if x < 0.

We can not define the arithmetic operations (∞) + (−∞), (−∞) + (∞) or any the fourratios of the form (±∞)/(±∞).

We can now define the Borel sigma - algebra in <. Let E be any Borel set in <. Let

E1 = E ∪ −∞, E2 = E ∪ +∞, and E3 = E ∪ +∞ ∪ +∞.

Then, we define

B = E,E1, E2, E3 | E ∈ B.

We leave to you the exercise of showing that B is a sigma - algebra in <.

Exercise 8.3.1 Prove that B is a sigma - algebra in <.

It is that open intervals in < are in B, but is it true that B contains arbitrary open sets? To see thatit does, we must prove a characterization for the open sets of <.

Theorem 8.3.1 Open Set Characterization Lemma

If U is an open set in <, then there is a countable collection of disjoint open intervalsC = (an, bn) so that U = ∪n(an, bn).

Proof 8.3.1Since U is open, if p ∈ U , there is an r > 0 so that B(p; r) ⊆ U . Hence, (p− r, p+ r) ⊆ U implyingboth (p, p+ r) ⊆ U and (p− r, p) ⊆ U . Let

Sp = y | (p, y) ⊆ U and Tp = x | (x, p) ⊆ U.


It is easy to see that both Sp and Tp are nonempty since U is open. Let bp = supSp and ap = inf Tp.Clearly, bp could be +∞ and ap could be −∞.

Consider u ∈ (ap, bp). From the Infimum and Supremum tolerance lemmas, we know there arepoints x∗ and y∗ so that

u < y∗ ≤ bp ≤ ∞ and (p, y∗) ⊆ U ,−∞ ≤ ap ≤ x∗ < u and (x∗, p) ⊆ U .

Hence, u ∈ (x∗, y∗) ⊆ U which implies u ∈ U . Thus, since u in (ap, bp) is arbitrary, we have(ap, bp) ⊆ U . If ap or bp were not finite, they can not be in < and can not be in U . However, what ifeither one was finite? Is it possible for the point to be in U? We will show that in this case, the pointsap and bp still can not lie in U . For concreteness, let us assume that ap is finite and in U . Then, apwould be an interior point of U . Hence, there would be a radius ρ > 0 so that (ap − ρ, ap) ⊆ Uimplying ap − ρ ∈ Tp. Thus, inf Tp = ap ≤ ap − ρ which is not possible. Hence, ap 6∈ U . A similarargument then shows that if bp is finite, bp is not in U .

Thus, we know that ap and bp are never in U and that p is always in the open interval (ap, bp) ⊆U . Let F = (ap, bp) | p ∈ U. We see immediately that

U = ∪F (ap, bp).

Let (a, b) and (c, d) be any two intervals from F which overlap. From the definition of F , we thenknow that a, b, c and d are not in U . Then, if a ≥ d, the two intervals would be disjoint; hence,we must have a < d. By the same sort of argument, it is also true that c < b. Hence, if c is in theintersection, we have a chain of inequalities like this:

a < c < b < d.

Next, since a 6∈ U , we see a ≤ c since (c, d) ⊆ U . Further, since c 6∈ U and (a, b) ⊆ U , it follows thatc ≤ a. Combining, we have a = c. A similar argument shows that b = d. Hence, (a, b) ∩ (c, d) 6= ∅implies that (a, b) = (c, d). Thus, two interval Ip and Iq in F are either the same or disjoint. Weconclude

U =⋃

(disjoint Ip∈F)

Ip.

Let F0 be this collection of disjoint intervals from F . Each Ip in F0 contains a rational numberrp. By definition, it then follows that if Ip and Iq are in F0, then rp 6= rq . The set of these rationalnumbers is countable and so we can label them using an enumeration rn. Label the interval Ip whichcontains rn as In. Then, we have

U =

∞⋃n=1

In,

which is the desired result.

8.4 Measurable Functions

Let f : < → < be a continuous function. let O be an open subset of <. By Theorem 8.3.1, we knowthat we can write

O =⋃n

(an, bn)

where the (an, bn) are mutually disjoint finite open intervals of <. It follows immediately that Ois in the Borel sigma - algebra B. Now consider the inverse image of O under f , f−1(O). If

8.4. MEASURABLE FUNCTIONS 139

p ∈ f−1(O), then f(p) ∈ O. Since O is open, f(p) must be an interior point. Hence, there is aradius r > 0 so that (f(p) − r, f(p) + r) ⊆ O. Since f is continuous at p, there then is a δ > 0 sothat f(x) ∈ ((f(p)− r, f(p) + r) if x ∈ (p− δ, p+ δ). This tells us that (p− δ, p+ δ) ⊆ f−1(O).Since p was arbitrarily chosen, we conclude that f−1(O) is an open set.

We see that if f is continuous on <, then f−1(O) is in the Borel sigma - algebra for any open setO in <. We can then say that f−1(α,∞) is in B for all α > 0. This suggests that an interesting wayto generalize the notion of continuity might be to look for functions f on an arbitrary nonempty setX with sigma - algebra S satisfying f−1(O) ∈ S for all open sets O. Further, by our last remark,it should be enough to ask that f−1((α,∞)) ∈ S for all α ∈ <. This is exactly what we will do.It should be no surprise to you that functions f satisfying this new definition will not have to becontinuous!

Definition 8.4.1 The Measurability of a Function

Let X be a nonempty set and S be a sigma - algebra of subsets of X . We say that f : X →< is a S - measurable function on X or simply S measurable if

∀α ∈ <, x ∈ X | f(x) > α ∈ S.

We can easily prove that there are equivalent ways of proving a function is measurable.

Lemma 8.4.1 Equivalent Conditions For The Measurability of a Function

Let X be a nonempty set and S be a sigma - algebra of subsets of X . The followingstatements are equivalent:

(i): ∀α ∈ <, Aα = x ∈ X | f(x) > α ∈ S,

(ii): ∀α ∈ <, Bα = x ∈ X | f(x) ≤ α ∈ S,

(iii): ∀α ∈ <, Cα = x ∈ X | f(x) ≥ α ∈ S,

(iv): ∀α ∈ <, Dα = x ∈ X | f(x) < α ∈ S.

Proof 8.4.1

(i)⇒ (ii):

Proof If Aα ∈ S, then its complement is in S also. Since Bα = ACα , (ii) follows.

(ii)⇒ (i):

Proof If Bα ∈ S, then its complement is in S also. Since Aα = BCα , (i) follows.

(iii)⇔ (iv):

Proof Since Cα = DCα and Dα = CCα , arguments similar to those of the previous cases can be

applied.

Hence, if we show (i)⇔ (iii), we will be done. (i)⇒ (iii):

Proof By (i), Aα−1/n ∈ S for all n. We know

Cα =⋂n

Aα−1/n =⋂n

x|f(x) > α− 1/n


We also know ACα−1/n is measurable and so ∪n ACα−1/n is also measurable. Thus, the complementof ∪n ACα−1/n is also measurable. Then, by De Morgan’s Laws, Cα = ∩n Aα−1/n is measurable.

(iii)⇒ (i):

Proof Note, Cα+1/n ∈ S for all n and so

Aα =⋃n

Cα+1/n =⋃n

x|f(x) ≥ α+ 1/n

is also measurable.

We conclude all four statements are equivalent.

8.4.1 Examples

Example 8.4.1 Any constant function f on a nonempty set X with given sigma - algebra S is mea-surable as if f(x) = c for some c ∈ <, then

x|f(x) > α =

∅ ∈ S α ≥ cX ∈ S α < c

Example 8.4.2 Let X be a nonempty set X with given sigma - algebra S. Let E ∈ S be given.Define

IE(x) =

1 if x ∈ E0 if x 6∈ E

Then IE is measurable. Note

x|IE(x) > α =

∅ ∈ S α ≥ 1E ∈ S 0 ≤ α < 1X ∈ S α < 0

Example 8.4.3 Let X = < and S = B. Then, if f : < → < is continuous, f is measurable by thearguments we made at the beginning of this section. More generally, let f : [a, b]→ < be continuouson [a, b]. Then, extend f to < as f defined by

f =

f(a) x < af(x) a ≤ x ≤ bf(b) x > b

Then f is continuous on < and measurable with f−1(α,∞) ∈ B for all α. It is not hard to show that

B ∩ [a, b] = E ⊆ [a, b] | E ∈ B

is a sigma -algebra of the set [a, b]. Further, the standard arguments for f continuous on [a, b]show us that f−1(α,∞) ∈ B ∩ [a, b] for all α. Hence, a continuous f on the interval [a, b] will bemeasurable with respect to the sigma - algebra B ∩ [a, b].

8.5. PROPERTIES 141

We can argue is a similar fashion for functions continuous on intervals of the form (a, b], [a, b)and (a, b) whether a and b is finite or not.

Example 8.4.4 If X = < and S = B, then any monotone function is Borel measurable. To see this,note we can restrict our attention to monotone increasing functions as the argument is quite similarfor monotone decreasing. It is enough to consider the cases where f takes on the value α without ajump at the point x0 or f has a jump across the value α at x0. In the first case, since f is monotoneincreasing and f(x0) = α, f−1(α,∞) = (x0,∞) ∈ B. On the other hand, if f has a jump at x0

across the value α, then f(x−0 ) 6= f(x+0 ) and α ∈ [f(x−0 ), f(x+

0 )]. there are three possibilities:

(i): f(x−0 ) = f(x0) < f(x+0 ): If α = f(x0), then since f is monotone, f−1(α,∞) = (x0,∞). If

f(x0) < α < f(x+0 ), we again have f−1(α,∞) = (x0,∞). Finally, if α = f(x+

0 ), we havef−1(α,∞) = [x0,∞). In all cases, these inverse images are in B.

(ii): f(x−0 ) < f(x0) < f(x+0 ): A similar analysis shows that all the possible inverse images are

Borel sets.

(iii): f(x−0 ) < f(x0) = f(x+0 ): we handle the arguments is a similar way.

We conclude that in all cases, f−1(α,∞) ∈ B and hence f is measurable.Note, the analysis of the previous example could be employed here also to show that a monotone

function defined on an interval such as [a, b], (a, b) and so forth is Borel measurable with respect tothe restricted sigma - algebra B ∩ [a, b] etc.

Exercise 8.4.1 Let f be piecewise continuous on [a, b]. Prove that f is measurable with respect tothe restricted Borel sigma - algebra B ∩ [a, b]. Recall, a function is piecewise continuous on [a, b] ifthere are a finite number of points xi in [a, b] where f is not continuous.

Comment 8.4.1 For convenience, we will start using a more abbreviated notation for sets like

x ∈ X | f(x) > α;

we will shorten this to f(x) > α or (f(x) > α) in our future discussions.

8.5 Properties of Measurable FunctionsWe now want to see how we can build new measurable functions from old ones we know.

Lemma 8.5.1 Properties of Measurable Functions

LetX be a nonempty set and S a sigma - algebra onX . Then if f and g are S measurable,so are

(i): cf for all c ∈ <.

(ii): f2.

(iii): f + g.

(iv): fg.

(v): | f |.

Proof 8.5.1

(i):


Proof If c = 0, cf = 0 and the result is clear. If c > 0, then (cf(x) > α) = (f(x) > α/c) which ismeasurable as f is measurable. If c < 0, a similar argument holds.

(ii):

Proof If α < 0, then (f2(x) > α) = X which is in S. Otherwise, if α ≥ 0, then

(f2(x) > α) = (f(x) >√α) ∪ (f(x) < −

√α),

and both of these sets are measurable since f is measurable. The conclusion follows.

(iii):

Proof If r ∈ Q, let Sr = (f(x) > r) ∩ (g(x) > α − r) which is measurable since f and g aremeasurable. We claim that

(f(x) + g(x) > α) =⋃r∈Q

Sr.

To see this, let x satisfy f(x) + g(x) > α. Thus, f(x) > α − g(x). Since the rationals are densein <, we see there is a rational number r so that f(x) > r > α − g(x). This clearly implies thatf(x) > r and g(x) > α− r and so x ∈ Sr. Since our choice of x was arbitrary, we have shown that

(f(x) + g(x) > α) ⊆⋃r∈Q

Sr.

The converse is easier as if x ∈ Sr, it follows immediately that f(x) + g(x) > α.Since Sr is measurable for each r and the rationals are countable, we see (f(x) + g(x) > α) is

measurable.

(iv):

Proof To prove this result, note that fg = (1/4)

((f +g)2− (f −g)2

)and all the individual pieces

are measurable by (iii) and (i).

(v):

Proof If α < 0, (f(x) > α) = X which is measurable. On the other hand, if α ≥ 0,

(| f | (x) > α) = (f(x) > α) ∪ (f(x) < −α),

which implies the measurability of | f |.

We can also prove another characterization of the measurability of f .

Lemma 8.5.2 A Function is Measurable If and Only If Its Positive and Negative Parts AreMeasurable

LetX be a nonempty set and S be a sigma - algebra onX . Then f : X → < is measurableif and only if f+ and f− are measurable, where

f+(x) = max f(x), 0, and f+(x) = −min f(x), 0.

8.6. EXTENDED VALUED 143

Proof 8.5.7We note f = f+ − f− and | f |= f+ + f−. Thus,

f+ = (1/2)

(| f | +f

)

f− = (1/2)

(| f | −f

).

Hence, if f is measurable, by Lemma 8.5.1 (i), (iii) and (v), f+ and f− are also measurable. Con-versely, if f+ and f− are measurable, f = f+ − f− is measurable as well.

8.6 Extended Valued Measurable Functions

We now extend these ideas to functions which are extended real valued.

Definition 8.6.1 The Measurability Of An Extended Real Valued Function

Let X be a nonempty set and S be a sigma - algebra on X . Let f : X → <. We say f is Smeasurable if (f(x) > α) is in S for all α in <.

Comment 8.6.1 If the extended valued function f is measurable, then (f(x) = +∞) = ∩n(f(x) >n) is measurable. Also, since

(f(x) = −∞) =

(∪n(f(x) > −n)

)C,

it is measurable also.

We can then prove an equivalence theorem just like before.

Lemma 8.6.1 Equivalent Conditions For The Measurability of an Extended Real Valued Func-tion

Let X be a nonempty set and S be a sigma - algebra of subsets of X . The followingstatements are equivalent:

(i): ∀α > 0, Aα = x ∈ X | f(x) > α ∈ S,

(ii): ∀α > 0, Bα = x ∈ X | f(x) ≤ α ∈ S,

(iii): ∀α > 0, Cα = x ∈ X | f(x) ≥ α ∈ S,

(iv): ∀α > 0, Dα = x ∈ X | f(x) < α ∈ S.

Proof 8.6.1The proof follows that of Lemma 8.4.1

The collection of all extended valued measurable functions is important to future work. We makethe following definition:

Definition 8.6.2 The Set of Extended Real Valued Measurable Functions


Let X be a nonempty set and S be a sigma - algebra of subsets of X . We denote byM(X,S) the set of all extended real valued measurable functions on X . Thus,

M(X,S) = f : X → < | f is S measurable.

It is also easy to prove the following equivalent definition of measurability for extended valuedfunctions.

Lemma 8.6.2 Extended Valued Measurability In Terms Of The Finite Part Of The Function

Let X be a nonempty set and S be a sigma - algebra of subsets of X . Then f ∈M(X,S)if and only if (i): (f(x) = +∞) ∈ S, (ii): (f(x) = −∞) ∈ S and (iii): f1 is measurablewhere

f1(x) =

f(x) x 6∈ (f(x) = +∞) ∪ (f(x) = −∞),0 x ∈ (f(x) = +∞) ∪ (f(x) = −∞).

Proof 8.6.2By Comment 8.6.1, if f is measurable, (i) and (ii) are true. Now, if α ≥ 0 is given, we see

(f1(x) > α) = (f(x) > α) ∩ (f(x) = +∞)C ,

which is a measurable set. On the other hand, if α < 0, then

(f1(x) > α) = (f(x) > α) ∪ (f(x) = −∞),

which is measurable as well. We conclude f1 is measurable. Conversely, if (i), (ii) and (iii) hold,then if α ≥ 0, we have

(f(x) > α) = (f1(x) > α) ∪ (f(x) = +∞),

and if α < 0,

(f(x) > α) = (f1(x) > α) ∩ (f(x) = −∞)C ,

implying both sets are measurable. Thus, f is measurable.

Example 8.6.1 Let X be a nonempty set X with given sigma - algebra S. Let E ∈ S be given.Define the extended value characteristic function

JE(x) =

∞ if x ∈ E0 if x 6∈ E

Then JE is measurable. Note

x|JE(x) > α =

E ∈ S α ≥ 0X ∈ S α < 0

8.7. EXTENDED PROPERTIES 145

Note also that if we define

JE(x) =

∞ if x ∈ E−∞ if x 6∈ E

Then JE is measurable. We have

x|JE(x) > α =

E ∈ S α ≥ 0E ∈ S α < 0

Finally, JE(x) = +∞) = E and (JE(x) = −∞) = EC are both measurable and the f1 typefunction used in Lemma 8.6.2 here is (JE)1(x) = 0 always.

8.7 Properties Of Extended Valued Measurable Functions

It is straightforward to prove these properties:

Lemma 8.7.1 Properties of Extended Valued Measurable Functions

Let X be a nonempty set and S a sigma - algebra on X . Then if f and g are in M(X,S),so are

(i): cf for all c ∈ <.

(ii): f2.

(iii): f + g, as long as we restrict the domain of f + g to be Efg where

ECfg =

((f(x) = +∞) ∩ (g(x) = −∞) ∪ (f(x) = −∞) ∩ (g(x) = +∞)

)C.

We usually define (f + g)(x) = 0 on Efg . Note Efg is measurable since f and gare measurable functions.

(iv): | f |, f+ and f−.

Proof 8.7.1These proofs are similar to those shown in the proof of Lemma 8.5.1. However, let’s look at thedetails of the proof of (ii). We see that our definition of addition of the extended real valued summeans that

(f + g)(x) =

(f + g

)IECfg .

Define h by

h(x) =

(f + g

)IECfg (x).


Let α be a real number. Then

(h(x) > α) =

(f(x) + g(x) > α)

⋂ECfg α ≥ 0(

(f(x) + g(x) > α)⋂ECfg

)∪ Efg α < 0

Similar to what we did in Lemma 8.5.1, for r ∈ Q, let

Sr = (f(x) > r) ∩ (g(x) > α− r) ∩ ECfg

which is measurable since f and g are measurable. We claim that

(f(x) + g(x) > α) ∩ ECfg =⋃r∈Q

Sr. (8.1)

To see this, let x be in the left hand side of Equation 8.1. There are several cases. First, neither f(x)or g(x) can be −∞ since α is a real number. Now, if f(x) = ∞, then g(x) > −∞ is and it is easyto see there is a rational number r satisfying f(x) = ∞ > r > α − g(x) and so x is in the righthand side. If g(x) = ∞, then f(x) > −∞ and again, we see there is a rational number so thatf(x) > r > α − g(x) = −∞. Thus, x is in the right hand side again. The case where both f(x)and g(x) are finite is then handled just like we did in the proof of Lemma 8.5.1. We conclude

(f(x) + g(x) > α) ⊆⋃r∈Q

Sr.

The converse is easier as if x ∈ Sr, it follows immediately that f(x) + g(x) is defined and f(x) +g(x) > α.

Since Sr is measurable for each r and the rationals are countable, we see (f(x) + g(x) >α) ∩ ECfg is measurable.

To prove that products of extended valued measurable functions are also measurable, we have touse a pointwise limit approach.

Lemma 8.7.2 Pointwise Infimums, Supremums, Limit Inferiors and Limit Superiors are Mea-surable

Let X be a nonempty set and S a sigma - algebra on X . Let (fn) ⊆M(X,S). Then

(i): If f(x) = infn fn(x), then f ∈M(X,S).

(ii): If F (x) = supn fn(x), then F ∈M(X,S).

(iii): If f∗(x) = lim infn fn(x), then f∗ ∈M(X,S).

(iv): If F ∗(x) = lim supn fn(x), then F ∗ ∈M(X,S).

Proof 8.7.2It is straightforward to see that (f(x) ≥ α) = ∩n (fn(x) ≥ α) and (F (x) ≥ α) = ∪n (fn(x) > α)and hence, are measurable for all α. It follows that f and F are in M(X,S) and so (i) and (ii) hold.Next, recall from classical analysis that at each point x,

lim inf(fn(x)) = supn

infk≥n

fk(x),

8.7. EXTENDED PROPERTIES 147

lim sup(fn(x)) = infn

supk≥n

fk(x).

Now let zn(x) = infk≥n fk(x) and wn(x) = supk≥n fk(x). Applying (i) to zn, we have zn ∈M(X,S) and applying (ii) to wn, we have wn ∈ M(X,S). Then apply (i) and (ii) to supn zn andinf wn, respectively, to get the desired result.

This leads to an important result.

Theorem 8.7.3 Pointwise Limits of Measurable Functions Are Measurable

Let X be a nonempty set and S a sigma - algebra on X . Let (fn) ⊆ M(X,S) and letf : X → < be a function such that fn → f pointwise on X . Then f ∈M(X,S).

Proof 8.7.3We know that lim infn fn(x) = lim supn fn(x) = limn fn(x). Thus, by Lemma 8.7.2, we knowthat f is measurable.

Comment 8.7.1 This is a huge result. We know from classical analysis that the pointwise limit ofcontinuous functions need not be continuous (e.g. let fn(t) = tn on [0, 1]). Thus, the closure ofa class of functions which satisfy a certain property (like continuity) under a limit operation is notalways guaranteed. We see that although measurable functions are certainly not as smooth as wewould like, they are well behaved enough to be closed under pointwise limits!

We now show that M(X,S) is closed under multiplication.

Lemma 8.7.4 Products of Measurable Functions Are Measurable

Let X be a nonempty set and S a sigma - algebra on X . Let f, g ∈ M(X,S). Thenfg ∈M(X,S).

Proof 8.7.4Let fn, the truncation of f , be defined by

fn(x) =

f(x) | f(x) |≤ nn f(x) > n−n f(x) < −n

We define the truncation of g, gn, is a similar way. We can easily show fn and gm are measurablefor any n and m. We only show the argument for fn as the argument for gm is identical. Let α be agiven real number. Then

(fn(x) > α) =

∅ α ≥ n,(f(x) > n) ∪ (α < f(x) ≤ n) 0 ≤ α < n,(f(x) > n) ∪ (α < f(x) ≤ n) −n < α < 0,X α ≤ −n.

It is easy to see all of these sets are in S since f is measurable. Thus, each real valued fn ismeasurable.

It then follows by Lemma 8.5.1 that fngm is also measurable. Note we are using the definition ofmeasurability for real valued functions here. Next, an easy argument shows that at each x,

f(x) = limnfn(x) and g(x) = lim

mgm(x)


It then follows that

f(x) gm(x) = limn

(fn(x)gm(x)

)

Using Theorem 8.7.3, we see fgm is measurable. Then, noting

f(x) g(x) = limm

(f(x)gm(x)

)

another application of Theorem 8.7.3 establishes the result.

8.8 Continuous Functions of Measurable Functions

We wish to explore what properties the composition of a continuous function and a measurablefunction might have.

8.8.1 The Composition With Finite Measurable Functions

We begin with the case of finite measurable functions.

Lemma 8.8.1 Continuous Functions Of Finite Measurable Functions Are Measurable

Let X be nonempty and (X,S, µ) be a measure space. Let f ∈ M(X,S) be finite. Letφ : < → < be continuous. Then φ f is measurable.

Proof 8.8.1Let α be in <. We claim (

φ f)−1

(α,∞) = f−1

(φ−1(α,∞)

).

First, let x be in the right hand side. Then,

f(x) ∈ φ−1(α,∞) ⇒ φ

(f(x)

)∈ (α,∞)

⇒ x ∈(φ f

)−1

(α,∞).

Conversely, if x is in the left hand side, then(φ f

)(x) ∈ (α,∞) ⇒ f(x) ∈ φ−1(α,∞)

⇒ x ∈ f−1

(φ−1(α,∞)

).

Since φ is continuous, G = φ−1(α,∞) is an open set. Finally, since f is measurable, f−1(G) is inS. We conclude that φ f is measurable, since our choice of α is arbitrary.

8.8. CONTINUOUS COMPOSITIONS 149

To handle the composition of a continuous function and an extended valued measurable function, weneed an approximation result.

8.8.2 The Approximation Of Non-negative Measurable Functions

Theorem 8.8.2 The Approximation Of Non negative Measurable Functions By Monotone Se-quences

Let X be a nonempty set and S a sigma - algebra on X . Let f ∈ M(X,S) which is nonnegative. Then there is a sequence (φn) ⊆M(X,S) so that

(i): 0 ≤ φn(x) ≤ φn+1 for all x and for all n ≥ 1.

(ii): φn(x) ≤ f(x) for all x and n and f(x) = limn φn(x).

(iii): Each φn has a finite range of values.

Proof 8.8.2Pick a positive integer n. Let

Ek,n =

x ∈ X | k2n ≤ f(x) < k+1

2n , for 0 ≤ k ≤ n2n − 1x ∈ X | n ≤ f(x), for k = n2n

You should draw some of these sets for a number of choices of non negative functions f to get a feelfor what they mean. Once you have done this, you will see that this definition slices the [0, n] range off into n2n slices each of height 2−n. The last set, En2n,n is the set of all points where f(x) exceedsn. This gives us a total of n2n + 1 sets. It is clear that X = ∪k Ek,n and that each of these sets aredisjoint from the others. Now define the functions φn by

φn(x) =k

2n, x ∈ Ek,n.

It is evident that φn only takes on a finite number of values and so (iii) is established. Also, since fis measurable, we know each Ek,n is measurable. Then, given any real number α, the set (φn(x) >α) is either empty or consists of a union of the finite number of sets Ek,n with the property thatα > (k/2n). Thus, (φn(x) > α) is measurable for all α. We conclude each φn is measurable. Iff(x) = +∞, then by definition, φn(x) = n for all n and we have f(x) = limn φn(x). Note, the φnvalues are strictly monotonically increasing which shows (i) and (ii) both hold in this case.

On the other hand, if f(x) is finite, let n0 be the first integer with n0− 1 ≤ f(x) < n0. Then, wemust have φ1(x) = 1, φ2(x) = 2 and so forth until we have φn0−1 = n0 − 1. These first values aremonotone increasing. We also know from the definition of φn0

that there is a k0 so that

k0

2n0≤ f(x) <

k0 + 1

2n0.

Thus, 0 ≤ f(x)− φn0(x) < 2−n0 . Now consider the function φn0+1. We know

f(x) ∈[k0

2n0,k0 + 1

2n0

)=

[2k0

2n0+1,

2k0 + 1

2n0+1

)∪[

2k0 + 1

2n0+1,

2k0 + 2

2n0+1

)


If f(x) lands in the first interval above, we have

φn0+1(x) =2k0

2n0+1=

k0

2n0= φn0(x)

and if f(x) is in the second interval, we have

φn0+1(x) =2k0 + 1

2n0+1>

k0

2n0= φn0

(x).

In both cases, we have φn0(x) ≤ φn0+1(x). We also have immediately that 0 ≤ f(x)−φn0+1(x) <

2−n0−1.The argument for n0+2 and so on in quite similar and is omitted. This establishes (i) for this case.

In general, we have 0 ≤ f(x)− φk(x) < 2−k for all k ≥ n0. This implies that f(x) = limn φn(x)which establishes (ii).

Finally, we can handle the case of the composition of a continuous function and an extendedvalued measurable function.

8.8.3 Continuous Functions of Extended Valued Measurable Functions

Lemma 8.8.3 Continuous Functions Of Measurable Functions Are Measurable

Let X be nonempty and (X,S, µ) be a measure space. Let f ∈M(X,S). Let φ : < → <be continuous and assume that limn φ(n) and limn φ(−n) are well defined extended valuenumbers. Then φ f is measurable.

Proof 8.8.3Assume first that f is non negative. Then by Theorem 8.8.2, there is a sequence of finite non negativeincreasing functions (fn) which are measurable and satisfy fn ↑ f . Let E be the set of points wheref is infinite. Then, fn(x) = n when x is in E and.

limn

fn(x) =

f(x) x ∈ EC∞ x ∈ E.

Thus, since φ is continuous on EC and limn φ(n) is a well-defined extended valued real number, wehave,

limn

φ

(fn(x)

)=

φ(f(x)) x ∈ EClimn φ(n) x ∈ E.

Let limn φ(n) = β in [∞,∞]. Thus, if β is finite, we have

limn

φ

(fn(x)

)= φ

(f IEC

)+ β IE

which is measurable since the first part is measurable by Lemma 8.8.1 and the second part is mea-surable since E is a measurable set by Lemma 8.6.2. If β =∞, we have

limn

φ

(fn(x)

)=

φ(f(x)) x ∈ EC∞ x ∈ E.

8.9. HOMEWORK 151

Now apply Lemma 8.6.2. Since E is measurable and f1 defined by

f1(x) =

φ(f(x)) x ∈ EC0 x ∈ E,

is measurable, we see limn φ

(fn(x)

)is measurable. A similar argument holds if β = −∞. We

conclude that if f is non negative, φ f interpreted as above is a measurable function.Thus, if f is arbitrary, the argument above shows that φ f+ and φ f− are measurable. This

implies that φ f = φ (f+ − f−) is measurable when interpreted right.

8.9 HomeworkExercise 8.9.1 If a, b and c are real numbers, define the value in the middle, mid(a, b, c) by

mid(a, b, c) = inf supa, b, supa, c, supb, c .

Let X be a nonempty set and S a sigma - algebra on X . Let f1, f2, f3 ∈ M(X,S). Prove thefunction h defined pointwise by h(x) = mid(f1(x), f2(x), f3(x)) is measurable.

Exercise 8.9.2 Let X be a nonempty set and S a sigma - algebra on X . Let f ∈ M(X,S) andA > 0. Define fA by

fA(x) =

f(x), | f(x) |≤ AA, f(x) > A−A, f(x) < −A

Prove fA is measurable.

Exercise 8.9.3 Let X be a nonempty set and S a sigma - algebra on X . Let f ∈ M(X,S) andassume there is a positive K so that 0 ≤ f(x) ≤ K for all x. Prove the sequence φn of functionsgiven in Theorem 8.8.2 converges uniformly to f on X .

Exercise 8.9.4 Let X and Y be nonempty sets and let f : X → Y be given. Prove that if T is asigma - algebra of subsets of Y , then f−1(E) | E ∈ T is a sigma - algebra of subsets of X .

Exercise 8.9.5 Let (X,S) be a measurable space. Let (µn) be a sequence of measures on S withµn(X) ≤ 1 for all n. Define λ on S by

λ(E) =∞∑n=1

1/2n µn(E)

for all measurable E. Prove λ is a measure on S.

Chapter 9

Measure And Integration

Once we have a nonempty set X with a given sigma - algebra S, we can develop an abstract versionof integration. To motivate this, consider the Borel sigma - algebra on <, B. We know how todevelop and use an integration theory that is based on finite intervals of the form [a, b] for boundedfunctions. Hence, we have learned to understand and perform integrations of the form

∫ baf(t)dt for

the standard Riemann integral. We could also write this as∫ b

a

f(t)dt =

∫[a,b]

f(t)dt

and we have learned that∫[a,b]

f(t)dt =

∫(a,b)

f(t)dt =

∫(a,b]

f(t)dt =

∫[a,b)

f(t)dt.

Note that we can thus say that we can compute∫Ef(t)dt for E ∈ B for sets E which are finite

and have the form [a, b], (a, b], [a, b) and (a, b). We can extend this easily to finite unions of disjointintervals of the form E as given above by taking advantage of Theorem 4.5.3 to see∫

∪nEnf(t)dt =

∑n

∫En

f(t)dt.

However, the development of the Riemann integral is closely tied to the interval [a, b] and so it isdifficult to extend these integrals to arbitrary elements F of B. Still, we can see that the Riemannintegral is defined on some subset of the sigma - algebra B.

From our discussions of the Riemann - Stieljes integral, we know that the Riemann integralcan be interpreted as a Riemann - Stieljes integral with the integrator given by the identity functionid(x) = x. Let’s switch to a new notation. Define the function µ(x) = x. Then for our allowable E,we can write

∫Ef(t)dt =

∫Ef(t)dµ(t) which we can further simplify to

∫Efdµ as usual. Note

that µ is a function which assigns a real value which we interpret as length to all of the allowable setsE we have been discussing. In fact, note µ is a mapping which satisfies

(i): If E is the empty set, then the length of E is 0; i.e. µ(∅) = 0.

(ii): If E is the finite interval [a, b], (a, b], [a, b) or (a, b), µ(E) = b− a.

(iii): If (En) is a finite collection of disjoint intervals, then the length of the union is clearly the sumof the individual lengths; i.e. µ(∪nEn) =

∑n µ(En).

153

154 CHAPTER 9. ABSTRACT INTEGRATION

However, µ is not defined on the entire sigma -algebra. Also, it seems that we would probably liketo extend (iii) above to countable disjoint unions as it is easy to see how that would arise in practice.If we could find a way to extend the usual length calculation of an interval to the full sigma -algebra,we could then try to extend the notion of integration as well.

It turns out we can do all of these things but we can not do it by reusing our development processfrom Riemann integration. Instead, we must focus on developing a theory that can handle integratorswhich are mappings µ defined on a full sigma - algebra. It is time to precisely define what we meanby such a mapping.

Definition 9.0.1 Measures

Let X be a nonempty set and S a sigma - algebra of subsets in X . We say µ : S → < is ameasure on S if

(i): µ(∅) = 0,

(ii): µ(E) ≥ 0, for all E ∈ S,

(iii): µ is countably additive on S; i.e. if (En) ⊆ S is a countable collection of disjointsets, then µ(∪nEn) =

∑n µ(En).

We also say (X,S, µ) is a measure space. If µ(X) is finite, we say µ is a finite measure.Also, even if µ(X) = ∞, the measure µ is “almost finite” if we can find a collection ofmeasurable sets (Fn) so that X = ∪nFn with µ(Fn) finite for all n. In this case, we saythe measure µ is σ - finite.

We can drop the requirement that the mapping µ be non negative. The resulting mapping is calleda charge instead of a measure. This will be important later.

Definition 9.0.2 Charges

Let X be a nonempty set and S a sigma - algebra of subsets in X . We say ν : S → < is acharge on S if

(i): ν(∅) = 0,

(ii): ν is countably additive on S; i.e. if (En) ⊆ S is a countable collection of disjointsets, then ν(∪nEn) =

∑n ν(En).

Note that we want the value of the charge to be finite on all members of S as otherwisewe could potentially have trouble with subsets having value∞ and−∞ inside a given set.That would then lead to undefined∞−∞ operations.

Let’s look at some examples:

Example 9.0.1 Let X be any nonempty set and let the sigma - algebra be S = P(X), the power setof X . Define µ1 on S by µ1(E) = 0 for all E. Then µ1 is a measure, albeit not very interesting!Another non interesting measure is defined by µ2(E) =∞ if E is not empty and 0 if E = ∅.

Example 9.0.2 Let X be any set and again let S = P(X). Pick any element p in X . Define µ byµ(E) = 0 if p 6∈ E and 1 if p ∈ E. Then µ is a measure.

Example 9.0.3 Let X be the counting numbers, N , and S = P(N). Define µ by µ(E) is thecardinality of E if E is a finite set and ∞ otherwise. Then µ is a measure called the countingmeasure. Note that N = ∪n 1, . . . , n for all n and µ(1, . . . , n) = n, which implies µ is a σ -finite measure.

9.1. PROPERTIES 155

Example 9.0.4 This example is just a look ahead to future material we will be covering. Let B be theextended Borel sigma - algebra. We will show later there is a measure λ : B → < that extends theusual idea of the length of an interval. That is, if E is a finite interval of the form (a, b), [a, b), (a, b]or [a, b], then the length of E is b− a and λ(E) = b− a. Further, if the interval has infinite length,(for example, E is (−∞, a)), then λ(E) =∞ also. The measure λ will be called Borel measure andsince < = ∪n[−n, n], we see Borel measure is a σ - finite measure. The sets in B are called Borelmeasurable sets.

Example 9.0.5 We will be able to show that there is a larger sigma - algebraM of subsets of < anda measure µ defined onM which also returns the usual length of intervals. Hence, B ⊆ M strictly(i.e. there are sets inM not in B) with µ = λ on B. This measure will be called Lebesgue measureand the sets in M will be called Lebesgue measurable sets. The proof that there are Lebesguemeasurable sets that are not Borel sets will require a non constructive argument using the Axiom ofChoice. Further, we will be able to show that the Lebesgue sigma - algebra is not the entire powerset as there are non Lebesgue measurable sets. The proof that such sets exist requires the use of theinteresting functions built using Cantor sets discussed in Chapter 13.

Example 9.0.6 In the setting of Borel measure on <, we will be able to show that if g is a continuousand monotone increasing function of <, then there is a measure, λg defined on B which satisfies

λg(E) =

∫E

dg

for any finite interval E. Here,∫Edg is the usual Riemann - Stieljes integral.

9.1 Some Basic Properties Of Measures

Lemma 9.1.1 Monotonicity

Let (X,S, µ) be a measure space. IfE,F ∈ S withE ⊆ F , then µ(E) ≤ µ(F ). Moreover,if µ(E) is finite, then µ(F \ E) = µ(F )− µ(E).

Proof 9.1.1We know F = E ∪ (F \ E) is a disjoint decomposition of F . By the countable additivity of µ, itfollows immediately that µ(F ) = µ(E) +µ(F \E). Since µ is non-negative, we see µ(F ) ≥ µ(E).Finally, if µ(E) is finite, then subtraction is allowed in µ(F ) = µ(E) + µ(F \ E) which leads toµ(F \ E) = µ(F )− µ(E).

Lemma 9.1.2 The Measure Of Monotonic Sequence Of Sets

Let (X,S, µ) be a measure space.

(i): If (En) is an increasing sequence of sets in S (i.e. En ⊆ En+1 for all n), thenµ(∪nEn) = limn µ(En).

(ii): If (Fn) is an decreasing sequence of sets in S (i.e. Fn+1 ⊆ Fn for all n) and µ(F1)is finite, then µ(∩nFn) = limn µ(Fn).

Proof 9.1.2To prove (i), if there is an index n0 where µ(En0 is infinite, then by the monotonicity of µ, we musthave∞ = µ(En0 ≤ µ(∪nEn). Hence, µ(∪nEn) = ∞. However, since En0 ⊆ En for all n ≥ n0,


again by monotonicity, n ≥ n0 implies µ(En) = ∞. Thus, limn µ(En) = µ(∪nEn) = ∞. On theother hand, if µ(En) is finite for all n, define the disjoint sequence of set (An) as follows:

A1 = E1

A2 = E2 \ E1

A3 = E3 \ E2

......

...

An = En \ En−1

We see ∪nAn = ∪nEn and since µ is countably additive, we must have µ(∪nAn) =∑n µ(An).

Since by assumption µ(En) is finite in this case, we know µ(An) = µ(En) − µ(En−1). It followsthat

n∑k=1

µ(Ak) = µ(E1) +

n∑k=2

(µ(Ek)− µ(Ek−1)

)= µ(E1) + µ(En)− µ(E1)

= µ(En).

We conclude

µ(∪nEn) = µ(∪nAn)

= limn

n∑k=1

µ(Ak)

= limnµ(En)

this proves the validity of (i). Next, for (ii), construct the sequence of sets (En) by

E1 = ∅E2 = F1 \ F2

E3 = F1 \ F3

......

...

En = F1 \ Fn.

Then (En) is an increasing sequence of sets and so by (i), µ(∪n En) = limn µ(En). Since µ(F1) isfinite, we then know that µ(En) = µ(F1)− µ(Fn). Hence, µ(∪n En) = µ(F1)− limn µ(Fn). Next,note by De Morgan’s Laws,

µ(∪n En) = µ

(∪n F1 ∩ FCn

)= µ

(F1 ∩ ∪nFCn

)= µ

(F1 ∩

(∩nFn

)C)

= µ

(F1 \

(∩nFn

)).

9.1. PROPERTIES 157

Thus, since µ(F1) is finite and ∩nFn ⊆ F1, we have µ(∪nEn) = µ(F1) − µ(∩nFn). Combiningthese results, we have

µ(F1)− limnµ(Fn) = µ(F1)− µ(∩nFn).

The result then follows by canceling µ(F1) from both sides which is allowed as this is a finite number.

We will now develop a series of ideas involving sequences of sets.

Definition 9.1.1 Limit Inferior And Superior Of Sequences Of Sets

Let X be a nonempty set and (An) be a sequence of subsets of X . The limit inferior of(An) is defined to be the set

lim inf = lim(An) =

∞⋃m=1

∞⋂n=m

An

while the limit superior of (An) is defined by

lim sup = lim(An) =

∞⋂m=1

∞⋃n=m

An

It is convenient to have a better characterization of these sets.

Lemma 9.1.3 Characterizing Limit Inferior And Superiors Of Sequences Of Sets

Let (An) be a sequence of subsets of the nonempty set X . Then we have

lim inf(An) = x ∈ X | x ∈ Ak for all but finitely many indices k = B

and

lim sup(An) = x ∈ X | x ∈ Ak for infinitely many indices k = C

Proof 9.1.3We will prove the statement about lim inf(An) first. Let x ∈ B. If there are no indices k so thatx 6∈ Ak, then x ∈ ∩∞n=1 telling us that x ∈ lim inf(An). On the other hand, if there are a finitenumber of indices k that satisfy x 6∈ Ak, we can label these indices as k1, . . . , kp for some positiveinteger p. Let k∗ be the maximum index in this finite list. Then, if k > k∗, x ∈ ∩∞n=k. This impliesimmediately that x ∈ lim inf(An). Conversely, if x ∈ lim inf(An), there is an index k0 so thatx ∈ ∩∞n=k0

. This implies that x can fail to be in at most a finite number of Ak where k < k0. Hence,x ∈ B.

Next, we prove that lim sup(An) = C. If x ∈ C, then if there were an index m0 so that x 6∈∪∞n=m0

, then x would belong to only a finite number of setsAk which contradicts the definition of theset C. Hence, there is no such index m0 and so x ∈ ∪∞n=m for all m. This implies x ∈ lim sup(An).On the other hand, if x ∈ lim sup(An), then x ∈ ∪∞n=m for all m. So, if x was only in a finitenumber of sets An, there would be a largest index m∗ satisfying x ∈ Am∗ but x 6∈ Am if m > m∗.But this then says x 6∈ lim sup(An). This is a contradiction. Thus, our assumption that x was onlyin a finite number of sets An is false. This implies x ∈ C.


Lemma 9.1.4 Limit Inferiors And Superiors Of Monotone Sequences Of Sets

Let X be a nonempty set. Then

(i): If (An) is an increasing sequence of subsets of X , then

lim inf(An) = lim sup(An) =

∞⋃n=1

An.

(ii): If (An) is a decreasing sequence of subsets of X , then

lim inf(An) = lim sup(An) =

∞⋂n=1

An.

(iii): If (An) is an arbitrary sequence of subsets of X , then

∅ ⊆ lim inf(An) ⊆ lim sup(An)

Proof 9.1.4

(i): If x ∈ lim sup(An), then x ∈ ∪∞n=1An. Conversely, if x ∈ ∪∞n=1An, there is an index n0 so thatx ∈ An0

. But since the sequence (An) is increasing, this means x ∈ An for all n > n0 also. Hence,x ∈ ∪∞n=mAn for all indices m ≥ n0. However, it is also clear that x is in any union that starts at nsmaller than n0. Thus, x must be in ∩∞m=1 ∪∞n=m An. But this is the set lim sup(An). We concludelim sup(An) = ∪∞n=1. Now look at the definition of lim inf(An). Since An is monotone increasing,∩∞n=mAn = Am. Hence, it is immediate that lim inf(An) = ∪∞n=1.

(ii): the argument for this case is similar to the argument for case (i) and is left to you.

(iii): it suffices to show that lim inf(An) ⊆ lim sup(An). If x ∈ lim inf(An), by Lemma 9.1.3, xbelongs to all but finitely many An. Hence, x belongs to infinitely many An. Then, applying Lemma9.1.3 again, we have the result.

There will be times when it will be convenient to write an arbitrary union of sets as a countableunion of disjoint sets. In the next result, we show how this is done.

Lemma 9.1.5 Disjoint Decompositions Of Unions

Let X be a nonempty set and let (An) be a sequence of subsets of X . Then there exists asequence of mutually disjoint set (Fn) satisfying ∪nAn = ∪nFn.

Proof 9.1.5Define sets En and Fn as follows:

E0 = ∅, F1 = A1 \ E0 = A1

E1 = A1, F2 = A2 \ E1 = A2 \A1

E2 = A1

⋃A2, F3 = A3 \ E2 = A3 \

(A1

⋃A2

)

9.1. PROPERTIES 159

E3 =

3⋃k=1

Ak, F4 = A4 \ E3 = A4 \( 3⋃k=1

Ak

)... =

...,...

En =

n⋃k=1

Ak, Fn+1 = An+1 \ En = A4 \( n⋃k=1

Ak

)

Note that (En) forms a monotonically increasing sequence of sets with cupnAn ∪n En. We claimthe sets Fn are mutually disjoint and ∪nj=1fj = ∪nj=1Aj . We do this by induction.

Proof Basis: It is clear that F1 and F2 are disjoint and F1 ∪ F2 = A1 ∪A2. Induction: We assumethat (Fk) are mutually disjoint for 1 ≤ k ≤ n and ∪kj=1fj = ∪kj=1Aj for 1 ≤ k ≤ n as well. Then

Fn+1 = An+1 \ En

= An+1

⋂( n⋃j=1

Aj

)C

=

n⋂j=1

(An+1

⋂ACj

).

Now, by construction, Fj ⊆ Aj for all j. However, from the above expansion of Fn+1, we seeFn+1 ⊆ ACj for all 1 ≤ j ≤ n. This tells us Fn+1 ⊆ FCj for these indices also. We concludeFn+1 is disjoint from all the previous Fj . This shows (Fj) is a collection of mutually disjoint sets for1 ≤ j ≤ n+ 1. This proves the first part of the assertion. To prove the last part, note

n+1⋃j=1

Fj =

n⋃j=1

Fj⋃

Fn+1

=

n⋃j=1

Aj⋃(

An+1 \( n⋃j=1

Aj

))

=

n+1⋃j=1

Aj .

This completes the induction step. We conclude that this proposition holds for all n.

Since the claim holds, it is then obvious that ∪nj=1fj = ∪nj=1Aj .

To finish this section on measures, we want to discuss the idea that a property holds except ona set of measure zero. Recall, this subject came up when we discussed the content of a subset of <earlier in Section 5.3. However, we can extend this concept of an arbitrary measure space (X,S, µ)as follows.

Definition 9.1.2 Propositions Holding Almost Everywhere

Let (X,S, µ) be a measure space. We say a proposition P holds almost everywhere on Xif x ∈ X | P does not hold has µ measure zero. We usually say the proposition holdsµ a.e. rather than writing out the phrase µ almost everywhere. Also, if the measure µis understood from context, we usually just say the proposition hold a.e. to make it eveneasier to write down.


Comment 9.1.1 Given the measure space (X,S, µ), if f and g are extended real valued functions onX which are measurable, we would say f = g µ a.e. if µ(x ∈ X | f(x) 6= g(x)) = 0.

Comment 9.1.2 Given the measure space (X,S, µ), If (fn) is a sequence of measurable extendedreal valued functions on the X , and f : X → < is another measurable function on X , we wouldsay fn converges pointwise a.e. to f if the set x ∈ X | fn(x) 6→ f(x) has measure 0. We wouldusually write fn → f pointwise µ a.e.

9.2 Integration

In this section, we will introduce an abstract notion of integration on the measure space (X,S, µ).Recall that M(X,S) denotes the class of extended real valued measurable functions f on X . Firstwe introduce a standard notation for some useful classes of functions. When we want to restrict ourattention to the non negative members of M(X,S), we will use the notation that f ∈M+(X,S).

To construct an abstract integration process on the measure space (X,S, µ), we begin by definingthe integral of a class of functions which can be used to approximate any function f in M+(X,S).

Definition 9.2.1 Simple Functions

Let (X,S, µ) be a measure space and let f : X → < be a function. We say f is a simplefunction if the range of f is a finite set and f is S measurable. This implies the followingstandard unique representation of f . Since the range is finite, there is an positive integerN and distinct numbers aj , 1 ≤ j ≤ N so that

(i): the sets Ej = f−1(aj) are measurable and mutually disjoint for 1 ≤ j ≤ N ,

(ii): X =⋃Nj=1 Ej ,

(iii): f has the characterization

f(x) =

N∑j=1

ajIEj (x).

We then define the integral of a simple function as follows.

Definition 9.2.2 The Integral Of A Simple Function

Let (X,S, µ) be a measure space and let φ : X → < be a simple function. Let

φ(x) =

N∑j=1

ajIEj (x),

be the standard representation of φ where the numbers aj are distinct and the sets Ej aremutually disjoint, cover X , and are measurable for 1 ≤ j ≤ N for some positive integerN . Then the integral of φ with respect to the measure µ is the extended real valued number∫

φ dµ =

N∑j=1

aj µ(Ej).

9.2. INTEGRATION 161

Comment 9.2.1 We note that∫φdµ can be +∞. Recall, our convention that 0 · ∞ = 0. Hence, if

one of the values aj is 0, the contribution to the integral is 0µ(Ej) which is 0 even if µ(Ej) = ∞.Further, note the 0 function onX can be defined as I∅ which is a simple function. Hence,

∫0dµ = 0.

Using this, we can define the integral of any function in M+(X,S).

Definition 9.2.3 The Integral Of A Non-negative Measurable Function

Let (X,S, µ) be a measure space and let f ∈M+(X,S), µ). For convenience of notation,let F+ denote the collection of all non negative simple functions on X . Then, the integralof f with respect to the measure µ is the extended value real number∫

f dµ = sup∫

φ dµ | φ ∈ F+, φ ≤ f.

If E ∈ S, we define the integral of f over E with respect to µ to be∫E

f dµ =

∫fIE dµ.

It is time to prove some results about this new abstract version of integration.

Lemma 9.2.1 Properties Of Simple Function Integrations

Let (X,S, µ) be a measure space and let φ, ψ ∈M+(X,S)) be simple functions. Then,

(i): If c ≥ 0 is a real number, then cφ is also a simple function and∫cφ dµ = c

∫φ dµ.

(ii): φ+ ψ is also a simple function and∫

(φ+ ψ) dµ =∫φ dµ +

∫ψ dµ.

(iii): The mapping λ : S → < defined by λ(E) =∫Eφ dµ for all E in S is a measure.

Proof 9.2.1Let φ have the standard representation

φ(x) =

N∑j=1

ajIEj (x),

where the numbers aj are distinct, the sets Ej are mutually disjoint, cover X , and are measurablefor 1 ≤ j ≤ N for some positive integer N . Similarly, let ψ have the standard representation

ψ(x) =

M∑k=1

bkIFk(x),

where the numbers bk are distinct, the sets Fk are mutually disjoint, cover X , and are measurablefor 1 ≤ k ≤M for some positive integer M . Now to the proofs of the assertions:(i):First, if c = 0, cφ = 0 and

∫0dµ = 0 ·

∫φdµ. Next, if c > 0, then it is easy to see cφ is a simple

function with representation

cφ(x) =

N∑j=1

cajIEj (x),


and hence, by the definition of the integral of a simple function∫cφ dµ =

N∑j=1

caj µ(Ej)

= c

(N∑j=1

aj µ(Ej)

)

= c

∫φ dµ.

(ii):This one is more interesting to prove. First, to prove φ + ψ is a simple function, all we have to dois find its standard representation. From the standard representations of φ and ψ, it is clear the setsFk ∩ Ej are mutually disjoint and since X = ∪Ej = ∪Fk, we have the identities

Fk =

N⋃j=1

Fk ∩ Ej , and Ej =

M⋃k=1

Fk ∩ Ej .

Now define h : X → < by

h(x) =

N∑j=1

M∑k=1

(aj + bk) IFk∩Ej (x).

Next, since X = ∪j ∪k Fk ∩ Ej , given x ∈ X , there are indices k0 and j0 so that x ∈ Fk0 ∩ Ej0 .Thus,

φ(x) + ψ(x) = aj0 IEj0 + bk0 IFk0 = aj0 + bk0 = h(x).

From the above argument, we see h(x) = φ(x) + ψ(x) for all x in X . It follows that the rangeof h is finite and hence it is a measurable simple function, but we still do not know its standardrepresentation.

To find the standard representation, let ci, 1 ≤ i ≤ P be the set of distinct numbers formed bythe collection aj + bk | 1 ≤ j ≤ N, 1 ≤ k ≤ M. Then let Ui be the set of index pairs (j, k) thatsatisfy ci = aj + bk. Finally, let

Gi =⋃

(j,k)∈Ui

Ej ∩ Fk.

Since the sets Fk ∩ Ej are mutually disjoint, we have

µ(Gi) =∑

(j,k)∈Ui

µ(Ej ∩ Fk).

It follows that

h(x) =

P∑i=1

ci IGi

9.2. INTEGRATION 163

is the standard representation of h = φ+ ψ. Thus∫h dµ =

∫(φ+ ψ) dµ =

P∑i=1

ci µ(Gi)

=

P∑i=1

ci

( ∑(j,k)∈Ui

µ(Ej ∩ Fk)

)

=

P∑i=1

∑(j,k)∈Ui

ci µ(Ej ∩ Fk).

But we know that

N∑j=1

M∑k=1

=

P∑i=1

∑(j,k)∈Ui

.

Hence, we can write∫(φ+ ψ) dµ =

N∑j=1

M∑k=1

(aj + bk) µ(Ej ∩ Fk)

=

N∑j=1

M∑k=1

aj µ(Ej ∩ Fk) +

N∑j=1

M∑k=1

bk µ(Ej ∩ Fk).

This can be reorganized as

N∑j=1

aj

M∑k=1

µ(Ej ∩ Fk) +

M∑k=1

bk

N∑j=1

µ(Ej ∩ Fk)

=

N∑j=1

aj µ

( M⋃k=1

Ej ∩ Fk)

+

M∑k=1

bk µ

( N⋃j=1

Ej ∩ Fk)

=

N∑j=1

aj µ(Ej) +

M∑k=1

bk µ(Fk)

=

∫φ dµ +

∫ψ dµ.

(iii):Given

φ(x) =

N∑j=1

ajIEj (x),

it is easy to see that

φ IE(x) =

N∑j=1

ajIE∩Ej (x).

Further, it is straightforward to show that the mappings µj : (S)→ < defined by µj(A) = µ(A∩Ej)


for all A in S are measures on the sigma - algebras S ∩ Ej for each 1 ≤ j ≤ N . It is also easy tosee that the finite linear combination of these measures given by ξ =

∑Nj=1 aj µj is a measure on S

itself. Thus, applying part (ii) of this lemma, we see

λ(E) =

∫φIE dµ =

∫φI∪Nj=1E∩Ej dµ

=

∫ ( N∑j=1

φIE∩Ej

)dµ =

N∑j=1

∫ajIE∩Ej dµ

=

N∑j=1

ajµ(E ∩ Ej) =

N∑j=1

aj µj(E) = ξ(E).

We conclude λ = ξ and λ is a measure on S.

Lemma 9.2.2 Monotonicity Of The Abstract Integral For Non Negative Functions

Let (X,S, µ) be a measure space and let f and g be in M+(X,S) with f ≤ g. Then,∫fdµ ≤

∫gdµ. Further, if E ⊆ F with E and F measurable sets, then

∫Efdµ ≤∫

Ffdµ.

Proof 9.2.2Let φ be a positive simple function which is dominated by f ; i.e., φ ≤ f . Then φ is also dominatedby g and so by the definition of the integral of f , we have∫

fdµ = sup ∫φdµ | 0 ≤ φ ≤ f

≤ sup ∫ψdµ | 0 ≤ ψ ≤ g

=

∫gdµ.

Next, if E ⊆ F with E and F measurable sets, then fIE ≤ fIF and from the first result, we have∫fIEdµ ≤

∫fIF dµ,

which implies the result we seek.

9.3 Complete Measures And Equality a.e.We know if a sequence of extended real - valued measurable functions (fn) converges pointwise toa function f , then the limit function is also measurable. But what if the convergence was pointwisea.e? Is it still true that the limit function is also measurable? In general, the answer is no. We haveto add an additional property to the measure. We will motivate this with an example that we are notreally fully prepared for, but it should make sense anyway. The argument below will also be placedin Chapter 13 for completeness of our exposition.

Let B ∩ [0, 1] denote the Borel sigma - algebra of subsets of [0, 1]. We will be able to show inlater chapters that there is a measure called Lebesgue measure, µL, defined on a sigma - algebra ofsubsets L, the Lebesgue sigma - algebra, which extends the usual meaning of length in the followingsense. If [a, b] is a finite interval then the length of [a, b] is the finite number b−a. Denote this length

9.3. EQUALITY A.E. 165

by `([a, b]). Then we can show that

µL([a, b]) = `([a, b]) = b− a.

We can show also that every subset in B is also in L. The restriction of µL to B is called Borelmeasure and we will denote it by µB .

We can argue that the Borel sigma - algebra is strictly contained in the Lebesgue sigma - algebraby using the special functions we construct in Chapter 13. In that Chapter, we show if C is a Cantorset constructed from the generating sequence (an) where lim 2nan = 0, the content of C is 0.Then if we let Ψ be the mapping discussed above for this C in Section 13.3, we define the mappingmapping g : [0, 1] → [0, 1] by g(x) = (Ψ(x) + x)/2. The mapping g is quite nice: it is 1 − 1,onto, strictly increasing and continuous. We also showed in the exercises in Section 13.3 that g(C)is another Cantor set with lim 2na′n = 1/2, where (a′n) is the generating sequence for g(C).

Now it turns out that the notion of content and Lebesgue measure coincide. Thus, we can saysince C is a Borel set,

µB(C) = µL(C) = 0.

Also, we can show that since lim 2na′n = 1/2,

µB(g(C)) = µL(g(C)) = 1/2.

A nonconstructive argument we will present later using the Axiom of Choice allows us to showthat any Lebesgue measurable set with positive Lebesgue measure must contain a subset which is notin the Lebesgue sigma - algebra. So since µL(g(C)) = 1/2, there is a set F ⊆ g(C) which is notis L. Thus, g−1(F ) ⊆ C which has Lebesgue measure 0. Lebesgue measure is a measure which hasthe property that every subset of a set of measure 0 must be in the Lebesgue sigma - algebra. Then,using the monotonicity of µL, we have µL(g−1(F )) is also 0. From the above remarks, we can infersomething remarkable.

Let the mapping h be defined to be g−1. Then h is also continuous and hence it is measurablewith respect to the Borel sigma-algebra. Note since B ⊆ L, this tells us immediately that h is alsomeasurable with respect to the Lebesgue sigma - algebra. Thus, h−1(U) is in the Borel sigma -algebra for all Borel sets U . But we know h−1 = g, so this tells us g(U) is in the Borel sigma-algebra if U is a Borel set. Hence, if we chose U = g−1(F ), then g(U) = F would have to be aBorel set if U is a Borel set. However, we know that F is not inL and so it is also not a Borel set. Wecan only conclude that g−1(F ) can not be a Borel set. However, g−1(F ) is in the Lebesgue sigma- algebra. Thus, there are Lebesgue measurable sets which are not Borel! Thus, the Borel sigma -algebra is strictly contained in the Lebesgue sigma - algebra!

We can use this example to construct another remarkable thing.

Comment 9.3.1 Using all the notations from above, note the indicator function of CC , the comple-ment of C, is defined by

ICC (x) =

1 x ∈ CC

0 x ∈ C.

We see f = ICC is Borel measurable. Next, define a new mapping like this:

φ(x) =

1 x ∈ CC

2 x ∈ C \ g−1(F )3 x ∈ g−1(F ).


Note that φ = f a.e. with respect to Borel measure. However, φ is not Borel measurable becauseφ−1(3) is the set g−1(F ) which is not a Borel set.

We conclude that in this case, even though the two functions were equal a.e. with respect toBorel measure, only one was measurable! The reason this happens is that even though C has Borelmeasure 0, there are subsets of C which are not Borel sets!

Hence, in some situations, we will have to stipulate that the measure we are working with hasthe property that every subset of a set of measure zero is measurable. We make this formal with adefinition.

Definition 9.3.1 Complete Measure

Let X be a nonempty set and (X,S, µ) be a measure space. If E ∈ S with µ(E) = 0 andF ⊆ E implies F ∈ S , we say µ is a complete measure. Further, it follows immediatelythat since µ(F ) ≤ µ(E) = 0, that µ(F ) = 0 also.

Comment 9.3.2 This example above can be used in another way. Consider the composition of themeasurable function IC and the function g defined above. For convenience, let W = g−1(F ) whichis Lebesgue measurable. Then IW is a measurable function. Consider(IW g−1

)(x) =

1 g−1(x) ∈W0 g−1(x) ∈WC =

1 x ∈ g(W )0 x ∈ g(WC)

=

1 x ∈ F0 x ∈ FC = IF .

But IF is not a measurable function as F is not a measurable set! Hence, the composition of themeasurable function IW and the continuous function g−1 is not measurable. This is why we can onlyprove measurability with the order of the composition reversed as we did in Lemma 8.8.1.

Theorem 9.3.1 Equality a.e. Implies Measurability If The Measure Is Complete

Let X be a nonempty set and (X,S, µ) be a complete measure space. Let f and g both beextended real valued functions on X with f = g a.e. Then, if f is measurable, so is g.

Proof 9.3.1We leave it as an exercise for you to show that an equivalent condition for measurability is to provef(x) ∈ G is measurable for all open sets G. Then to prove this result, let G be open in < and letE = (f(x) 6= g(x)). Then, by assumption, E is measurable and µ(E) = 0. Then, we claim

g−1(G) =

(g−1(G) ∩ E

)∪(f−1(G) \ E

).

If x is in g−1(G), then g(x) is inG∩E or it is inG∩EC . Now if g(x) ∈ E, g(x) 6= f(x), but if g(x)is in the complement of E, g and f must match. Thus, we see x is in the right hand side. Conversely,if x is in g−1(G)∩E, x is clearly in g−1(G). Finally, if x is in f−1(G) \E, then since x is not in E,f(x) = g(x). Thus, x ∈ g−1(G) also. We conclude x ∈ g−1(G). This shows the right hand side iscontained in the left hand side. Combining these arguments, we conclude the two sets must be equal.

Since g−1(G)∩E is a subset of E, the completeness of µ implies that g−1(G)∩E is measurableand has measure 0. The measurability of f tells us that f−1(G) \ E is also measurable. Hence,g−1(G) is measurable implying g is measurable.

If the measure µ is not complete, we can still prove the following.

Theorem 9.3.2 Equality a.e. Can Imply Measurability Even If The Measure Is Not Complete

9.4. CONVERGENCE THEOREMS 167

Let X be a nonempty set and (X,S, µ) be a measure space. Let f and g both be extendedreal valued functions on X with f = g on the measurable set EC with µ(E) = 0. Then, iff is measurable and g is constant on E, g is measurable.

Proof 9.3.2We will repeat the notation of the previous theorem’s proof. As before, if G is open, we can write

g−1(G) =

(g−1(G) ∩ E

)∪(f−1(G) \ E

).

Then, since g is constant on E with value say c, we have

g−1(G) =

(E c ∈ G∅ c 6∈ G

) ⋃(f−1(G) \ E

)=

E ∪

(f−1(G) \ E

)c ∈ G(

f−1(G) \ E)

c 6∈ G.

In both cases, the resulting set is measurable. Hence, we conclude g is measurable.

Comment 9.3.3 In Comment 9.3.1, we set

φ(x) =

1, x ∈ CC

2, x ∈ C \ g−1(F )3, x ∈ g−1(F ).

and since φ was not constant on E = C, φ was not measurable. However, if we had defined

φ(x) =

1 x ∈ CC

c x ∈ C,

then φ would have been measurable!

9.4 Convergence Theorems

We are now ready to look at various types of interchange theorems for abstract integrals. We will beable to generalize the results of Chapter 5 substantially. There are three basic results: (i) The Mono-tone Convergence Theorem, (ii) Fatou’s Lemma and (iii) The Lebesgue Dominated ConvergenceTheorem. We will examine each in turn.

Theorem 9.4.1 The Monotone Convergence Theorem

Let (X,S, µ) be a measure space and let (fn) be an increasing sequence of functions inM+(X,S). Let f : X → < be an extended real valued function such that fn → fpointwise on X . Then f is also in M+(X,S) and

limn

∫fndµ =

∫fdµ.

Proof 9.4.1Since fn converges to f pointwise, we know that f is measurable by Theorem 8.7.3. Further, sincefn ≥ 0 for all n on X , it is clear that f ≥ 0 also. Thus, f ∈M+(X,S). Since fn ≤ fn+1 ≤ f , the


monotonicity of the integral tells us∫fndµ ≤

∫fn+1dµ ≤

∫fdµ.

Hence,∫fndµ is an increasing sequence of real numbers bounded above by

∫fdµ. Of course, this

limit could be∞. Thus, we have the inequality∫fndµ ≤

∫fdµ.

We now show the reverse inequality,∫fdµ ≤

∫fndµ. Let α be in (0, 1) and choose any non

negative simple function φ which is dominated by f . Let

An = x | fn(x) ≥ α φ(x).

We claim that X = ∪nAn. If this was not true, then there would be an x which is not in any An.This implies x is in ∩nACn . Thus, using the definition of An, fn(x) < α φ(x) for all n. Since fn isincreasing and converges pointwise to f , this tells us

f(x) ≤ α φ(x) ≤ α f(x).

We can rewrite this as (1− α) f(x) ≤ 0 and since 1− α is positive by assumption, we can concludef(x) ≤ 0. But f is non negative, so combining, we see f(x) = 0. Since f dominates φ, we musthave φ(x) = 0 too. However, if this is true, fn(x) must be 0 also. Hence, fn(x) = 0 ≥ αφ(x) = 0for all n. This says x ∈ An for all n. This is a contradiction; thus, X = ∪nAn.

Next, since f and αφ are measurable, so is f − αφ. This implies x | f(x) − αφ(x) ≥ 0 is ameasurable set. Therefore, An is measurable for all n. Further, it is easy to An ⊆ An+1 for all n;hence, (An) is an increasing sequence of measurable sets. Then, we know by the monotonicity of theintegral, that ∫

An

αφdµ ≤∫An

fndµ ≤∫

fdµ.

Next, we know that λ(E) =∫Eφdµ defines a measure. Thus,

limnλ(An) = λ(∪nAn) = λ(X).

Replacing λ by its meaning in terms of φ, we have∫φdµ = lim

n

∫An

φdµ.

Multiplying through by the positive number α, we have

α

∫φdµ = lim

n

∫An

α φdµ ≤ limn

∫An

fndµ.

Thus, for all α ∈ (0, 1), we have

α

∫φdµ ≤ lim

n

∫An

fndµ.

Letting α→ 1, we obtain ∫φdµ ≤ lim

n

∫An

fndµ.


Since the above inequality is valid for all non negative simple functions dominated by f , we haveimmediately ∫

fdµ ≤ limn

∫An

fndµ,

which provides the other inequality we need to prove the result.

This has an immediate extension to series of non-negative functions.

Theorem 9.4.2 The Extended Monotone Convergence Theorem

Let (X,S, µ) be a measure space and let (gn) be a sequence of functions in M+(X,S, µ).Then, the sequence of partial sums,

Sn =

n∑k=1

gn

converges pointwise on X to the extended real valued non negative valued function S =∑∞k=1 gn. Further, S is also in M+(X,S) and

limn

∫Sndµ =

∫S dµ.

Comment 9.4.1 Once we establish Theorem 9.4.3 (below), we can rewrite this in series notation as

∞∑k=1

∫gk dµ =

∫ ∞∑k=1

gk dµ.

Proof 9.4.2To prove this result, just apply the Monotone Convergence Theorem to the sequence of partial sums(Sn).

The Monotone Convergence Theorem allows us to prove that this notion of integration is additiveand linear for positive constants.

Theorem 9.4.3 Abstract Integration Is Additive

Let (X,S, µ) be a measure space and let f and g be functions in M+(X,S). Further, letα be a non negative real number. Then

(i): α f is in M+(X,S) and ∫α fdµ = α

∫fdµ.

(ii): Also, f + g is in M+(X,S) and∫(f + g)dµ =

∫fdµ +

∫gdµ.


Proof 9.4.3

(i): The case α = 0 is clear, so we may assume without loss of generality that α > 0. We know fromTheorem 8.8.2 that there is a sequence of non negative simple functions (φn) which are increasingand converge up to f on X . Hence, since α > 0, we also know that α φn ↑ α f . Thus, by theMonotone Convergence Theorem, αf is in M+(X,S) and

limn

∫α φndµ =

∫α fdµ.

From Lemma 9.2.1, we know that∫αφndµ = α

∫φndµ. Thus,∫

α fdµ = α limn

∫φndµ = α

∫fdµ.

(ii): If we apply Theorem 8.8.2 to f and g, we find two sequences of increasing simple functions(φn) and (ψn) so that φn ↑ f and ψn ↑ g. Thus, (φn + ψn) ↑ (f + g). Hence, by the MonotoneConvergence Theorem, f + g is in M+(X,S) and∫

(f + g)dµ = limn

∫(φn + ψn)dµ = lim

n

∫φndµ + lim

n

∫ψndµ

=

∫fdµ +

∫gdµ.

Comment 9.4.2 The conclusion of Theorem 9.4.2 can now be restated. By Theorem 9.4.3, each Snis measurable and ∫

Sndµ =

n∑k=1

∫gk dµ.

Thus, the left hand side can be written as

limn

∫Sndµ = lim

n

n∑k=1

∫gk dµ

=

∞∑k=1

∫fk dµ.

The limit function S can clearly be written as an infinite series giving

∫S dµ =

∫ ∞∑k=1

fk dµ.

Combining these statements, we get the result.

Theorem 9.4.4 Fatou’s Lemma


Let (X,S, µ) be a measure space and let (fn) be a sequence of functions in M+(X,S).Then ∫

lim inf fn dµ ≤ lim inf

∫fn dµ.

Proof 9.4.4Recall

lim inf fn(x) = supm

infn≥m

fn(x)

= limm

infn≥m

fn(x),

lim sup fn(x) = infm

supn≥m

fn(x)

= limm

supn≥m

fn(x).

Further, if we define gm = infn≥m fn(x), we know

gm ↑ lim inf fn(x).

It follows immediately that gm is measurable for all m and by the monotonicity of the integral∫gm dµ ≤

∫fn(x) dµ ∀ n ≥ m.

This implies that∫gm dµ is a lower bound for the set of numbers

∫fn(x)dµ and so by definition

of the infimum, ∫gm dµ ≤ inf

n≥m

(∫fn(x) dµ

).

Let αm denote the number infn≥m

(∫fn(x) dµ

). Then, αm ↑ lim inf

∫fndµ. We see

limm

∫gm dµ ≤ lim

minfn≥m

(∫fn(x) dµ

)= lim

mαm = lim inf

∫fndµ.

But since gm ↑ lim inf fn(x), this implies∫lim inf fn(x) dµ ≤ lim inf

∫fn dµ.

These results allow us to construct additional measures.

Theorem 9.4.5 Constructing Measures From Non Negative Measurable Functions


Let (X,S, µ) be a measure space and let f be a function in M+(X,S). Then λ : S → <defined by

λ(E) =

∫E

f dµ, E ∈ S

is a measure.

Proof 9.4.5It is clear λ(∅) is 0 and that λ(E) is always non negative. To show that λ is countably additive, let(En) be a sequence of disjoint measurable sets in S and let E = ∪nEn, be their union. Then E ismeasurable. Define

fn =

n∑k=1

fIEk = fI∪nk=1 Ek.

We note that fn ↑ fIE and so by the Monotone Convergence Theorem,

λ(E) =

∫E

f dµ =

∫fIEdµ = lim

n

∫fn dµ.

But, ∫fn dµ =

n∑k=1

∫fIEk dµ

=

n∑k=1

∫Ek

f dµ =

n∑k=1

λ(Ek).

Combining, we have

λ(E) = limn

n∑k=1

λ(Ek) =

∞∑k=1

λ(Ek),

which proves that λ is countably additive.

Once we can construct another measure λ from a given measure µ , it is useful to think abouttheir relationship. One useful relationship is that of absolute continuity.

Definition 9.4.1 Absolute Continuity Of A Measure

Let (X,S, µ) be a measure space and let λ be another measure defined on S. We say λ isabsolutely continuous with respect to the measure µ if given E in S with µ(E) = 0, thenλ(E) = 0 also. This is written as λ µ.

We can also now prove an important result set within the framework of functions which are equala.e.

Lemma 9.4.6 Function f Zero a.e. If and Only If Its Integral Is Zero

Let (X,S, µ) be a measure space and let f be a function in M+(X,S). Then f = 0 a.e.if and only if

∫fdµ = 0.


Proof 9.4.6

(⇐): If∫fdµ = 0, then let En = (f(x) > 1/n). Note En ⊆ En+1 so that (En) is an increasing

sequence. Since (En) is an increasing sequence, we also know limn µ(En) = µ(∪nEn). Further,

∪nEn = x | f(x) > 0.

From the definition of En, we have

f(x) ≥ 1

nIEn ,

which implies

0 =

∫fdµ ≥

∫1

nIEn =

1

nµ(En).

We see µ(En) = 0 for all n which implies

µ(f(x) > 0) = limn

µ(En) = 0.

Hence, f is zero a.e.(⇒): If f is zero a.e., let E be the set where f(x) > 0. Let fn = nIE . Note that

lim inf fn(x) = supm

infn≥m

n f(x) > 00 f(x) = 0

= supm

m f(x) > 00 f(x) = 0

=

∞ f(x) > 00 f(x) = 0.

Clearly, f(x) ≤ lim inf fn(x) which implies∫f dµ ≤

∫lim inf fn dµ. Finally, by Fatou’s Lemma,

we find ∫f dµ ≤

∫lim inf fn dµ ≤ lim inf

∫fn dµ = lim inf nµ(E) = 0.

We conclude∫f dµ = 0.

Comment 9.4.3 Given f in M+(X,S), Theorem 9.4.5 allows us to construct the new measure λ byλ(E) =

∫Efdµ. If E has µ measure 0, we can use Lemma 9.4.6 to conclude that λ(E) = 0. Hence,

a measure constructed in this way is absolutely continuous with respect to µ.

We can now extend the Monotone Convergence Theorem slightly. It is often difficult to knowthat we have pointwise convergence up to a limit function on all of X . The next theorem allowsus to relax the assumption to almost everywhere convergence as long as the underlying measure iscomplete.

Theorem 9.4.7 The Extended Monotone Convergence Theorem Two

Let (X,S, µ) be a measure space with complete measure µ and let (fn) be an increasingsequence of functions in M+(X,S). Let f : X → < be an extended real valued functionsuch that fn → f pointwise a.e. on X . Then f is also in M+(X,S) and

limn

∫fndµ =

∫fdµ.

Proof 9.4.7Let E be the set of points where fn does not converge to f . Then by assumption E has measure 0


and fn ↑ f on EC . Thus,fn IEC ↑ f IEC

and applying the Monotone Convergence Theorem, we have

limn

∫fn IEC =

∫f IEC

and we can say f IEC is in M+(X,S). Now f is equal to fIEC a.e. and so although in general,f need not be measurable, since µ is a complete measure, we can invoke Theorem 9.3.1 to concludethat f is actually measurable. Hence, fIE is measurable too. Since µ(E) = 0, we thus know that∫

fIE dµ =

∫fIE dµ =

∫fnIE dµ = 0.

Therefore, we have∫f dµ =

∫E

f dµ +

∫EC

f dµ =

∫EC

f dµ

= limn

∫EC

fn dµ = limn

(∫EC

fn dµ +

∫E

fn dµ

)= lim

n

∫fn dµ.

Now to develop the Dominated Convergence Theorem, we need a few more concepts.

9.5 Extending Integration To Extended Real Valued Functions

The results of the previous sections can now be used to extend the notion of integration to generalextended real valued functions f in M(X,S).

Definition 9.5.1 Summable Functions

Let (X,S, µ) be a measure space and f be in M(X,S). We say f is summable or inte-grable on X if

∫f+dµ and

∫f−dµ are both finite. In this case, we define the integral of

f on X with respect to the measure µ to be∫f dµ =

∫f+ dµ −

∫f− dµ.

Also, if E is a measurable set, we define∫E

f dµ =

∫E

f+ dµ −∫E

f− dµ.

We let L1(X,S, µ) be the collection of summable functions on X with respect to the mea-sure µ.

Comment 9.5.1 If f can be decomposed into two non negative measurable functions f1 and f2 asf = f1 − f2 a.e. with

∫f1dµ and

∫f2dµ both finite, then note since f = f+ − f− also, we have

f1 + f− = f2 + f+.

9.5. EXTENDED INTEGRANDS 175

Thus, since all functions involved are summable,∫f1dµ +

∫f−dµ =

∫f2dµ +

∫f+dµ.

This implies that ∫(f2 − f1)dµ =

∫(f+ − f−)dµ =

∫fdµ.

Hence, the value of the integral of f is independent of the decomposition.

There are a number of results that follow right away from this definition.

Theorem 9.5.1 Summable Implies Finite a.e.

Let (X,S, µ) be a measure space and f be in L1(X,S). Then the set of points where f isnot finite has measure 0.

Proof 9.5.1Let En = (f(x) > n). Then it is easy to see that (En) is a decreasing sequence of sets and so

µ

(⋂n

En

)= lim

nµ(En).

It is also clear that(f(x) =∞) =

⋂n

En.

Next, note ∫f+ dµ =

∫En

f+ dµ +

∫ECn

f+ dµ

≥∫En

f+ dµ > n µ(En).

Thus, µ(En) < (∫f+dµ)/n. Since, the integral is a finite number, this tells us that limn µ(En) = 0.

This immediately implies that µ(E) = 0.

A similar argument shows that the set (f(x) = −∞) which is the same as the set (f−(x) =∞)has measure 0.

Theorem 9.5.2 Summable Function Equal a.e. To Another Measurable Function Implies TheOther Function Is Also Summable

Let (X,S, µ) be a measure space and f be in L1(X,S). Then if g ∈M(X,S) with f = ga.e., g is also summable.

Proof 9.5.2Let E be the set of points in X where f and g are not equal. Then E has measure zero. We then havefIEC = gIEC and so gIEC must be summable. Further, f+IEC = g+IEC and f−IEC = g−IEC .We then note that ∫

g+IECdµ =

∫g+IECdµ +

∫g+IEdµ


because∫g+IEdµ = 0 since E has measure zero. But then we see∫

g+ dµ =

∫g+IECdµ +

∫g+IEdµ =

∫f+IECdµ +

∫f+IEdµ =

∫f+ dµ.

Thus, we can see that∫g+ dµ is finite. A similar argument shows

∫g+ dµ is finite and so g is

summable.

Theorem 9.5.3 Summable Function Equal a.e. To Another Function With Measure CompleteImplies The Other Function Is Also Summable

Let (X,S, µ) be a measure space with µ complete and f be in L1(X,S). Then if g is afunction equal a.e. to f , g is also summable.

Proof 9.5.3First, the completeness of µ implies that g is measurable. The argument to show g is summable isthen the same as in the previous theorem’s proof.

We can extend the Monotone Convergence a bit more and actually construct a summable limitfunction in certain instances. This is known as Levi’s Theorem.

Theorem 9.5.4 Levi’s Theorem

Let (X,S, µ) be a measure space and let (fn) be a sequence of functions in L1(X,S, µ)which satisfy fn ≤ fn+1 a.e. Further, assume

limn

∫fn dµ <∞.

Then, there is a summable function f on X so that fn ↑ f a.e. and∫fndµ ↑

∫fdµ.

Proof 9.5.4Define the new sequence of functions (gn) by gn = fn− f1. Then, since (fn) is increasing a.e., (gn)is increasing and non negative a.e. By assumption, limn

∫gndµ is then finite. Call its value I for

convenience of exposition. Now define the function g pointwise on X by

g(x) = limngn(x).

This limit always exists as an extended real number in [0,∞] and since each gn is measurable, so isg. Let E = (g(x) =∞). Note that

E =⋂i

(⋃n

(gn(x) > i

)),

and so we know that E is measurable.For each non-negative measurable function gi, there is an increasing sequence of simple functions

(φin) such that φin ↑ gi. For each n, define (recall the binary operator∨means a pointwise maximum)

Ψn = φ1n ∨ φ2

n ∨ · · · ∨ φnn.

Then it is clear that Ψn is measurable. Given any x in X , we have that

Ψn+1(x) = φ1n+1 ∨ φ2

n+1 ∨ · · · ∨ φn+1n+1

9.5. EXTENDED INTEGRANDS 177

≥ φ1n+1 ∨ φ2

n+1 ∨ · · · ∨ φn+1n

≥ φ1n ∨ φ2

n ∨ · · · ∨ φnn

= Ψn(x).

Hence, (Ψn) is an increasing sequence. Moreover, it is straightforward to see that

Ψn(x) ≤ g1(x) ∨ g2(x) ∨ · · · ∨ gn(x) ≤ gn(x) = g(x).

Hence, we know that limn Ψn(x) ≤ g(x). If this limit was strictly less than g(x), let r denote halfof the gap size; i.e., r = (1/2)(g(x) − limn Ψn(x). Then, since Ψn(x) ≥ φin where i is an indexbetween 1 and n, we would have

φin < g(x) − r, 1 ≤ i ≤ n.

This implies that φnn ≤ g(x)− r for all n. In particular, fixing the index i, we see that φin ≤ g(x)− rfor all n. But since φin ↑ gi, this says gi(x) ≤ g(x) − r. Since, we can do this for all indices i, wehave limi gi(x) ≤ g(x) − r or g(x) ≤ g(x) − r which is not possible. We conclude limn Ψn = gpointwise on X .

Next, we claim∫

Ψndµ = limn

∫gndµ. To see this, first notice that

∫Ψndµ ≥

∫φindµ for all

1 ≤ i ≤ n. In fact, for any index j, there is an index n∗ so that n∗ > j. Hence,∫

Ψn∗dµ ≥∫φjndµ.

This still holds for any n > n∗ as well. Thus, for any index j, we can say

limn

∫Ψn dµ ≥ lim

n

∫φjn dµ =

∫gj dµ.

This implies that

limn

∫Ψn dµ ≥ sup

j

∫gj dµ = lim

j

∫gj dµ = I.

Also, since Ψn ≤ gn(x),

limn

∫Ψndµ ≤ lim

n

∫gndµ = I.

This completes the proof that∫

Ψndµ = limn

∫gndµ.

We now show the measure of E is zero. To do that, we start with the functions Ψn ∧ kIE forany positive integer k, where the wedge operation ∧ is simply taking the minimum. If g(x) is finite,then IE(x) = 0 and since Ψn is non negative, Ψn ∧ kIE = 0. On the other hand, if g(x) = ∞,then x ∈ E and so kIE(x) = k. Since Ψn ↑ g, eventually, Ψn(x) will exceed k and we will haveΨn ∧ kIE = k. These two cases allow us to conclude

Ψn ∧ kIE ↑ kIE

for all x. Thus,∫kIE dµ =

∫Ψn ∧ kIE dµ ≤

∫Ψn dµ ≤ lim

n

∫gndµ = I.

We conclude k µ(E) ≤ I for all k which implies that µ(E) = 0.

Finally, to construct the summable function f we need, define h = gIEC . Clearly, gn ↑ h onEC , that is, a.e. Also, since Ψn ↑ g on EC , the Monotone Convergence Theorem tells us that

limn

∫EC

Ψn dµ ↑∫EC

g dµ.


But, ∫EC

g dµ =

∫h dµ.

Hence, h is summable and so f1 + h is also summable. Define f = f1 + h on X and we have f issummable and

fn ↑ f1 + h∫fn dµ =

∫f1 dµ +

∫h dµ

=

∫f dµ.

Each summable function can also be used to construct a charge.

Theorem 9.5.5 Integrals Of Summable Functions Create Charges

Let (X,S, µ) be a measure space and let f be a functions in L1(X,S, µ). Then the map-ping λ : S → < defined by

λ(E) =

∫E

f dµ

for all E in S defines a charge on S. The integral∫Ef dµ is also called the indefinite

integral of f with respect to the measure µ.

Proof 9.5.5Since f is summable, note that the mappings λ+ and λ− defined by

λ+(E) =

∫E

f+ dµ, λ−(E) =

∫E

f− dµ

both define measures. It then follows immediately that λ is countably additive and hence is a charge.

Comment 9.5.2 Since∫Ef dµ defines a charge and is countably additive, we see that if (En) is a

collection of mutually disjoint measurable subsets, then∫∪nEn

f dµ =∑n

∫En

f dµ.

9.6 Properties Of Summable Functions

We need to know if L1(X,S, µ) is a linear space under the right interpretation of scalar multipli-cation and addition. To do this, we need some fundamental inequalities and conditions that forcesummability.

Theorem 9.6.1 Fundamental Abstract Integration Inequalities

9.6. SUMMABLE PROPERTIES 179

Let (X,S, µ) be a measure space.

(i): f ∈ L1(X,S, µ) if and only if | f |∈ L1(X,S, µ).

(ii): f ∈ L1(X,S, µ) implies |∫f dµ | ≤

∫| f | dµ.

(iii): f measurable and g ∈ L1(X,S, µ) with | f |≤| g | implies f is also summable and∫| f | dµ ≤

∫| g | dµ.

Proof 9.6.1

(i): If f is summable, f+ and f− are in M+(X,S) with finite integrals. Since | f |= f+ + f−, wesee | f |+=| f | and | f |−= 0. Thus,

∫| f |+ dµ =

∫(f+ + f−)dµ which is finite. Also, since∫

| f |− dµ = 0, we see that | f | is summable.

Conversely, if | f | is summable, then∫| f |+ dµ =

∫(f+ + f−)dµ is finite. This, in turn, tells

us each piece is finite and hence f is summable too.(ii): If f is summable, then

|∫

f dµ | = |∫

f+ dµ −∫

f− dµ |

≤∫

f+ dµ +

∫f− dµ

=

∫(f+ + f−) dµ =

∫| f | dµ.

(iii): Since g is summable, so it | g | by (i). Also, because | f |≤| g |, each function is in M+(X,S)and so

∫| f |+ dµ ≤

∫| g |+ dµ which is finite. Hence, | f | is summable. Then, also by (i), f is

summable.

We can now tackle the question of the linear structure of L1(X,S, µ).

Theorem 9.6.2 The Summable Function Form A Linear Space


Let (X,S, µ) be a measure space. We define operations on L1(X,S, µ) as follows:

• scalar multiplication: for all α in <, αf is the function defined pointwise by(αf)(x) = αf(x).

• addition of functions: for any two functions f and g the sum of f and g is the newfunction defined pointwise on ECfg by (f + g)(x) = f(x) + g(x), where, recall,

Efg =

((f =∞) ∩ (g = −∞)

⋃(f = −∞) ∩ (g =∞)

).

This is equivalent to defining f + g to be the function h where

h = (f + g)IECfg

This is a measurable function as we discussed in the proof of Lemma 8.7.1.

Then, we have

(i): αf is summable for all real α if f is summable and∫αfdµ = α

∫fdµ.

(ii): f + g is summable for all f and g which are summable and∫

(f + g)dµ =∫fdµ +∫

gdµ.

Hence, L1(X,S, µ) is a vector space over <.

Proof 9.6.2

(i): If α is 0, this is easy. Next, assume α > 0. Then, (αf)+ = αf+ and (αf)− = αf− and thesetwo functions are clearly summable since f+ and f− are. Thus, αf is summable. Then, we have∫

αf dµ =

∫(αf)+ dµ −

∫(αf)− dµ

= α

(∫f+ dµ −

∫f− dµ

= α

∫f dµ.

Finally, if α < 0, we have (αf)+ = −αf− and (αf)− = −αf+. Now simply repeat the previousarguments making a few obvious changes.(ii): Since f and g are summable, we know that µ(Efg = 0. Further, we know | f | and | g | aresummable. Since

| f + g | IECfg ≤(| f | + | g |

)IECfg ≤ | f | + | g |,

we see | f + g | IECfg is summable by Theorem 9.6.1, part (iii). Hence, (f + g) IECfg is summable

also. Now decompose f + g on ECfg as

f + g = (f+ + g+) − (f− + g−).

9.7. THE DCT 181

Then, note ∫ECfg

(f + g) dµ =

∫ECfg

(f+ + g+) dµ −∫

(f− + g−) dµ

=

∫ECfg

(f+ − f−) dµ +

∫(g+ − g−) dµ,

where we are permitted to manipulate the terms in the integrals above because all are finite in value.However, we can rewrite this as∫

ECfg

(f + g) dµ =

∫ECfg

f dµ +

∫g dµ.

Since we define the sum of f and g to be the function (f + g) IECfg , we see f + g is in L1(X,S, µ).

9.7 The Dominated Convergence Theorem

We can now complete this chapter by proving the important limit interchange called the LebesgueDominated Convergence Theorem.

Theorem 9.7.1 Lebesgue’s Dominated Convergence Theorem

Let (X,S, µ) be a measure space, (fn) be a sequence of functions in L1(X,S, µ) andf : X → < so that fn → f a.e. Further, assume there is a summable g so that | fn |≤ gfor all n. Then, suitably defined, f is also measurable and summable with limn

∫fn dµ =∫

f dµ.

Proof 9.7.1LetE be the set of points inX where the sequence does not converge. Then, by assumption, µ(E) = 0and

fnIEC → fIEC , and | fnIEC | ≤ gIEC .

Hence, fIEC is measurable and satisfies | fIEC |≤ gIEC . Therefore, since g is summable, we havethat fIEC is summable too.

We can write out our fundamental inequality as follows

−gIEC ≤ fn IEC ≤ gIEC . (α)

This implies that hn = fn IEC + gIEC is non negative and hence, we can apply Fatou’s lemmato find ∫

lim inf hn dµ ≤ lim inf

∫hn dµ.

However, we know

lim inf hn = lim inf

(fn IEC + gIEC

)= gIEC + lim inf fn IEC


= gIEC + f IEC ,

because fn converges pointwise to f on EC . It then follows that∫ (gIEC + f IEC

)dµ ≤ lim inf

∫ (fn IEC + gIEC

)dµ

=

∫gIEC dµ + lim inf

∫fn IEC dµ.

Since g is summable, we also know∫(g IEC + f IEC ) dµ =

∫(g IEC dµ +

∫f IEC ) dµ.

Using this identity, we have∫(g IEC dµ +

∫f IEC ) dµ. ≤

∫g IEC dµ + lim inf

∫fn IEC dµ.

The finiteness of the integral of the g term then allows cancellation so that we obtain the inequality∫f IEC dµ. ≤ lim inf

∫fn IEC dµ.

Since the integrals of f and fn are all zero on E, we have shown∫f dµ. ≤ lim inf

∫fn dµ.

We now show the reverse inequality holds. Using Equation α, we see zn = gIEC − fnIEC isalso non negative for all n. Applying Fatou’s Lemma, we find∫

lim inf zn dµ ≤ lim inf

∫zn dµ.

Then, we note

lim inf zn = lim inf

(−fn IEC + gIEC

)= gIEC + lim inf

(−fn IEC

)= gIEC − f IEC ,

because fn converges pointwise to f on EC . It then follows that∫ (gIEC − f IEC

)dµ ≤ lim inf

∫ (−fn IEC + gIEC

)dµ

=

∫gIEC dµ + lim inf

∫ (−fn IEC

)dµ.

Now,

lim inf

∫ (−fn IEC

)dµ = sup

minfm≥n

∫ (−fn IEC

)dµ

9.8. HOMEWORK 183

= supm

(−∑m≥n

∫fn IEC dµ

)= − inf

msupm≥n

∫fn IEC dµ

= − lim sup

∫fn IEC dµ.

Thus, we can conclude∫ (gIEC − f IEC

)dµ ≤

∫gIEC dµ − lim sup

∫fn IEC dµ.

Again, since g is summable, we can write∫gIEC dµ −

∫f IEC dµ ≤

∫gIEC dµ − lim sup

∫fn IEC dµ.

After canceling the finite value∫gIEC dµ, we have∫

f IEC dµ ≥ lim sup

∫fn IEC dµ.

This then implies, using arguments similar to the ones used in the first case, that∫f dµ ≥ lim sup

∫fn dµ.

However, limit inferiors are always less than limit superiors and so we have

lim sup

∫fn dµ ≤

∫f dµ ≤ lim inf

∫fn dµ ≤ lim sup

∫fn dµ.

It follows immediately that limn

∫fndµ =

∫fdµ.

Finally, we can now see how to define f in a suitable fashion. The function fIEC is measurableand is 0 on E. Hence, the limit function f can has the form

f(x) =

limn fn(x) when the limit exists, i.e. when x ∈ EC0 when the limit does not exist, i.e. when x ∈ E.

9.8 Homework

Exercise 9.8.1 Assume f ∈ L1(X,S, µ) with f(x) > 0 on X . Further, assume there is a positivenumber α so that α < µ(X) <∞. Prove that

inf ∫E

f dµ | µ(E) ≥ α > 0.

Exercise 9.8.2 Assume f ∈ L1(X,S, µ). Let α > 0. Prove that

µ(x | | f(x) |≥ α)


is finite.

Exercise 9.8.3 Assume (fn) ⊆ L1(X,S, µ). Let f : X → < be a function. Assume fn →f [ptws ae]. Prove ∫

| fn − f | dµ → 0 ⇒∫| fn | dµ →

∫| f | dµ

Exercise 9.8.4 Let (X,S) be a measurable space. Let C be the collection of all charges on S . Provethat C is a Banach Space under the operations(

c µ

)(E) = c µ(E), ∀ c ∈ <, ∀ µ(

µ + ν

)(E) = µ(E) + ν(E), ∀ µ, ν,

with norm || µ ||= |µ|(X)

9.9 Alternative Abstract Integration SchemesIt is possible to define abstract integration in other ways. Let (X,S, µ) be a measure space withµ(X) finite. If f : X → < is bounded and S measurable, let m denote the lower bound of f on Xand M , its upper bound. If [a, b] is a finite interval in < (a < b, of course), let pi denote a partitionof [a, b] as defined in Definition 3.1.1. For convenience of exposition, we will let pi be the set ofpoints a = y0 < y1 < · · · < yn = b. In our discussion of Riemann integration and functionsof bounded variation, we used the variable x because we often think of the symbol x as a domainvariable; here, we use the variable y because it is often used as a range variable in many settings. Theimportant thing to remember is we are partitioning the range of f now, rather than the its domain.We define the norm of pi as usual following Definition 3.1.4. Further, we define refinements andcommon refinements of a partition as in Definition 3.1.1.

Now, letM0 > M be chosen. Let π be any partition of [m,M0] and label its points asm = y0 <y1 < · · · < yn = M0. Define the following measurable subsets of X:

Ej = x : yj−1 ≤ f(x) < yj, 1 ≤ j ≤ n.

It is clear the sets Ej are disjoint and X = ∪nj=1 Ej . Thus,

µ(X) =

n∑j=1

µ(Ej).

Define Lower and Upper sums as follows: the Lower sum is

L(f,π) =∑π

yj−1 µ(Ej)

and the upper sum is

U(f,π) =∑π

yj µ(Ej).

9.9. ALTERNATE INTEGRATION 185

It is clear

mµ(X) ≤ L(f,π) ≤ U(f,π) ≤ M0µ(X). (9.2)

It is clear then that

supπ

L(f,π) < ∞

infπU(f,π) < ∞.

We can thus define abstract Lower and Upper Darboux integrability as usual (see Theorem 4.2.3)for our choice of M0.

Theorem 9.9.1 The Upper And Lower Abstract Darboux Integrals Are Finite

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Then the Lower Integral L(f,M0) and Upper Integral U(f,M0) definedby

L(f,M0) = supπ

L(f,π)

U(f,M0) = infπU(f,π).

are both finite.

Proof 9.9.1This is a consequence of Equation 9.2.

We can then define our new abstract integral by

Definition 9.9.1 Abstract Darboux Integrability

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. We say f is Abstract Darboux Integrable if L(f,M0) = U(f,M0). Thecommon value is then called the Abstract Darboux Integral of f and is denoted by thesymbol DAI(f,M0).

We then prove the following results.

Theorem 9.9.2 π′ refines π Implies L(f,π) ≤ L(f,π′) and U(f,π) ≥ U(f,π′)

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded andS measurable. If π′ refines π on [m,M0], then L(f,π) ≤ L(f,π′) and U(f,π) ≥U(f,π′).

Proof 9.9.2Mimic the proof of Theorem 4.2.1, mutatis mutandi.

Theorem 9.9.3 L(f,π1) ≤ U(f,π2)


Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Let π1 and π2 be any two partitions of [m,M0]. Then L(f,π1) ≤ U(f,π2).

Proof 9.9.3See the proof of Theorem 4.2.2

This implies

Theorem 9.9.4 L(f,M0) ≤ U(f,M0)

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Then L(f,M0) ≤ U(f,M0).

From Equation 9.2 and Theorem 9.9.3, it immediately follows that for any partition π that

0 ≤ S(f, π) − L(f, π) ≤ ‖π‖µ(X).

We can use this inequality to prove

Theorem 9.9.5 U(f,M0) ≤ L(f,M0)

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Then U(f,M0) ≤ L(f,M0).

We conclude that the Abstract Darboux Integral of f exists and DAI(f,M0) has the common valueL(f,M0) = U(f,M0).We can do a similar analysis for partitions of the form [m0,M0] for any m0 < m. We canprove that the resulting abstract Darboux integrals always exist: hence, we have finite numbersDAI(f,m0,M0) for many choices of m0 and M0. Clearly, our development of this abstract in-tegral is without much application if these numbers depend on our choice of m0 and M0. This is notthe case. We can prove

Theorem 9.9.6 DAI(f,m0,M0) is independent of the choice of m0 and M0

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Then DAI(f,m0,M0) is independent of the choice of m0 and M0

Proof 9.9.6It is enough to do the argument for the upper bound side, M0 as the arguments for the m0 case aresimilar. Here is a sketch of the argument. LetM < M0 < M ′0 be given. Show that each partition of[m,M0] corresponds to a partition of [m,M ′0] for which the lower and upper sums for the partitionof [m,M0] are the same as the ones for the partition of [m,M ′0]. Then, with that done, show thisimplies the desired result.

With Theorem 9.9.6 established, we can choose any value of m0 and M0 useful in our calculations.This can be of great help in some arguments. Since the value of the abstract Darboux integral is nowwell-defined, we will begin using the standard notation,

∫Xf dµ or simply

∫f dµ for this value.

Also, sometimes, we will continue to refer to this value as DAI(f) where we no longer need addthe argument m0 or M0. Note this is the same symbol we use for the other abstract integral we havediscussed. We also define the symbol

∫Ef dµ = DAI(f,E) as usual for any measurable set E.

You can then prove the following theorems.


Theorem 9.9.7 Abstract Darboux Integral Lower and Upper Bounds

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Then

mµ(X) ≤ DAI(f) ≤ M µ(X).

Theorem 9.9.8 DAI(f) = 0 if µ(X) = 0

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Then DAI(f) = 0 if µ(X) = 0.

Theorem 9.9.9 DAI(f(x) ≡ c) is cµ(X)

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded, S mea-surable and have constant value c. Then DAI(f) = cµ(X).

Theorem 9.9.10 Abstract Darboux Integral Measures

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded, S measur-able and non negative. Then λ : S → < defined by λ(E) = DAI(f,E) defines a measureon S.

Proof 9.9.10We will provide a sketch of the proof. First, note our argument for a non negative f can be used toshow a finite valued summable f can be used to create a charge. We will not show that argumenthere, but it should be straightforward for you to do.

• First, prove λ(A∪B) = λ(A) +λ(B) when A and B are disjoint and measurable. Let π be apartition consisting of points (yj) and corresponding sets (Fj). Then for some set D and thefunction fID, each set Fj has the form

Fj = (yj−1 ≤ f(x)ID(x) < yj)

.

This is the same as

F1 = (0 = y0 ≤ f(x) < y1) ∩D ∪ [0, y1) ∩DC

and for j larger than 1,

Fj = (0 = y0 ≤ f(x) < y1) ∩D

.

If we let Ej = (yj−1 ≤ f(x) < yj), we see Fj = Ej ∩ D if j > 1 and F1 = E0 ∩ D ∪[0, y1) ∩DC .

Hence, since the first term vanishes, we find the lower sum is

∑j>1

yj−1 µ(Ej ∩D)


.

Hence, if we apply this idea to the set D = A ∪B for A and B disjoint, we find

L(fIA∪B , π) = L(fIA, π) + L(fIB , π)

which leads to the inequality λ(A∪B) ≤ λ(A)+λ(B). A similar argument using upper sumsgives us the other inequality which proves this first result. We will the details for you in anexercise below.

• It then follows immediately that the result above holds for finite unions of disjoint measurablesets via a standard induction argument. Again, the details are for you.

• Now let (An) be a countable union of disjoint measurable sets. Let A be the full countableunion, VN be the union of the first N sets and RN be X \ VN . We can then apply what weknow about finite unions to find∫

A

f dµ =

∫VN

f dµ+

∫RN

f dµ

which further expands to

∫A

f dµ =

N∑i=1

∫Ai

f dµ+

∫RN

f dµ

.

It is then not hard to show∫RN

f dµ→ 0, which proves countable additivity.

Theorem 9.9.11 Abstract Darboux Integral Is Monotone

Let (X,S, µ) be a measure space with µ(X) finite and f, g : X → < be bounded and Smeasurable. Then if f ≤ g, DAI(f) ≤ DAI(g).

Proof 9.9.11It is easy to see that g determines a charge. Hence, for any partition π with points (yj) and associatedsets (Ej), we have ∫

g dµ =∑j

∫Ej

g dµ

.On each Ej , g dominates f implying

∫g dµ dominates the lower sum for this partition. The result

then follows.

Theorem 9.9.12 Abstract Darboux Integral Is Additive

Let (X,S, µ) be a measure space with µ(X) finite and f, g : X → < be bounded and Smeasurable. Then DAI(f + g) = DAI(f) +DAI(g).

Proof 9.9.12We prove additivity in two steps.


• We prove for the case that g is a constant c on X . Let π be a partition with points (yj)and associated sets (Ej). Then, it is easy to see yj−1 ≤ f(x) < yj is the same set asyj−1 + c ≤ f(x) + c < yj + c. Hence, the points (yj + c and the sets (Ej) give a partitionπ′ for f + g. Thus, the lower sum is

L(f + g, π′) =∑j

(yj−1 + c)µ(Ej)

which can be broken into two sums giving

L(f + g, π′) =∑j

yj−1 µ(Ej) +∑j

c µ(Ej)

.

The first sum on the right is L(f, π) and the second is cµ(X) which is also∫g dµ. Hence, we

have shown additivity for this case.

• We now handle the case of an arbitrary g. Since f + g defines a charge, for any partition forf with points (yj) and associated sets (Ej), we have

∫(f + g) =

∑j

∫Ej

(f + g) dµ

.

However, on Ej , we know f(x) + g(x) ≥ yj−1 + g(x). Hence,∫Ej

(f + g) dµ ≥∫Ej

(yj−1 +

g)dµ. The additivity result holds for one of the functions a constant and so∫Ej

(yj−1 +g)dµ =

yj−1 µ(Ej) +∫Ej

g dµ. We conclude

∫(f + g) ≥ L(f, π) +

∫Ej

g dµ

.

Since the partition π is arbitrary, this shows∫

(f + g) ≥∫f dµ +

∫Ej

g dµ. A similarargument using upper sums completes this proof.

Theorem 9.9.13 Abstract Darboux Integral Is Scalable

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Then for all real numbers c, DAI(c f) = c DAI(f).

Proof 9.9.13The case c = 0 is easy. First prove the case c > 0 using a partition argument. Then, if c < 0, wecan write cf as |c|(−f) and write immediately

∫cfdµ = |c|

∫(−f)dµ. However, additivity tells us∫

(−f)dµ = −∫fdµ which completes the argument.

Theorem 9.9.14 Abstract Darboux Integral Absolute Inequality


Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded and Smeasurable. Then ∣∣∣∣DAI(f)

∣∣∣∣ ≤ DAI(|f |).

Theorem 9.9.15 Abstract Darboux Integral Zero Implies f = 0 a.e.

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded, S mea-surable and non negative. Then DAI(f) = 0 implies f = 0 a.e.

Finally, it is easy to see

Theorem 9.9.16 Abstract Darboux Integral Measures Are Absolutely Continuous

Let (X,S, µ) be a measure space with µ(X) finite and f : X → < be bounded, S mea-surable and non negative. Then λ : S → < defined by λ(E) = DAI(f,E) defines anabsolutely continuous measure on S .

9.9.1 HomeworkExercise 9.9.1 Prove Theorem 9.9.2.

Exercise 9.9.2 Prove Theorem 9.9.3.













Chapter 10

The Lp Spaces

In mathematics and other fields, we often group objects of interest into sets and study the propertiesof these sets. In this book, we have been studying a set X with a sigma - algebra of subsets con-tained within it, the collection of functions which are measurable with respect to the sigma - algebraand recently, the set of functions which are summable. In addition, we have noted that the sets ofmeasurable and summable functions are closed under scalar multiplication and addition as long aswe interpret addition in the right way when the functions are extended real - valued.

We can do more along these lines. We will now study the sets of summable functions as vectorspaces with a suitable norm. We begin with a review.

Definition 10.0.2 The Norm On A Vector Space

Let X be a non empty vector space over <. We say ρ : X → < is a norm on X if

(N1): ρ(x) is non negative for all x in X ,

(N2): ρ(x) = 0 ⇔ x = 0,

(N3): ρ(αx) = |α|ρ(x), for all α in < and for all x in X ,

(N4): ρ(x + y) ≤ ρ(x) + ρ(y), for all x and y in X .

If ρ satisfies only N1, N3 and N4, we say ρ is a semi-norm or pseudo-norm. We will usuallydenote a norm of x by the symbol || x ||.The pair (X, ‖ ‖) is called a Normed Linear Space or NLS.

If a set X has no linear structure, we can still have a notion of the distance between objects in theset, if the set is endowed with a metric. This is defined below.

Definition 10.0.3 The Metric On A Set

191

192 CHAPTER 10. THE LP SPACES

Let X be a non empty set. We say d : X ×X → < is a metric if

(M1): d(x, y) is non negative for all x and y in X ,

(M2): d(x, y) = 0 ⇔ x = y,

(M3): d(x, y) = d(y, x), for all for all x and y in X ,

(M4): d(x, y) ≤ d(x, z) + d(y, z), for all x, y and z in X .

If d satisfies only M1, M2 and M4, we say d is a semi-metric or pseudo-metric. The pair(X, d) is called a metric space. Note that in a metric space, there is no notion of scalingor adding objects because there is no linear structure.

Comment 10.0.1 It is a standard result from a linear analysis course, that the norm in a NLS (X, ‖‖)induces a metric on X by defining

d(x, y) = || x− y ||, ∀ x, y ∈ X.

Given a sequence (xn) in a NLS (X, ‖ ‖), we can define what we mean by the convergence ofthis sequence to another object x in X .

Definition 10.0.4 Norm Convergence

Let (X, ‖ ‖) be a non empty NLS. Let (xn) be a sequence in X . We say the sequence (xn)converges to x in X if

∀ ε > 0, ∃N 3 n > N ⇒ || xn − x ||< ε.

Now let (X,S, µ) be a nonempty measurable space. We are now ready to discuss the spaceL1(X,S, µ). By Theorem 9.6.2, we know that this space is a vector space with suitably definedaddition. We can now define a semi-norm for this space.

Theorem 10.0.17 The L1 Semi-norm

Let (X,S, µ) be a nonempty measurable space. Define || x ||1 on L1(X,S, µ) by

|| f ||1 =

∫|f | dµ, ∀ f ∈ L1(X,S, µ).

Then, || x ||1 is a semi-norm. Moreover, property N3 is almost satisfied: instead of N3, wehave

|| f ||1 = 0 ⇔ f = a.e.

Proof 10.0.14

(N1): || f ||1 is clearly non negative.(N2): This proof is an easy calculation.

|| αf ||1 =

∫|α f | dµ =

∫|α| |f | dµ

= |α|∫|f | dµ = |α| || f ||1 .

193

(N4): To prove this, we start with the triangle inequality for real numbers. We know that if f and gare summable, then the sum of f + g is defined to be h = (f + g)IECfg . Let A be the set of pointswhere this sum is∞ and B be the set where the sum is −∞. Then µ(Efg ∪ A ∪ B) = 0 and on(Efg ∪ A ∪ B)C , h is finite. For convenience of exposition, we will simply write h as f + g fromnow on. So f + g is finite off a set of measure 0. At the points where f + g is finite, we can apply thestandard triangle inequality to f(x) + g(x). We have

|f(x) + g(x)| ≤ |f(x)| + |g(x)|, a.e.

This implies ∫|f + g| dµ ≤

∫|f | dµ +

∫|g| dµ.

At the risk of repeating ourselves too much, let’s go through the integral on the left hand side again.We actually have ∫

|f + g| IECfg ∩ AC ∩ BC dµ =

∫h IAC ∩ BC dµ

=

∫h dµ

since µ(AC ∩ BC) = 0. Now the above inequality estimates clearly tell us

|| f + g ||1 ≤ || f ||1 + || g ||1 .

Finally, we look at what is happening in condition N2. Since |f | is inM+(X,S, µ), by Lemma 9.4.6,we know

|f | = 0 a.e. ⇔∫|f | dµ = 0.

Hence, || f ||1= 0 if and only if f = 0 a.e.

Although || x ||1 is only a semi-norm, there is a way to think of this class of functions as a normedlinear space. Let’s define two functions f and g in L1(X,S, µ) to be equivalent or to be precise, µ- equivalent if f = g except of a set of µ measure 0. We use the notation f ∼ g to indicate thisequivalence. It is easy to see that ∼ defines an equivalence relation on L1(X,S, µ). We will let [f ]denote the equivalence class defined by f :

[f ] = g ∈ L1(X,S, µ) | g ∼ f.

Any g in [f ] is called a representative of the equivalence class [f ]. A straightforward argument showsthat two equivalence classes [f1] and [f2] are either equal as sets or disjoint. The collection of alldistinct equivalence classes of L1(X,S, µ) under a.e. equivalence will be denoted by L1(X,S, µ).

Theorem 10.0.18 L1 Is A Normed Linear Space


L1(X,S, µ) is a vector space over< with scalar multiplication and object addition definedas

α [f ] = [α f ], ∀ [f ]

[f ] + [g] = [f + g], ∀[f ] and [g].

Further, || [f ] ||1 defined by

|| [f ] ||1 =

∫|g| dµ,

for any representative g of [f ], is a norm on L1(X,S, µ).

Proof 10.0.15The definition of scalar multiplication is clear. However, as usual, we can spend some time withaddition. We know f + g is defined on ECfg and that Efg has measure 0. Hence, if u ∈ [f ] andv ∈ [g], then u = f and v = g except on sets A and B of measure 0. Also, as usual, the sum u + vis defined on ECuv . Hence,

u + v = f + g, x ∈ ECuv ∩ ECfg ∩ AC ∩ BC .

which is the complement of a set of measure 0. Hence, u+ v ∈ [f + g]. Thus, [f ] + [g] ⊆ [f + g].Conversely, let h ∈ [f + g]. Now

(f + g) IEfgC = f IEfgC + g IEfgC .

Hence, if we letu = f IEfgC and v = g IEfgC ,

we see h ∼ (u+v), with u ∈ [f ] and v ∈ [g]. We conclude [f + g] ⊆ [f ]+ [g]. Hence, the additionof equivalence classes makes sense.

We now turn our attention to the possible norm || [f ] ||1. First, we must show that our definitionof norm is independent of the choice of representative chosen from [f ]. If g ∼ f , then g = f excepton a set A of measure 0. Thus, we know the integral of f and g match by Lemma 9.4.6. Here are thedetails: ∫

|g| dµ =

∫A

|g| dµ +

∫AC|g| dµ

= 0 +

∫AC|f | dµ

=

∫A

|f | dµ +

∫AC|f | dµ

=

∫|f | dµ.

We conclude the value of || [f ] ||1 is independent of the choice of representative from [f ]. Now weprove this is a norm.(N1): || [f ] ||1=

∫|g|dµ ≥ 0.

(N2): If || [f ] ||1= 0, then for any representative g of [f ], we have∫|g|dµ = 0. By Lemma 9.4.6, this

implies that g = 0 a.e. and hence, g ∈ [0] (we abuse notation here by simply writing the zero functionh(x) = 0, ∀x as 0 ). But since g ∈ [f ] also, this means [f ] ∩ [0] is nonempty. This immediatelyimplies that [f ] = [0]. Conversely, if [f ] = [0], the result is clear.

10.1. THE GENERAL LP SPACES 195

(N3): Let α be a real number. Then, if g is any representative of [f ], we have α g is a representativeof [α f ]. We find

|| [α f ] ||1 =

∫|α g| dµ = |α|

∫|g| dµ

= |α| || [f ] || .

(N4): The triangle inequality follows from the triangle inequality that holds for the representatives.

10.1 The General Lp spaces

We can construct additional spaces of summable functions. Let p be a real number satisfying 1 ≤ p <∞. Then the function φ(u) = up is a continuous function on [0,∞) that satisfies limn φ(n) = ∞.Thus, by Lemma 8.8.3, if f is an extended real - valued function on X , then the composition φ |f |or |f |p is also measurable. Hence, we know the integral

∫|f |p dµ exists as an extended real - valued

number. The class of measurable functions that satisfy∫|f |p dµ <∞ is another interesting class of

functions.We begin with some definitions.

Definition 10.1.1 The Space Of p Summable Functions

(X,S, µ) be a nonempty measurable space. Let p be a real number satisfying 1 ≤ p < ∞.Then, |f |p is a measurable function. We let

Lp(X,S, µ) = f ∈M(X,S, µ) |∫|f |p dµ <∞.

For later use, we will also define what are called conjugate index pairs.

Definition 10.1.2 Conjugate Index Pairs

Let p be a real number satisfying 1 ≤ p ≤ ∞. If 1 < p is finite, the index conjugate to pis the real number q satisfying

1

p+

1

q= 1,

while if p = 1, the index conjugate to p is q =∞.

We will be able to show that Lp(X,S, µ) is a vector space under the usual scalar multiplica-tion and addition operations once we prove some auxiliary results. These are the Holder’s andMinkowski’s Inequality. First, there is a standard lemma we will call the Real Number ConjugateIndices Inequality.

Lemma 10.1.1 Real Number Conjugate Indices Inequality

Let 1 < p <∞ and q be the corresponding conjugate index. Then if α and β are positivenumbers,

AB ≤ Ap

p+

Bq

q.


Proof 10.1.1This proof is standard in any Linear Analysis book and so we will not repeat it here.

Theorem 10.1.2 Holder’s Inequality

Let 1 < p < ∞ and q be the index conjugate to p. Let f ∈ Lp(X,S, µ) and g ∈Lq(X,S, µ). Then f g ∈ L1(X,S, µ) and∫

|fg| dµ ≤(∫

|f |p dµ)1/p (∫

|g|q dµ)1/q

.

Proof 10.1.2The result is clearly true if f = g = 0 a.e. Also, if

∫|f |pdµ = 0, then |f |p = 0 a.e. which tells us

f = 0 a.e. and the result follows again. We handle the case where∫|g|qdµ = 0 in a similar fashion.

Thus, we will assume both Ip =∫|f |pdµ > 0 and Jq =

∫|g|qdµ > 0.

Let Ef and Eg be the sets where f and g are not finite. By our assumption, we know the measureof these sets is 0. Hence, for all x in ECf ∩ECg , the values f(x) and g(x) are finite. We apply Lemma10.1.1 to conclude

|f(x)|I

|g(x)|J

≤ (1/p)|f(x)|p

Ip+ (1/q)

|g(x)|q

Jq.

holds on ECf ∩ECg . Off of this set, we have that the left hand side is∞ and so is the right hand side.Hence, even on Ef ∪ Eg , the inequality is satisfied. Thus, since the function on the right hand sideis summable, we must have the left hand side is a summable function too by Theorem 9.6.1. Hence,f g ∈ L1(X,S, µ). We then have∫

|f(x)|I

|g(x)|J

dµ ≤∫

(1/p)|f(x)|p

Ipdµ +

∫(1/q)

|g(x)|q

Jqdµ

=1

pIp

∫|f(x)|p dµ +

1

qJq

∫|g(x)|q dµ

=1

p+

1

q= 1.

Thus, ∫|f g| dµ ≤ I J =

(∫|f |p dµ

)1/p (∫|g|q dµ

)1/p

.

The special case of p = q = 2 is of great interest. The resulting Holder’s Inequality is oftencalled the Cauchy - Schwartz Inequality. We see

Theorem 10.1.3 Cauchy - Bunyakovskii - Schwartz Inequality

Let f, g ∈ L2(X,S, µ). Then f g ∈ L1(X,S, µ) and∫|fg| dµ ≤

(∫|f |2 dµ

)1/2 (∫|g|2 dµ

)1/2

.

Theorem 10.1.4 Minkowski’s Inequality


Let 1 ≤ p <∞ and let f, g ∈ Lp(X,S, µ). Then f + g is in Lp(X,S, µ) and(∫|f + g|p dµ

)1/p

≤(∫

|f |p dµ)1/p

+

(∫|g|p dµ

)1/p

.

Proof 10.1.4If p = 1, this is property N4 of the semi-norm || · ||1. Thus, we can assume 1 < p <∞. Since f andg are measurable, we define the sum of f + g as h = (f + g) IA where A = ECfg with µ(Efg = 0.Then as discussed h is measurable. We see on A,

|f(x) + g(x)| ≤ |f(x)| + |g(x)| ≤ 2 max|f(x)|, |g(x)|

even when function values are∞. Hence,

|f(x) + g(x)|p ≤ 2p(

max|f(x)|, |g(x)|)p≤ 2p

(|f(x)|p + |g(x)|p

).

Then, since the right hand side is summable, so is the left hand side. We conclude f + g is inLp(X,S, µ). Further,

|f(x) + g(x)|p = |f + g| |f + g|p−1 ≤ |f | |f + g|p−1 + |g| |f + g|p−1.

We have the identity

|f(x) + g(x)|p ≤ |f | |f + g|p−1 + |g| |f + g|p−1. (∗)

Now since p and q are conjugate indices, we know

(1/p) + (1/q) = 1 ⇒ p + q = pq

⇒ p = q(p− 1).

Thus, the function (∣∣∣∣f + g

∣∣∣∣p−1)q

= |f + g|p,

and so this function is summable implying |f+g|p−1 ∈ Lq(X,S, µ). Now apply Holder’s Inequalityto the two parts of the right hand side of Equation ∗. We find

∫|f | |f + g|p−1 dµ ≤

(∫|f |p dµ

)1/p (∫ (|f + g|p−1

)qdµ

)1/q

.

and ∫|g| |f + g|p−1 dµ ≤

(∫|g|p dµ

)1/p (∫ (|f + g|p−1

)qdµ

)1/q

.

But we have learned we can rewrite the second terms of the above inequalities to get

∫|f | |f + g|p−1 dµ ≤

(∫|f |p dµ

)1/p (∫ (|f + g|p dµ

)1/q

.


and ∫|g| |f + g|p−1 dµ ≤

(∫|g|p dµ

)1/p (∫ (|f + g|p dµ

)1/q

.

Thus, combining, we have

∫|f + g|p dµ ≤

((∫|f |p dµ

)1/p

+

(∫|g|p dµ

)1/p)(∫ (

|f + g|p dµ

)1/q

.

We can rewrite this as(∫|f + g|p dµ

)1−1/q

≤(∫

|f |p dµ)1/p

+

(∫|g|p dµ

)1/p

.

Since 1− 1/q = 1/p, we have established the desired result.

Holder’s and Minkowski’s Inequalities allow us to prove that the Lp spaces are normed linearspaces.

Theorem 10.1.5 Lp Is A Vector Space

Let (X,S, µ) be a measure space and let 1 ≤ p < ∞. Then, if scalar multiplication andobject addition are defined pointwise as usual, Lp(X,S, µ) is a vector space.

Proof 10.1.5The only thing we must check is that if f and g are in Lp(X,S, µ), then so is f + g. This followsfrom Minkowski’s inequality.

Since Lp(X,S, µ) is a vector space, the next step is to find a norm for the space.

Theorem 10.1.6 The Lp Semi-Norm

Let (X,S, µ) be a measure space and let 1 ≤ p <∞. Define || · ||p on Lp(X,S, µ) by

|| f ||p =

(∫|f |p dµ

)1/p

.

Then, || · ||p is a semi-norm.

Proof 10.1.6Properties N1 and N3 of a norm are straightforward to prove. To see that the triangle inequalityholds, simply note that Minkowski’s Inequality can be rewritten as

|| f + g ||p ≤ || f ||p + || g ||p,

for arbitrary f and g in Lp(X,S, µ).

If we use the same notion of equivalence a.e. as did earlier, we can define the the collectionof all distinct equivalence classes of Lp(X,S, µ) under a.e. equivalence. This will be denoted byLp(X,S, µ). We can prove that this space is a normed linear space using the norm || [·] ||p.

Theorem 10.1.7 Lp Is A Normed Linear Space


Let 1 ≤ p <∞. Then Lp(X,S, µ) is a vector space over < with scalar multiplication andobject addition defined as

α [f ] = [α f ] ∀ [f ]

[f ] + [g] = [f + g],∀[f ] and [g].

Further, || [f ] ||p defined by

|| [f ] ||p =

(∫|g|p dµ

)1/p

,

for any representative g of [f ] is a norm on Lp(X,S, µ).

Proof 10.1.7The proof of this is quite similar to that of Theorem 10.0.18 and so we will not repeat it.

We will now show that Lp(X,S, µ) is a complete NLS. First, recall what a Cauchy Sequencemeans.

Definition 10.1.3 Cauchy Sequence In Norm

Let (X, || · ||) be a NLS. We say the sequence (fn) of X is a Cauchy Sequence, if givenε > 0, there is a positive integer N so that

|| fn − fm || < ε, ∀ n, m > N.

This leads to the definition of a complete NLS or Banach space.

Definition 10.1.4 Complete NLS

Let (X, || · ||) be a NLS. We say the X is a complete NLS if every Cauchy sequence in Xconverges to some object in X .

It is a standard proof to show that any sequence in a NLS that converges must be a Cauchysequence. Let’s prove that in the context of the Lp(X,S, µ) space to get some practice.

Theorem 10.1.8 Sequences That Converge in Lp Are Cauchy

Let ([fn]) be a sequence in Lp(X,S, µ) which converges to [f ] in Lp(X,S, µ) in the|| · ||p norm. Then, ([fn]) is a Cauchy sequence.

Proof 10.1.8Let ε > 0 be given. Then, there is a positive integer N so that if n > N , then

|| [fn − f ] ||p < ε/2.

Thus, if n and m are larger than N , we by property N4 that

|| [fn − fm] ||p = || [(fn − f) + (f − fm)] ||p ≤ || [fn − f ] ||p + || [fm − f ] ||p < ε.

This shows the sequence in a Cauchy sequence.

We will now show the Lp(X,S, µ) is a Banach space.


Theorem 10.1.9 Lp Is A Banach Space

Let 1 ≤ p <∞. Then Lp(X,S, µ) is complete with respect to the norm || · ||p.

Proof 10.1.9Let [fn] be a Cauchy sequence. These are the steps of the proof.

(Step 1): we find a subsequence ([gk]) so that for all k,∫|gk+1 − gk|p dµ < (1/2k)p (α)

(Step 2): Define the function g by

g(x) = |g1(x)| +

∞∑k=1

|gk+1(x)− gk(x)|. (β)

We show that g satisfies

|| g ||p ≤ || g1 ||p + 1. (γ)

This implies that g, defined by Equation β, converges and is finite a.e.

(Step 3): Then, we show

f(x) = g1(x) +

∞∑k=1

(gk+1(x)− gk(x)

)is defined a.e. and is in Lp(X,S, µ). This is our candidate for the convergence of the Cauchy se-quence.

(Step 4): We show gk converge to f in || · ||p.

(Step 5): We show [fn] converges to [f ] in || · ||p. This last step will complete the proof of complete-ness.

Now to the proof of these steps.

(Proof Step 1): For ε = (1/2), since [fn] is a Cauchy sequence, there is a positive integer N1 sothat n,m > N1 implies ∫

|fn − fm|p dµ < (1/2).

Note we use representative fn ∈ [fn] for simplicity of exposition since the norms are independent ofchoice of representatives. Define g1 = fN1+1.

Next, for ε = (1/2)2, there is a positive integer N2, which we can always choose so that N2 >N1, so that n,m > N2 implies ∫

|fn − fm|p dµ <

(1/(22)

)p.


Let g2 = fN2+1. Then, again by our choice of indices, we have∫|g2 − g1|p dµ < (1/2)p.

The next step is similar. For ε = (1/2)3, there is a positive integer N3, which we can alwayschoose so that N3 > N2, so that n,m > N3 implies∫

|fn − fm|p dµ <

(1/(23)

)p.

Let g3 = fN3+1. Then, we have ∫|g3 − g2|p dµ <

(1/(22)

)p.

An induction argument thus shows that there is a subsequence [gk] that satisfies∫|gk+1 − gk|p dµ <

(1/(2k)

)p.

for all k ≥ 1. This establishes Equation α.

(Proof Step 2): Define the non negative sequence (hn) by

hn(x) = |g1(x)| +

n∑k=1

|gk+1(x)− gk(x)|.

In this definition, there is the usual messiness of where all the differences are defined. Let’s clear thatup. Each pair (gk, gk+1) has a potential set Ek of measure zero where the subtraction is not defined.Thus, we need to throw away the set E = ∪k Ek which also has measure 0. Thus, it is clear thatall of the hn are defined on EC . Now they may take on the value∞, but that is acceptable. We seehpn ↑ gp on EC . Apply Fatou’s Lemma to (hn). We find∫ (

lim inf hpn IEC

)dµ ≤ lim inf

∫hpn IEC dµ.

But, lim inf hpn = gp and so ∫gp IEC dµ ≤ lim inf

∫hpn IEC dµ.

The pth root function is continuous and so

lim inf

(∫hpn IEC dµ

)1/p

=

(lim inf

∫hpn IEC dµ

)1/p

.

Then, since the pth root function is increasing, we have(∫gp IEC dµ

)1/p

≤ lim inf

(∫hpn IEC dµ

)1/p

.


Next, applying Minkowski’s Inequality to a finite sum, we obtain

(∫hpn IEC

)1/p

=

(∫ (|g1|+

n∑k=1

|gk+1 − gk|)p

IEC

)1/p

≤ || g1IEC ||p +

n∑k=1

|| (gk+1 − gk)IEC ||p .

Since the finite sum is monotonic increasing, we have immediately that the series

∞∑k=1

|| (gk+1 − gk)IEC ||p

is a well defined extended real - valued number. Thus, we have(∫hpn IEC

)1/p

≤ || g1IEC ||p +

∞∑k=1

|| (gk+1 − gk)IEC ||p .

By Equation α, we also know that

∞∑k=1

|| (gk+1 − gk)IEC ||p ≤∞∑k=1

1/(2)k = 1.

Hence, we can actually say (∫g IEC

)1/p

≤ || g1IEC ||p +1

We conclude g IEC is in Lp(X,S, µ). Next, note if F = x | g(x)IEC (x) = ∞, we know F hasmeasure 0. Hence, g IEC∩FC is finite. This completes Step 2.

(Proof Step 3): Now define the function f by

f(x) =

g1(x) +∑∞k=1

(gk+1(x)− gk(x)

), x ∈ EC ∩ FC

0 x ∈ E ∪ F.

Note, for x ∈ EC ∩ FC ,

|gk| = |g1 + (g2 − g1) + (g3 − g2) + . . . + (gk − gk−1|

≤ |g1| +

k∑i=1

|gk+1 − gk| = hk.

However, we already seen that on this set hk ↑ g. Hence, we can say

|gk| ≤ g.

This tells us that the partial sum expansion of gk converges absolutely on EC ∩ FC and thus, gkconverges to g. But g = f on this set, so we have shown that gk converges to f a.e. We can now


apply the Lebesgue Dominated Convergence Theorem to say

limn

∫gn dµ =

∫f dµ.

Since |gk| ≤ g for all k, it follows |f |p ≤ |g|p. Since g is p summable, we have established that f isin Lp(X,S, µ).

(Proof Step 4): Now we show gk converges to f in || · ||p. To see this, let zk = f − gk on EC ∩FC .From the definition of f , we can write this as

∑∞j=k (gj+1 − gj). The rest of the argument is very

similar to the one used in Step 2. Consider the partial sums of this convergent series

znk =

n∑j=k

|gj+1 − gj|.

Minkowski’s Inequality then gives for all n,

|| znk ||p ≤n∑j=k

|| gj+1 − gj ||p .

Using Equation α, it follows that the right hand side is bounded above by∑nj=k 1/2j which sums

to 1/2n−1. Now apply Fatou’s Lemma to find∫lim inf |znk |p ≤ lim inf

∫|znk |p

or ∫|zk|p ≤ lim inf

∫|znk |p.

The continuity and increasing nature of the pth root then give us(∫|zk|p

)1/p

≤ lim inf

(∫|znk |p

)1/p

≤ lim inf (1/2n−1) = 0.

Thus, || f − gk ||→ 0.

(Proof Step 5): Finally, given ε > 0, since [fn] is a Cauchy sequence, there is an N so that

|| fn − fm ||p< ε/2, ∀ n,m > N.

Since ([gk]) is a subsequence of ([fn]), there is a K1 so that if k > K1, we have

|| fm − gk ||p< ε/2, ∀m > N, k > K1.

Also, since gk → f in p - norm, there is a K2 so that

|| gk − f ||p< ε/2, ∀ k > K2.

We conclude for any given k > max(K1,K2), we have

|| fm − f ||p ≤ || fm − gk ||p + || gk − f ||p < ε

if m > N . Thus, [fn]→ [f ] in p - norm.


The proof of the theorem above has buried in it a powerful result. We state this below.

Theorem 10.1.10 Sequences That Converge In p - Norm Possess Subsequences ConvergingPointwise a.e.

Let 1 ≤ p < ∞. Let ([fn]) be a sequence in Lp(X,S, µ) which converges in norm to [f ]in Lp(X,S, µ). Then, there is a subsequence ([f1

n]) of ([fn]) which converges pointwisea.e. to f .

Proof 10.1.10The sequence we seek is the sequence (gn) as defined in the proof of Theorem 10.1.9; see the discus-sion for the proof of Step (3).

10.2 The World Of Counting Measure

Let’s see what the previous material means when we use counting measure, µC , on the set of naturalnumbers N. In this case, the sigma - algebra is the power set of N. Note if f : N → <, then f isidentified with a sequence of extended real - valued numbers, (an) so that f(n) = an. It is thereforepossible for f(n)∞ or f(n) = −∞ for some n. Let

φN (n) =

|f(n)|, 1 ≤ n ≤ N0, n > N

Then, φN ↑ f and so by the Monotone Convergence Theorem,∫|f | dµC = lim

N

∫φN (n) dµC .

Now the simple functions φN are not in their standard representation. Let c1, . . . , cM be thedistinct elements of |a1|, . . . , |aN |. Then we can write

φN =

M∑i=1

ci IEi ,

whereEI is the pre-image of each distinct element ci. The setsEi are clearly disjoint by construction.It is a straightforward matter to see that∫

φN dµC =

M∑i=1

ci µCEi =

N∑i=1

|ai|.

Thus, we have ∫|f | dµC = lim

N

N∑i=1

|f(i)|.

Since all the terms |f(i)| are non negative, we see the sequence of partial sums converges to someextended real - valued number (possibly∞). For counting measure, the only set of measure 0 is ∅,so measurable functions can not differ on a set of measure 0 in this case. We see for 1 ≤ p <∞,

Lp(N,P(N), µC) = Lp(N,P(N), µC).

10.3. ESSENTIALLY BOUNDED FUNCTIONS 205

Further,

Lp(N,P(N), µC) = sequences (an) |∞∑i=1

|ai|p converges .

We typically use the notation

`p = Lp(N,P(N), µC) = sequences (an) |∞∑i=1

|ai|p converges .

and we call this a sequence space. Note in all cases, summability implies the sequence involved mustbe finite everywhere.

In this context, Holder’s Inequality becomes:

Theorem 10.2.1 Holder’s Inequality: Sequence Spaces

Let 1 < p < ∞ and q be the index conjugate to p. Let (an) ∈ `p and (bn) ∈ ellq . Then(an bn) ∈ `1 and

∑n

|an bn| ≤(∑

n

|an|p)1/p (∑

n

|bn|q)1/q

.

and Minkowski’s Inequality becomes

Theorem 10.2.2 Minkowski’s Inequality: Sequence Spaces

Let 1 ≤ p <∞ and let (an), (bn) ∈ `p. Then (an + bn) is in `p and(∑n

|an + bn|p)1/p

≤(∑

n

|an|p)1/p

+

(∑n

|bn|p)1/p

.

Finally, the special case of p = q = 2 should be mentioned. The sequence space version of theresulting Holder’s Inequality Cauchy - Schwartz Inequality has this form:

Theorem 10.2.3 Cauchy - Bunyakovskii - Schwartz Inequality: Sequence Spaces

Let (an), (bn) ∈ `2. Then (an bn) ∈ `1 and

∑n

|an bn| ≤(∑

n

|an|2)1/2 (∑

n

|bn|2)1/2

.

10.3 Equivalence Classes of Essentially Bounded Functions

There is one more space to define. This will be the analogue of the space of bounded functions weuse in the definition of the Riemann Integral.

Definition 10.3.1 Essentially Bounded Functions


Let (X,S, µ) be a measure space and let f be measurable. If E is a set of measure 0, let

ξ(E) = supx∈EC

|f(x)|

andρ∞(f) = inf ξ(E) | E ∈ S, µ(E) = 0.

If ρ∞(f) is finite, we say f is an essentially bounded function.

There are then two more spaces to consider:

Definition 10.3.2 The Spaces of Essentially Bounded Functions

Let (X,S, µ) be a measure space. Then we define

L∞(X,S, µ) = f : X → < | f ∈M(X,S), ρ∞(f) <∞.

and defining equivalence classes using a.e. equivalence,

L∞(X,S, µ) = [f ] | ρ∞(f) <∞.

There is an equivalent way of characterizing an essentially bounded function. This requiresanother definition.

Theorem 10.3.1 Alternate Characterization Of Essentially Bounded Functions

Let (X,S, µ) be a measure space and f be a measurable function. Define q∞(f) by

q∞(f) = inf a | µ(x | |f(x)| > a) = 0.

Then, ρ∞(f) = q∞(f).

Proof 10.3.1Let Ea = x | |f(x)| > a. If a is a number so that µ(Ea) = 0, then for any other measurable set Awith measure 0, we have

AC = AC ∩ Ea ∪ AC ∩ ECa .

Thus,supAC|f | ≥ sup

AC∩Ea|f | ≥ a.

because if x ∈ AC ∩ Ea, then |f(x)| > a. Since we can do this for such a, it follows that

supAC|f | ≥ q∞(f).

Further, since the measurable set A with measure zero is arbitrary, we must have

ρ∞(f) ≥ q∞(f).

Next, we prove the reverse inequality. If µ(Ea) = 0, then by the definition of ρ∞(f), we have

ρ∞(f) ≤ supECa

|f | = sup|f(x)|≤a

|f(x)| ≤ a.


But this is true for all such a. Thus, ρ∞(f) is a lower bound for the set a|µ(Ea) = 0 and we cansay

ρ∞(f) ≤ q∞(f).

We need to know that if two functions are equivalent with respect to the measure µ, then theirρ∞ values agree.

Lemma 10.3.2 Essentially Bounded Functions That Are Equivalent Have The Same EssentialBound

Let (X,S, µ) be a measure space and f and g be equivalent measurable functions suchthat ρ(f) is finite. Then ρ(g) = ρ(f).

Proof 10.3.2Let E be the set of points where f and g are not equal. Then µ(E) = 0. Now,

0 ≤ µ

((|g(x)| > a) ∩ E

)≤ µ(E) = 0.

Thus,

µ

((|g(x)| > a)

)= µ

((|g(x)| > a) ∩ E

)+ µ

((|g(x)| > a) ∩ EC

)= µ

((|g(x)| > a) ∩ EC

).

But on EC , f and g match, so we have

µ

((|g(x)| > a)

)= µ

((|f(x)| > a) ∩ EC

)= µ

((|f(x)| > a)

),

by the same sort of argument we used on µ(

(|g(x)| > a)

). Hence, if µ

((|f(x)| > a)

)= 0, then

µ

((|g(x)| > a)

)= 0 as well. This immediately implies q∞(g) = q∞(f). The result then follows

because q∞ = ρ∞.

Finally, we can show that essentially bounded functions are bounded above by their essentialbound a.e.

Lemma 10.3.3 Essentially Bounded Functions Bounded Above By Their Essential Bound a.e

Let (X,S, µ) be a measure space and f be a measurable functions such that ρ(f) is finite.Then |f(x)| ≤ ρ(f) a.e.

Proof 10.3.3Let E = (|f(x) > ρ∞(f)). It is easy to see that

E =

∞⋃k=1

(|f(x)| > ρ∞(f) + 1/k

).


If you look at how q∞ is defined, if µ(|f(x)| > ρ∞(f) + 1/k) > 0, that would force q∞(f) =ρ∞(f) ≥ ρ∞(f) + 1/k which is not possible. Hence, µ(|f(x)| > ρ∞(f) + 1/k) = 0 for all k. Thismeans E has measure 0 also. It is then clear from the definition of the set E that |f(x)| ≤ ρ∞(f) onEC .

We can now prove that L∞(X,S, µ) is a vector space with norm || [f ] ||∞= ρ∞(f).

Theorem 10.3.4 The L∞ Semi-Norm

Let (X,S, µ) be a measure space Define || · ||∞ on L∞(X,S, µ) by

|| f ||∞ = ρ∞(g),

where g is any representative of [f ]. Then, || · ||∞ is a semi-norm.

Proof 10.3.4We show ρ∞(·) satisfies all the properties of a norm except N2 and hence it is a semi - norm.(N1): It is clear the N1 is satisfied because ρ∞(·) is always non negative.(N2): Let 0X is the function defined to be 0 for all x and let Ea = x | |0X(x)| > a. It is clearEa = ∅ for all a > 0. Thus, since ρ∞ = q∞,

q∞(0X) = inf a | µ(Ea) = 0 = 0.

However, if q∞(f) = 0, let Fn = (|fn(x)| > 1/n). Then, by definition of q∞(f), it follows thatµ(Fn) = 0 and |f(x)| ≤ 1/n on the complement FCn . Let F = ∪ Fn. Then, µ(F ) = 0 and

FC =⋂n

FCn =⋂n

(|f(x)| ≤ 1/n

)=

(f(x) = 0

).

Thus, f is 0 on FC and non zero on F which has measure 0. All that we can say then is that f = 0a.e. and hence, || · ||∞ does not satisfy N2.(N3): If α is 0, the result is clear. If α is a positive number, then

q∞(αf) = inf a | µ(x | |α f(x)| > a) = 0= inf a | µ(x | |f(x)| > a/α) = 0.

Let β = a/α and we have

q∞(αf) = inf α β | µ(x | |f(x)| > β) = 0= α inf β | µ(x | |f(x)| > β) = 0= α q∞(f).

If α is negative, simply write αf as |α| (−f) and apply the result for a positive α.(N4): Now let f and g be in L∞(X,S, µ) with the sum f + g defined in the usual way on ECfg withµ(Efg) = 0. Note on Efg itself, f(x) + g(x) = 0, so the sum is bounded above by ρ∞(f) + ρ∞(g)there. Now by Lemma 10.3.3, there are sets F and G of measure 0 so that

|f(x)| ≤ ρ∞(f), ∀x ∈ FC ,|g(x)| ≤ ρ∞(g), ∀x ∈ GC .

Thus,

|f(x) + g(x)| ≤ ρ∞(f) + ρ∞(g), ∀x ∈ FC ∩ GC ,


Thus, the measure of the set of points where |f(x)+g(x)| > ρ∞(f)+ρ∞(g) is zero as µ(F∪G) = 0.By definition of q∞, it then follows that

q∞(f + g) ≤ ρ∞(f) + ρ∞(g).

which implies the result.

Theorem 10.3.5 L∞ Is A Normed Linear Space

Then L∞(X,S, µ) is a vector space over < with scalar multiplication and object additiondefined as

α [f ] = [α f ] ∀ [f ]

[f ] + [g] = [f + g],∀[f ] and [g].

Further, || [f ] ||∞ defined by || [f ] ||∞= ρ∞(g), for any representative g of [f ] is a normon L∞(X,S, µ).

Proof 10.3.5The argument that the scalar multiplication and addition of equivalence classes is the same as theone we used in the proof of Theorem 10.1.5 and so we will not repeat it here. From Lemma 10.3.2we know that any two functions which are equivalent a.e. will have the same value for ρ∞ and so|| [f ] ||∞ is independent of the choice of representative from [f ]. The proofs that properties N1,N3 and N4 hold follow immediately from the fact that they hold for representatives of equivalenceclasses. It remains to show that if || [f ] ||∞= 0, then [f ] = [0X ] where 0X is the zero function onX . However, we have already established in the proof of Theorem 10.3.4 that such an f is 0 a.e. Thistells us f ∈ [0X ]; thus, [f ] = [0X ].

Theorem 10.3.6 L∞ Is A Banach Space

Then L∞(X,S, µ) is complete with respect to the norm || · ||∞.

Proof 10.3.6Let ([fn] be a Cauchy sequence of objects in L∞(X,S, µ). Now everything is independent of thechoice of representative of an equivalence class, so for convenience, we will use as our representa-tives the functions fn themselves. Then, by Lemma 10.3.3, there are sets En of measure 0 so that

|fn(x)| ≤ ρ∞(fn), ∀ x ∈ ECn .

Also, there are sets Fnm of measure 0 so that

|fn(x) − fm(x)| ≤ ρ∞(fn − fm), ∀ x ∈ FCnm.

Hence, both of the equations above hold on

U =

∞⋂m=1

∞⋂n=1

(ECn ∩ FCnm

).

We then use De Morgan’s Laws to rewrite U as follows:

U =

∞⋂m=1

∞⋂n=1

(En ∪ Fnm

)C


=

∞⋂m=1

( ∞⋃n=1

(En ∪ Fnm

))C

=

( ∞⋃m=1

∞⋃n=1

(En ∪ Fnm

))C.

Clearly, the measure of U is 0 and

|fn(x) − fm(x)| ≤ ρ∞(fn − fm), ∀ x ∈ UC . (α)

Now since ([fn] is a Cauchy sequence with respect to || · ||∞, given ε > 0, there is a positiveinteger N so that

|fn(x) − fm(x)| ≤ ρ∞(fn − fm) < ε/4, ∀ x ∈ UC , ∀ n, m > N. (β)

Equation β implies that at each x in UC , the sequence (fn(x)) is a Cauchy sequence of realnumbers. By the completeness of <, it then follows that limn fn(x) exists on UC . Define the functionf : X → < by

f(x) =

limn fn(x), x ∈ UC ,0 x ∈ U.

From Equation β, we have that

limn|fn(x) − fm(x)| ≤ ε/4, ∀ x ∈ UC ,∀m > N.

As usual, since the absolute value function is continuous, we can let the limit operation pass into theabsolute value function to obtain

|f(x) − fm(x)| ≤ ε/4, ∀ x ∈ UC , ∀m > N. (γ)

From the backwards triangle inequality, we then find

|f(x)| ≤ ε/4 + |fm(x)| < ε/4 + ρ∞(fm), ∀ x ∈ UC , ∀m > N.

Now fix M > N + 1. Then

|f(x)| < ε/2 + ρ∞(fM ), ∀ x ∈ UC .

Since the measure of the set (|f(x)| > ε/4 + ρ∞(fM )) is 0, from the definition of q∞(f), it thenfollows that

q∞(f) ≤ ε/4 + ρ∞(fM ).

We have thus shown f is essentially bounded and since f equals f a.e., we have f is in L∞(X,S, µ).It remains to show that [fn] converges to [f ] in norm. Note that Equation γ implies that (fn)converges uniformly on UC . Further, the measure of the set (|fn(x) − f(x)| > ε/4) is 0. Thus, wecan conclude

q∞(f − fm) ≤ ε/4 < ε, ∀m > N.

This shows the desired convergence in norm.Thus, we have shown L∞(X,S, µ) is complete.

From the proofs above, we see Minkowski’s Inequality holds for the case p =∞ because || · ||∞is a norm. Finally, we can complete the last case of Holder’s Inequality: the case of the conjugate

10.4. THE HILBERT SPACE L2 211

indices p = 1 and q =∞. We obtain

Theorem 10.3.7 Holder’s Inequality: p = 1

Let p = 1 and q = ∞ be the index conjugate to 1. Let [f ] ∈ L1(X,S, µ) and [g] ∈L∞(X,S, µ). Then [f g] ∈ L1(X,S, µ) and∫

|fg| dµ ≤ || [f ] ||1 || [g] ||∞.

Proof 10.3.7It is enough to prove this result for the representatives of the equivalence classes f ∈ [f ] and g ∈ [g].We know the product fg is measurable. It remains to show that fg is summable. Since g is essentiallybounded, by Lemma 10.3.3, there is a sets E of measure 0 so that

|g(x)| ≤ ρ∞(g), ∀ x ∈ EC .

Thus, |f(x)g(x)| ≤ |f(x)|ρ∞(g) a.e. and since the right hand side is summable, by Theorem 9.6.1,we see fg is also summable and∫

|f g| dµ ≤∫|f | ρ∞(g) dµ = ρ∞(g)

∫|f | dµ

10.4 The Hilbert Space L2

The space L2(X,S, µ) is a Normed linear space with norm || [·] ||2. This space is also an innerproduct space which is complete. Such a space is called a Hilbert space.

Definition 10.4.1 Inner Product Space

Let X be a non empty vector space over <. Let ω X × X → < be a mapping whichsatisfies

(IP1:) ω(x + y, z) = ω(x, z) + ω(y, x), ∀ x, y, z ∈ X,(IP2:) ω(α x, y) = α ω(x, y), ∀ α ∈ <, ∀ x, y ∈ X,

(IP3:) ω(x, y) = ω(y, x), ∀ x, y ∈ X,(IP4:) ω(x, x) ≥ 0, ∀ x, ∈ X, and ω(x, x) = 0 ⇔ x = 0.

Such a mapping is called an real inner product on the real vector space X . It is easyto define a similar mapping on complex vector spaces, but we will not do that here. Wetypically use the symbol < ·, · > to denote the value ω(·, ·).

There is much more we could say on this subject, but instead we will focus on how we can definean inner product on L2(X,S, µ).

Theorem 10.4.1 The Inner Product on The Space of Square Summable Equivalence Classes


For brevity, let L2 denote L2(X,S, µ). The mapping < ·, · > on L2 × L2 defined by

< [f ], [g] > =

∫u v dµ, ∀ u ∈ [f ], v ∈ [g]

is an inner product on L2. Moreover, it induces the norm || [·] ||2 by

|| [f ] ||2 =

√∫|f |2 dµ

=√< [f ], [f ] >.

Proof 10.4.1The proof of these assertions is immediate as we have already shown || · ||2 is a norm and theverification of properties IP1 to IP4 is straightforward.

Finally, from our general Lp results, we know L2 is complete. However, for the record, we statethis as a theorem.

Theorem 10.4.2 The Space of Square Summable Equivalence Classes Is A Hilbert Space

For brevity, let L2 denote L2(X,S, µ). Then L2 is complete with respect to the norminduced by the inner product < [·], [·] >. The inner product space (L2, < ·, · >) is oftendenoted by the symbolH.

10.5 HomeworkExercise 10.5.1 Let (X,S, µ) be a measure space. Let f be in Lp(X,Sµ) for 1 ≤ p < ∞. LetE = x | |f(x)| 6= 0. Prove E is σ - finite.

Hint 10.5.1 Divide the indicated set into (1 ≤ |f(x)|) and ∪n (1/n ≤ |f(x)| < 1).

Exercise 10.5.2 Let (X,S, µ) be a finite measure space. If f is measurable, let

En = x | n− 1 ≤ |f(x)| < n

.Prove f is in L1(X,Sµ) if and only if

∑∞n=1 nµ(En) <∞.

More generally, prove f is in Lp(X,Sµ), 1 ≤ p <∞, if and only if∑∞n=1 n

pµ(En) <∞.

Hint 10.5.2 Note E1 has finite measure because X does.

Part V

Constructing Measures

213

Chapter 11

Constructing Measures

Although you now know quite a bit about measures, measurable functions, associated integrationand the like, you still do not have many concrete and truly interesting measures to work with. In thischapter, you will learn how to construct interesting measures using some simple procedures. A verygood reference for this material is (Bruckner et al. (1) 1997). Another good source is (Taylor (12)1985). We begin with a definition.

11.1 Measures From Outer MeasureDefinition 11.1.1 Outer Measure

Let X be a non empty set and let µ∗ be an extended real valued mapping defined on allsubsets of X that satisfies

(i): µ∗(∅) = 0.

(ii): If A and B are subsets of X with A ⊆ B, then µ∗(A) ≤ µ∗(B).

(iii): If (An) is a sequence of subsets of X , then µ∗( ∪nAn) ≤∑n µ∗(An).

Such a mapping is an outer measure on X and condition (iii) is called the countable sub-additivity (CSA) condition if the sets are disjoint.

Comment 11.1.1 Since ∅ ⊆ A for all A in X , condition (ii) tells us µ∗(∅) ≤ µ∗(A). Hence, bycondition (i), we have µ∗(A) ≥ 0 always. Thus, the outer measure is non negative.

The outer measure is defined on all the subsets of X . In Chapter 9, we defined the notion of ameasure on a σ - algebra of subsets of X . Look back at Definition 9.0.1 again. Recall, the mappingµ : S → < is a measure on S if

(i): µ(∅) = 0,


(iii): µ is countably additive on S; i.e. if (En) ⊆ S is a countable collection of disjoint sets, thenµ(∪nEn) =

∑n µ(En).

The third condition says the mapping µ is countably additive and hence, we label this condition ascondition (CA). The collection of all subsets of X is the largest σ - algebra of subsets of X , so toconvert the outer measure µ∗ into a measure, we have to convert the countable subadditivity condition

215

216 CHAPTER 11. BUILDING MEASURES

to countable additivity. This is not that easy to do! Now if T and E are any subsets of X , then weknow

T =

(T ∩ E

) ⋃(T ∩ EC

).

The outer measure µ∗ is subadditive on finite disjoint unions and so we always have

µ∗(T ) ≤ µ∗(T ∩ E

)+ µ∗

(T ∩ EC

).

To have equality, we need to have

µ∗(T ) ≥ µ∗(T ∩ E

)+ µ∗

(T ∩ EC

),

also. So, as a first set towards the countable additivity condition we need, why don’t we look at allsubsets E of X that satisfy the condition

µ∗(T ) ≥ µ∗(T ∩ E

)+ µ∗

(T ∩ EC

), ∀ T ⊆ X.

We don’t know how many such sets E there are at this point. But we certainly want finite additivityto hold. Therefore, it seems like a good place to start. This condition is called the CaratheodoryCondition.

Definition 11.1.2 Caratheodory Condition

Let µ∗ be an outer measure on the non empty set X . A subset E of X satisfies theCaratheodory Condition if for all subsets T ,

µ∗(T ) ≥ µ∗(T ∩ E

)+ µ∗

(T ∩ EC

).

Such a set E is called µ∗ measurable. The collection of all µ∗ measurable subsets of Xwill be denoted byM.

We will first prove that the collection of µ∗ measurable sets is an algebra of sets.

Definition 11.1.3 Algebra Of Sets

Let X be a non empty set and let A be a nonempty family of subsets of X . We say A is analgebra of sets if

(i): ∅ is in A.

(ii): If A and B are in A, so is A ∪B.

(iii): if A is in A, so is AC = X \A.

Theorem 11.1.1 The µ∗ Measurable Sets Form An Algebra

Let X be a non empty set, µ∗ an outer measure on X and M be the collection of µ∗

measurable subsets of X . ThenM is a algebra.

11.1. VIA OUTER MEASURE 217

Proof 11.1.1For the empty set,

µ∗(T ∩ ∅

)+ µ∗

(T ∩ ∅C

)= µ∗

(∅)

+ µ∗(T ∩ X

)= 0 + µ∗

(T

).

Hence ∅ satisfies the Caratheodory condition and so ∅ ∈ M.

Next, if A ∈ M, we note the Caratheodory condition is symmetric with respect to complementationand so AC ∈M also.

To showM is closed under countable unions, we will start with the union of just two sets and thenproceed by induction. Let E1 and E2 be inM. Let T be in X . Then, since E1 and E2 both satisfyCaratheodory’s condition, we know

µ∗(T ) = µ∗(T ∩ E1) + µ∗(T ∩ EC1 ) (a)

and

µ∗(T ) = µ∗(T ∩ E2) + µ∗(T ∩ EC2 ). (b)

In Equation b, let “T” be “T ∩ EC1 ”. This gives

µ∗(T ∩ EC1 ) = µ∗(T ∩ EC1 ∩ E2) + µ∗(T ∩ EC1 ∩ EC2 ). (c)

We also know that

T ∩ E1 = T ∩ (E1 ∪ E2) ∩ E1, T ∩ EC1 ∩ E2 = T ∩ (E1 ∪ E2) ∩ EC1 . (d)

Now replace the term “µ∗(T ∩ EC1 )” in Equation a by the one in Equation c. This gives

µ∗(T ) = µ∗(T ∩ E1) + µ∗(T ∩ EC1 ∩ E2) + µ∗(T ∩ EC1 ∩ EC2 ).

Next, replace the sets in the first two terms on the right side in the equation above by what is shownin Equation d. We obtain

µ∗(T ) = µ∗(T ∩ (E1 ∪ E2) ∩ E1) + µ∗(T ∩ (E1 ∪ E2) ∩ EC1 ) + µ∗(T ∩ EC1 ∩ EC2 ).

But E1 is inM and so

µ∗(T ∩ (E1 ∪ E2)) = µ∗(T ∩ (E1 ∪ E2) ∩ E1) + µ∗(T ∩ (E1 ∪ E2) ∩ EC1 ).

Using this identity, we then have

µ∗(T ) = µ∗(T ∩ (E1 ∪ E2)) + µ∗(T ∩ EC1 ∩ EC2 )

= µ∗(T ∩ (E1 ∪ E2)) + µ∗(T ∩ (E1 ∪ E2)C),

using De Morgan’s laws. Since the set T is arbitrary, we have shown E1 ∪ E2 is also inM.

Since,E1 andE2 are inM, we now knowEC1 ∪EC2 is inM too. But this set is the same asE1 ∩E2.Thus,M is closed under intersection.


It then follows that E1 \E2 = E1 ∩EC2 is inM. SoM is also closed under set differences. Hence,M is an algebra.

Theorem 11.1.2 µ∗ Measurable Sets Properties


measurable subsets of X . Then, if (En) is a countable disjoint sequence fromM, ∪nEnis inM and

µ∗(T ∩ ∪∞i=1 Ei) =

∞∑i=1

µ∗(T ∩ Ei)

).

for all T in X .

Proof 11.1.2Let “T” be “T ∩ (E1 ∪ E2) in the Caratheodory condition of E2. Then, we have

µ∗(T ∩ (E1 ∪ E2)) = µ∗(T ∩ (E1 ∪ E2) ∩ E2) + µ∗(T ∩ (E1 ∪ E2) ∩ EC2 ).

This simplifies to

µ∗(T ∩ (E1 ∪ E2)) = µ∗(T ∩ E2) + µ∗(T ∩ E1 ∩ EC2 ).

But E1 and E2 are disjoint. Hence, E1 is contained in EC2 . Hence, we can further simplify to

µ∗(T ∩ (E1 ∪ E2)) = µ∗(T ∩ E2) + µ∗(T ∩ E1).

Let’s do another step. Since E3 is inM, we have

µ∗(T ∩ (E1 ∪ E2 ∪ E3)) = µ∗(T ∩ (E1 ∪ E2 ∪ E3) ∩ E3)

+ µ∗(T ∩ (E1 ∪ E2 ∪ E3) ∩ EC3 ).

This can be rewritten as

µ∗(T ∩ (E1 ∪ E2 ∪ E3)) = µ∗(T ∩ E3) + µ∗(T ∩ E1 ∩ EC3 ∪ T ∩ E2 ∩ EC3 )

= µ∗(T ∩ E3) + µ∗(T ∩ E1 ∪ T ∩ E2)

= µ∗(T ∩ E3) + µ∗(T ∩ (E1 ∪ E2)),

because E1 ⊆ EC3 and E2 ⊆ EC3 since all the En are disjoint. Then, we can apply the first step toconclude

µ∗(T ∩ (E1 ∪ E2 ∪ E3)) = µ∗(T ∩ E3) + µ∗(T ∩ E2) + µ∗(T ∩ E1).

We have therefore shown

µ∗(T ∩ (∪3i=1 Ei)) =

3∑i=1

µ∗(T ∩ Ei).

It is now clear, we can continue this argument by induction to show

µ∗(T ∩ (∪ni=1 Ei)) =

n∑i=1

µ∗(T ∩ Ei). (a)

11.1. VIA OUTER MEASURE 219

for any positive integer n. Further, sinceM is an algebra, induction also shows ∪ni−1 Ei is inMfor any such n. It then follows that for any T in X ,

µ∗(T ) = µ∗(T ∩ (∪ni=1 Ei)) + µ∗(T ∩ (∪ni=1 Ei)C).

Using Equation a, we then have

µ∗(T ) =

n∑i=1

µ∗(T ∩ Ei) + µ∗(T ∩ (∪ni=1 Ei)C). (b)

Next, note for all n

T ∩ (∪ni=1 Ei)C ⊇ T ∩ (∪∞i=1 Ei)

C ,

and hence

µ∗(T ∩ (∪∞i=1 Ei)

C

)≤ µ∗

(T ∩ (∪ni=1 Ei)

C

).

Using this in Equation b, we find

µ∗(T ) ≥n∑i=1

µ∗(T ∩ Ei) + µ∗(T ∩ (∪∞i=1 Ei)C). (c)

Since this holds for all n, letting n→∞, we obtain

µ∗(T ) ≥∞∑i=1

µ∗(T ∩ Ei) + µ∗(T ∩ (∪∞i=1 Ei)C). (d)

Finally, since

∞⋃i=1

(T ∩ Ei) = T⋂

(∪∞i=1Ei),

by the countable subadditivity of µ∗, it follows that

µ∗(T⋂

(∪∞i=1Ei)

)= µ∗

( ∞⋃i=1

(T ∩ Ei)

)≤

∞∑i=1

µ∗(T ∩ Ei)

).

Using this in Equation c, we have

µ∗(T ) ≥ µ∗(T⋂

(∪∞i=1Ei)

)+ µ∗

(T ∩ (∪∞i=1 Ei)

C

). (e)

Since this holds for all subsets T , this tells us ∪n En is inM. This proves thatM is a σ - algebra.

However, with all this work already done, we can also derive a very nice result which will help uslater. Countable subadditivity of µ∗ gives us

µ∗(T ) ≤ µ∗(T⋂

(∪∞i=1Ei)

)+ µ∗

(T ∩ (∪∞i=1 Ei)

C

).


Hence, using countable subadditivity again,

µ∗(T ) ≤∞∑i=1

µ∗(T ∩ Ei)

)+ µ∗

(T ∩ (∪∞i=1 Ei)

C

). (f )

Combining Equation d and Equation f , we find

µ∗(T ) =

∞∑i=1

µ∗(T ∩ Ei)

)+ µ∗

(T ∩ (∪∞i=1 Ei)

C

).

Thus, letting “T” be “T ∩ (∪nEn)”, we find

µ∗(T ∩ ∪∞i=1 Ei) =

∞∑i=1

µ∗(T ∩ Ei)

). (g)

Theorem 11.1.3 The Measure Induced By An Outer Measure


measurable subsets of X . Then,M is a σ - algebra and µ∗ restricted toM is a measurewe will denote by µ.

Proof 11.1.3Recall thatM is a σ - algebra if

(i) ∅, X ∈ M.

(ii) If A ∈M, so is AC .

(iii) If An∞n=1 ∈ M, then ∪∞n=1 An ∈ M.

Since we know M is an algebra of sets, all that remains is to show it is closed under countableunions. We have already shown all the properties of a σ - algebra except closure under arbitrarycountable unions. The previous theorem, however, does give us closure under countable disjointunions. So, let (An) be a countable collection of sets inM. Letting

E1 = A1

E2 = A2 \ A1

... =...

En = An \(∪n−1i=1 Ai

)... =

...,

we see each En is inM by Theorem 11.1.1. Further, they are pairwise disjoint and so by Theorem11.1.2, we can conclude ∪n En is inM. But it is easy to see that ∪n En = ∪nAn. Thus,M is a σ -algebra.

To show µ∗ restricted toM, µ, is a measure, we must show

(i): µ(∅) = 0,

11.2. VIA METRIC OUTER MEASURE 221


(iii): µ is countably additive on S; i.e. if (En) ⊆ S is a countable collection of disjoint sets, thenµ(∪nEn) =

∑n µ(En).

Since µ∗(∅) = 0, condition (i) follows immediately. Also, we know µ∗(E) ≥ 0 for all subsets E, andso condition (ii) is valid. It remains to show countable additivity. Let (Bn) be a countable disjointfamily inM. We can apply Equation g to conclude, using T = X , that

µ∗(∪∞i=1 Bi) =

∞∑i=1

µ∗(Bi).

Finally, since µ∗ = µ on these sets, we have shown µ is countably additive and so is a measure.

It is also true that the measure constructed from an outer measure in this fashion is a completemeasure.

Theorem 11.1.4 The Measure Induced By An Outer Measure Is Complete

If E is a subset of X satisfying µ∗(E) = 0, then E ∈ M. Also, if F ⊆ E, then F ∈ Mas well, with µ∗(F ) = 0. Note, this tells us that if µ(E) = 0, then subsets of E are also inM with µ(F ) = 0; i.e., µ is a complete measure.

Proof 11.1.4We know µ∗(T ∩ E) ≤ µ∗(E) for all T ; hence, µ∗(T ∩ E) = 0 here. Thus, for any T ,

µ∗(T ∩ E) + µ∗(T ∩ EC) = µ∗(T ∩ EC) ≤ µ∗(T ).

This tells us E satisfies the Caratheodory condition and so is inM. Thus, we have µ(E) = 0. Now,let F ⊆ E. Then, µ∗(F ) = 0 also; hence, by the argument above, we can conclude F ∈ M withµ(F ) = 0.

11.2 Measures From Metric Outer Measure

Definition 11.2.1 Metric Outer Measure

Let (X, d) be a non empty metric space and for two subsets A and B of X , define thedistance between A and B by

D(A,B) = inf d(a, b) | a ∈ A, b ∈ B .

If µ∗ is an outer measure on X which satisfies

µ∗(A ∪ B) = µ∗(A) + µ∗(B)

whenever D(A,B) > 0, we say µ∗ is metric outer measure.

The σ algebra of open subsets of X is called the Borel σ algebra B. We can use the constructionprocess in Section 11.1 to construct a σ algebra of subsets, M, which satisfy the Caratheodorycondition for this metric outer measure µ∗. This gives us a measure onM. We would like to be ableto say that open sets in the metric space are µ∗ measurable. Thus, we want to prove B ⊆ M. This iswhat we do in the next theorem. It is becoming a bit cumbersome to keep saying µ∗ measurable for


the sets inM. We will make the following convention for later use: a set inM will be called OMImeasurable, where OMI stands for outer measure induced.

Theorem 11.2.1 Open Sets in a Metric Space Are OMI Measurable

Let (X, d) be a non empty metric space and µ∗ a metric outer measure on X . Then opensets are OMI measurable.

Proof 11.2.1let E be open in X . To show E is µ∗ measurable we must show

µ∗(T ) ≥ µ∗(T ∩ E) + µ∗(T ∩ EC)

for all subsets T in X . Since this is true for all subsets with µ∗(T ) = ∞, it suffices to prove the in-equality is valid for all subsets with µ∗(T ) finite. Also, we already know ∅ andX are µ∗ measurable,so we can further restrict our attention to nonempty strict subsets E of X . We will prove this in aseries of steps:

Step (i): Let En be defined for each positive integer n by

En = x |D(x,EC) >1

n.

It is clear En ⊆ E and that En ⊆ En+1.

Note, if y ∈ En and x ∈ Ec, we have d(y, x) > 1/n and so

infy∈En, x∈Ec

d(y, x) ≥ 1

n

and so D(En, EC) ≥ 1/n. This immediately tells us

D(T ∩ En, T ∩ EC) ≥ 1/n

also for all T .

Since µ∗ is a metric outer measure, we then have

µ∗(

(T ∩ En) ∪ (T ∩ EC)

)= µ∗(T ∩ En) + µ∗(T ∩ EC).

However, we also know En is a subset of E and so

(T ∩ En) ∪ (T ∩ EC) ⊆ (T ∩ E) ∪ (T ∩ EC) = T.

We conclude then

µ∗(

(T ∩ En) ∪ (T ∩ EC)

)≤ µ∗(T ).

Hence, for all T , we have

µ∗(T ∩ En)

)+ µ∗

(T ∩ EC)

)≤ µ∗(T ). (∗)


Step (ii): If limn µ∗(T ∩ En) = µ∗(T ), then letting n go to infinity in Equation ∗, we would find

µ∗(T ∩ E)

)+ µ∗

(T ∩ EC)

)≤ µ∗(T ).

This means E satisfies the Caratheodory condition and so is µ∗ measurable.

To show this limit acts in this way, we will construct a new sequence of sets (Wn) that are disjointfrom one another with E = ∪nWn so that the new sets Wn have useful properties. Since E is open,every point p in E is an interior point. Thus, there is a positive r so thatB(p; r) ⊆ E. So, if z ∈ EC ,we must have and d(p, z) ≥ r. It follows that D(p,EC) ≥ r > r/2. We therefore know that p ∈ Enfor some n. Since our choice of p is arbitrary, we have shown

E ⊆ ∪n En.

It was already clear that ∪nEn ⊆ E; we conclude E = ∪nEn. We then define the needed disjointcollection (Wn) as follows

W1 = E1

W2 = E2 \ E1

W2 = E3 \ E2

......

...

Wn = En \ En−1

(It helps to draw a picture here for yourself in terms of the annuli En \ En−1. We can see that forany n, we can write

T ∩ E = (T ∩ En)⋃∪∞k=n+1 (T ∩ Wk)

as the terms T ∩ Wk give the contributions of each annuli or strip outside of the core En. Hence,

µ∗(T ∩ E) ≤ µ∗(T ∩ En) +

∞∑k=n+1

(T ∩ Wk) (∗∗)

because µ∗ is subadditive. At this point, the series sum∑∞k=n+1 (T ∩ Wk)could be∞; we haven’t

determined if it is finite yet.

For any k > 1, if x ∈Wk, then x ∈ Ek \ Ek−1 and so

1

k≤ D(x, EC) ≤ 1

k − 1.

Next, if x ∈ Wk and y ∈ Wk+p for any p ≥ 2, we can use the triangle inequality with an arbitraryz ∈ EC to conclude

d(x, z) ≤ d(x, y) + d(y, z).

But, this says

d(x, y) ≥ d(x, z) − d(y, z)

≥ D(x,EC) − d(y, z) >1

k− d(y, z).


We have shown the fundamental inequality

d(x, y) >1

k− d(y, z), ∀ x ∈ Wk,∀ y ∈ Wk+p (α)

holds for p ≥ 2. The definition of the set Ek+p then implies for these p,

1

k + p< D(y,EC) ≤ 1

k + p− 1. (β)

Now consider how D(y,EC) is defined. Since this is an infimum, by the Infimum Tolerance Lemma,given a positive ε, there is a zε ∈ EC so that

D(y,EC) ≤ d(y, zε) < D(y,EC) + ε.

Hence, using Equation β, we have

−d(y, zε) > −D(y,EC) − ε

> − 1

k + p− 1− ε.

Also, using Equation α, we find

d(x, y) >1

k− d(y, zε)

>1

k− − 1

k + p− 1− ε

=p− 1

k(k + p− 1)− ε.

Since ε > 0 is arbitrary, we conclude

d(x, y) ≥ p− 1

k(k + p− 1)> 0

for all x ∈Wk and y ∈Wk+p with p ≥ 2. Hence,

D(Wk, Wk+p) ≥ p− 1

k(k + p− 1)> 0

It follows that

D(W1, W3) > 0

and, in general, we find this is true for the successive odd integers

D(W2k+1, W2k+3) > 0.

Since µ∗ is a metric outer measure, this allows us to say

n∑k=0

µ∗(T ∩ W2k+1) = µ∗(∪nk=0 T ∩ W2k+1

)


≤ µ∗(∪∞k=0 T ∩ W2k+1

)≤ µ∗(T ).

A similar argument shows that successive even integers satisfy

D(W2k, W2k+2) > 0.

Again, as µ∗ is a metric outer measure, this allows us to say

n∑k=0

µ∗(T ∩ W2k) = µ∗(∪nk=0 T ∩ W2k

).

Therefore, we have

n∑k=0

µ∗(T ∩ W2k) ≤ µ∗(∪∞k=0 T ∩ W2k

)≤ µ∗(T ).

We conclude

n∑k=0

µ∗(T ∩ Wk) =∑k even

µ∗(T ∩ Wk) +∑k odd

µ∗(T ∩ Wk)

≤ 2 µ∗(T )

for all n. This implies the sum∑k µ∗(T ∩Wk) converges to a finite number.

Since the series converges, we now know given ε > 0, there is an N so that

∞∑k=n

µ∗(T ∩Wk) < ε,

for all n > N . Now go back to Equation ∗∗. We have for any n > N ,

µ∗(T ∩ E) ≤ µ∗(T ∩ En) + ε.

This tells us µ∗(T ∩ En)→ µ∗(T ∩ E). By our earlier remark, this completes the proof.

We can even prove more.

Theorem 11.2.2 Open Sets In A Metric Space Are µ∗ Measurable If and Only If µ∗ Is A MetricOuter Measure

Let X be a non empty metric space. Then Open sets are µ∗ measurable if and only if µ∗ isa metric outer measure.

Proof 11.2.2If we assume µ∗ is a metric outer measure, then opens sets are µ∗ measurable by Theorem 11.2.1.

On the other hand if we know that all the open sets of µ∗ measurable, this implies all Borel sets areµ∗ measurable as well. Let A and B be any two sets with D(A,B) = r > 0. For each x ∈ A, let

G(x) = u | d(x, u) < r/2


and

G =⋃x∈A

G(x).

Then G is open, A ⊆ G and G ∩ B = ∅. Since G is measurable, it satisfies the Caratheodorycondition using test set T = A ∪B; thus,

µ∗(A ∪ B

)= µ∗

((A ∪ B) ∩ G

)+ µ∗

((A ∪ B) ∩ GC

).

But (A ∪ B) ∩ G is simplified to A because A ⊆ B and B is disjoint from G. Further since A isdisjoint from GC and B ⊆ GC , we have (A ∪ B) ∩ GC = B. We conclude

µ∗(A ∪ B) = µ∗(A) + µ∗(B).

This shows µ∗ is a metric outer measure.

11.3 Constructing Outer MeasureWe still have to find ways to construct outer measures. We want the resulting OMI measure weinduce have certain properties useful to us. Let’s discuss how to do this now.

Definition 11.3.1 Premeasures and Covering Families

Let X be a nonempty set. Let T be a family of subsets of X that contains the empty set.This family is called a covering family for X . Let τ be a mapping on T so that τ(∅) = 0.The mapping τ is called a premeasure.

It is hard to believe, but even with virtually no restrictions on τ and T , we can build an outermeasure.

Theorem 11.3.1 Constructing Outer Measures Via Premeasures

Let X be a nonempty set. Let T be a covering family of subsets of X and τ : T → [0,∞]be a premeasure. For any A in X , define

µ∗(A) = inf ∑n

τ(Tn) | Tn ∈ T , A ⊆ ∪n Tn

where the sequence of sets (Tn) from T is finite or countably infinite. Such a sequenceis called a cover. In the case where there are no sets from T that cover A, we define theinfimum over the resulting empty set to be∞. Then µ∗ is an outer measure on X .

Proof 11.3.1To verify the mapping µ∗ is an outer measure on X , we must show

(i): µ∗(∅) = 0.

(ii): If A and B are subsets of X with A ⊆ B, then µ∗(A) ≤ µ∗(B).

(iii): If (An) is a sequence of disjoint subsets of X , then µ∗( ∪nAn) ≤∑n µ∗(An).

It is straightforward to see condition (i) and (ii) are true. It suffices to prove condition (iii) is valid.Let (An) be a countable collection, finite or infinite, of subsets of X . If there is an index n with

11.3. BUILDING OUTER MEASURE 227

τ(An) infinite, then since µ∗(∪nAn) ≤ ∞ anyway, it is clear

µ∗(∪∞i=1 Ai) ≤∞∑i=1

µ∗(Ai) = ∞.

On the other hand, if µ∗(An) is finite for all n, given any ε > 0, we can use the Infimum ToleranceLemma to find a sequence of families (Tn k) in T so that

∞∑k=1

τ(Tn k) < µ∗(An) +ε

2n.

We also know that

∞⋃n=1

An ⊆∞⋃n=1

∞⋃k=1

Tn k.

Hence, the collection ∪n ∪k Tn k is a covering family for ∪n An) and so by the definition of µ∗, wehave

µ∗( ∞⋃n=1

An

)≤

∞∑n=1

∞∑k=1

µ∗(Tn k

)

≤∞∑n=1

µ∗(An) +ε

2n

≤∞∑n=1

µ∗(An) + ε.

Since ε is arbitrary, we see µ∗ is countable subadditive and so is an outer measure.

There is so little known about τ and T , that it is not clear at all that

(i): T ⊆ M, whereM is the σ - algebra of sets that satisfy the Caratheodory condition for theouter measure µ∗ generated by τ . If this is true, we will callM an OMI-F σ - algebra, wherethe “F” denotes the fact that the covering family is inM.

(ii): If A ∈ T , then τ(A) = µ(A) where µ is the measure obtained by restricting µ∗ to M. Ifthis is true, we will call the constructed σ - algebra, an OMI-FE σ - algebra, where the “E”indicates the fact the µ restricted to T recovers τ .

If τ represents some primitive notion of size of special sets, like length of intervals on the realline, we normally want both condition (i) and (ii) above to be valid. We can obtain these results ifwe add a few more properties to τ and T . First, T needs to be an algebra (which we have alreadydefined) and τ needs to be additive on the algebra.

Definition 11.3.2 Additive Set Function

Let A of subsets of the set X be an algebra. Let ν be an extended real valued functiondefined on A which satisfies

(i): ν(∅) = 0.

(ii): If A and B in A are disjoint, then ν(A ∪B) = ν(A) + ν(B).

Then ν is called an additive set function on A.


We also need a property of outer measures called regularity.

Definition 11.3.3 Regular Outer Measures

Let X be a nonempty set, µ∗ be an outer measure on X and M be the set of all µ∗

measurable sets of X . The outer measure µ∗ is called regular if for all E in X there isa µ∗ measurable F ∈ M so that E ⊆ F with µ∗(E) = µ(F ), where µ is the measureinduced by µ∗ onM. The set F is often called a measurable cover for E.

We begin with a technical lemma.

Lemma 11.3.2 Condition For Outer Measure To Be Regular

LetX be a nonempty set, T a covering family and τ a premeasure. Then if the σ - algebra,M, generated by τ using T contains T , µ∗ is regular.

Proof 11.3.2Let A be a subset in X . We need to show there is measurable set B containing A so that µ∗(A) =µ(B). If the µ∗(A) = ∞, then we can choose X as the needed set. Otherwise, we have µ∗(A) isfinite. Applying the Infimum Tolerance Lemma, for each m, there is a family of sets (Emn ) so thatA ⊆ ∪n Emn and ∑

n

τ(Emn ) < µ∗(A) +1

m.

Let

Em =⋃n

Emn

H =⋂m

Em;

these sets are measurable by assumption. Also, A ⊆ H and H ⊆ Em. Hence, µ∗(A) ≤ µ(H). Wenow show the reverse inequality. For each m, we have

µ∗(Em) ≤∑n

µ∗(Emn ) ≤∑n

τ(Emn )

≤ µ∗(A) +1

m.

Further, since H ⊆ Em for each m, we find

µ(H) ≤ µ∗(Em) ≤ µ∗(A) +1

m.

This is true for all m; hence, it follows that µ(H) ≤ µ∗(A). Combining inequalities, we haveµ(H) = µ∗(A) and so H is a measurable cover. Thus, µ∗ is regular.

Theorem 11.3.3 Conditions For OMI-F Measures

Let X be a nonempty set, T a covering family which is an algebra and τ an additive setfunction on T . Then the σ - algebra,M, generated by τ using T contains T and µ∗ isregular.

11.3. BUILDING OUTER MEASURE 229

Proof 11.3.3By Lemma 11.3.2, it is enough to show each member of T is measurable. So, let A be in T . Asusual, it suffices to show that

µ∗(T ) ≥ µ∗(T ∩ A) + µ∗(T ∩ AC)

for all sets T of finite outer measure. This will show A satisfies the Caratheodory condition andhence, is measurable. Let ε > 0 be given. By the Infimum Tolerance Lemma, there is a family (An)from T so that T ⊆ ∪nAn and ∑

n

τ(An) < µ∗(T ) + ε.

since τ is additive on T , we know

τ(An) = τ(A ∩ An) + τ(AC ∩ An).

Also, we have

A⋂

T ⊆⋃n

(A ∩ An), and AC⋂

T ⊆⋃n

(AC ∩ An).

Hence,

µ∗(A ∩ T ) ≤∑n

µ∗(A ∩ An), µ∗(AC ∩ T ) ≤∑n

µ∗(AC ∩ An). (α)

µ∗(T ) + ε >∑n

τ(An) =∑n

τ(An ∩ A) +∑n

τ(An ∩ AC)

≥∑n

µ∗(An ∩ A) +∑n

µ∗(An ∩ AC)

≥ µ∗(A ∩ T ) + µ∗(AC ∩ T ),

by Equation α. Thus, A satisfies the Caratheodory condition and is measurable.

In order for condition (ii) to hold, we need to add one more additional property to τ : it needs tobe a pseudo-measure.

Definition 11.3.4 Pseudo-Measure

Let the mapping τ : A → [0,∞] be additive on the algebraA. Assume whenever (Ai) is acountable collection of disjoint sets in A whose union is also in A (note this is not alwaystrue because A is not a σ - algebra), then it is true that

τ(∪i Ai) =∑i

τ(Ai).

Such a mapping τ is called a pseudo-measure on A.

Theorem 11.3.4 Conditions For OMI-FE Measures

Let X be a nonempty set, T a covering family which is an algebra and τ an additive setfunction on T which is a pseudo-measure. Then the σ - algebra,M, generated by τ usingT contains T , µ∗ is regular and µ(T ) = τ(T ) for all T in T .


Proof 11.3.4see Bruckner.

Comment 11.3.1 The results above tell us that we can construct measures so that T is containedinM and the measure recovers τ as long as the premeasure is a pseudo-measure and the coveringfamily is an algebra. This means the covering family must be closed under complementation. Hence,if we a covering family such as the collection of all open intervals ( which we do when we constructLebesgue measure later) these theorems do not apply.

11.4 Worked Out Problems

Let’s work out a specific examples of this process to help the ideas sink in. Note the covering familieshere do not simply contain open intervals!

Example 11.4.1 Let U be the family of subsets of< of the form (a, b], (−∞, b], (a,∞) and (−∞,∞)and the empty set. It is easy to show that F , the collection of all finite unions of sets from U is analgebra of subsets of <.Let τ be the usual length of an interval. and extend τ toF additively. This extended τ is a premeasureon F . τ can then be used to define an outer measure as usual µ∗(τ). There is then an associated σ -algebra of µ∗τ measurable sets of <,Mτ , and µ∗τ restricted toMτ is a measure is a measure, µτ .We will now prove F is contained inMτ . Let’s consider the set I from U . Let T be any subset of <and let ε > 0 be given. It is enough to consider sets T with µ∗(F ) finite. Then there is a cover (An)of sets from the algebra F so that ∑

n

τ(An) ≤ µ∗τ (T ) + ε.

Now I ∩ T ⊆ ∪n (An ∩ I) and IC ∩ T ⊆ ∪n (An ∩ IC). So because F is an algebra, this means(An ∩ I) covers I ∩ T and (An ∩ IC) covers IC ∩ T . Hence,

µ∗τ (T ∩ I) ≤∑n

τ(An ∩ I),

µ∗τ (T ∩ IC) ≤∑n

τ(An ∩ IC).

Combining, we see

µ∗τ (T ∩ I) + µ∗τ (T ∩ IC) ≤∑n

(τ(An ∩ I) + τ(An ∩ IC)

).

But τ is additive on F , and hence∑n

(τ(An ∩ I) + τ(An ∩ IC)

)=

∑n

τ(An).

Thus,

µ∗τ (T ∩ I) + µ∗τ (T ∩ IC) ≤ µ∗τ (T ) + ε.

Since ε > 0 is arbitrary, we have shown I satisfies the Caratheodory condition. This shows that I isOMI measurable and so F ⊆Mτ .

11.4. EXAMPLES 231

Example 11.4.2 Let U be the family of subsets of< of the form (a, b], (−∞, b], (a,∞) and (−∞,∞)and the empty set. It is easy to show that F , the collection of all finite unions of sets from U is analgebra of subsets of <. Let g be the monotone increasing function on < defined by g(x) = x3. Noteg is right continuous which means

limh→0+

g(x+ h) exists , ∀ x,

limx→−∞

g(x) exists,

limx→∞

g(x) exists.

where the last two limits are −∞ and∞ respectively. Define the mapping τg on U by

τg

((a, b]

)= g(b) − g(a),

τg

((−∞, b)

)= g(b) − lim

x→−∞g(x),

τg

((a,∞)

)= lim

x→∞g(x) − g(a),

τg

((−∞,∞)

)= lim

x→∞g(x) − lim

x→−∞g(x).

Extend τg to F additively as usual. This extended τg is a premeasure on F . τg can then be used todefine an outer measure as usual µ∗(g). There is then an associated σ - algebra of µ∗g measurablesets of <,Mg , and µ∗g restricted toMg is a measure, µg .

We will now prove F is contained inMg . Let’s consider the set I from U . Let T be any subset of< and let ε > 0 be given. Again, it is enough to consider sets T with µ∗(F ) finite. Then there is acover (An) of sets from the algebra F so that∑

n

τg(An) ≤ µ∗g(T ) + ε.

Now I ∩ T ⊆ ∪n (An ∩ I) and IC ∩ T ⊆ ∪n (An ∩ IC). So

µ∗g(T ∩ I) ≤∑n

τg(An ∩ I),

µ∗g(T ∩ IC) ≤∑n

τg(An ∩ IC).

Combining, we see

µ∗g(T ∩ I) + µ∗g(T ∩ IC) ≤∑n

(τg(An ∩ I) + τg(An ∩ IC)

).

But τg is additive on F , and hence

∑n

(τg(An ∩ I) + τg(An ∩ IC)

)=

∑n

τg(An).

Thus,

µ∗g(T ∩ I) + µ∗g(T ∩ IC) ≤ µ∗g(T ) + ε.


Since ε > 0 is arbitrary, we have shown I satisfies the Caratheodory condition. This shows that I isOMI measurable and so F ⊆Mg .

11.5 HomeworkExercise 11.5.1 Let X = (0, 1]. Let A consist of the empty set and all finite unions of half- openintervals of the form (a, b] from X . Prove A is an algebra of sets of (0, 1].

Exercise 11.5.2 Let A be the algebra of subsets of (0, 1] given in Exercise 11.5.1. Let f be anarbitrary function on [0, 1]. Define νf on A by

νf

((a, b]

)= f(b) − f(a).

Extend νf to be additive on finite disjoint intervals as follows: if (Ai) = (ai, bi]) is a finite collectionof disjoint intervals of (0, 1], we define

νf

(∪ni=1 (ai, bi]

)=

n∑i=1

f(bi) − f(ai).

1. Prove that νf is additive on A.

Hint 11.5.1 It is enough to show that the value of νf (A) is independent of the way in which wewrite A as a finite disjoint union.

2. Prove νf is non negative if and only if f is non decreasing.

Exercise 11.5.3 If λ is an additive set function on an algebra of subsets A, prove that λ can not takeon both the value∞ and −∞.

Hint 11.5.2 If there is a set A in the algebra with λ(A) =∞ and there is a set B in the algebra withλ(B) = −∞, then we can find disjoint sets A′ and B′ in A so that λ(A′) = ∞ and λ(B′) = −∞.But this is not permitted as the value of λ(A′ ∪ B′) must be a well - defined extended real value notthe undefined value∞−∞.

Exercise 11.5.4 Let T be a covering family for a nonempty set X . Let τ be a non negative, possiblyinfinite valued premeasure. For any A in X , define

µ∗(A) = inf ∑n

τ(Tn) | Tn ∈ T , A ⊆ ∪n Tn

where the sequence of sets (Tn) from T is finite or countably infinite. In the case where there are nosets from T that cover A, we define the infimum over the resulting empty set to be∞.

Prove µ∗ is an outer measure on X .

Exercise 11.5.5 Let X = 1. 2, 3 and T consist of ∅, X and all doubleton subsets x, y of X .Let τ satisfy

(i): τ(∅) = 0.

(ii): τ(x, y

)= 1 for all x 6= y in X .

(iii): τ(X) = 2.

11.5. HOMEWORK 233

(a): Prove the method of Exercise 11.5.4 gives rise to an outer measure µ∗ defined by µ∗(∅) = 0,µ∗(X) = 2 and µ∗(A) = 1 for any other subset A of X .

(b): Now do the construction process again letting τ(X) = 3. What changes?

Exercise 11.5.6 Let X be the natural numbers N and let τ consist of ∅, N and all singleton sets.Define τ(∅) = 0 and τ(x) = 1 for all x in N.

(a): Let τ(N) = 2. Prove the method of Exercise 11.5.4 gives rise to an outer measure µ∗. De-termine the family of measurable sets (i.e., the sets that satisfy the Caratheodory Condition).

(b): Let τ(N) =∞ and answer the same questions as in Part (a).

(c): Let τ(N) = 2 and set τ(x) = 2−(x−1). Now answer the same questions as in Part (a).

(d): Let τ(N) = ∞ and again set τ(x) = 2−(x−1). Now answer the same questions as inPart (a). You should see N is measurable but τ(N) 6= µ(N), where µ denotes the measureconstructed in the process of Part (a).

(e): Let τ(N) = 1 and again set τ(x) = 2−(x−1). Now answer the same questions as in Part(a). What changes?

Chapter 12

Lebesgue Measure

We will now construct Lebesgue measure on <k. We will begin by defining the mapping µ∗ on thesubsets of <k which will turn out to be an outer measure. The σ - algebra of subsets that satisfythe Caratheodory condition will be called the σ - algebra of Lebesgue measurable subsets. We willdenote this σ - algebra byM as usual. We will usually be able to tell from context what σ - algebra ofsubsets we are working with in a given study area or problem. The primary references here are again(Bruckner et al. (1) 1997). and (Taylor (12) 1985). We like the development of Lebesgue measurein (Taylor (12) 1985) better than that of (Bruckner et al. (1) 1997) and so our coverage reflects that.In all cases, we have added more detail to the proofs of propositions to help you build your analysisskills by looking hard at many interesting and varied proof techniques.

12.1 Lebesgue Outer Measure

We will be working in <k for any positive integer k. We have to work our way through a fair bit ofdefinitional material; so be patient while we set the stage. We let x = (x1, x2, . . . , xk) denote a pointin the Euclidean space <k. An open interval in <k will be denoted by I and it is determined by thecross - product of k intervals of the form (ai, bi) where each ai and bi is a finite real number. Hence,the interval I has the form

I = Πki=1 (ai, bi).

The interval (ai, bi) is called the ith edge of I and the number ì = bi − ai is the length of the ith

edge. The content of the open interval I is the product of the edge lengths and is denoted by |I|; i.e.

|I| = Πki=1

(bi − ai

).

We need additional terminology. The center of I is the point

p = (a1 + b1

2,a2 + b2

2, . . . ,

ak + bk2

);

if the interval J has the same center as the interval I , we say the intervals are concentric.If I and J are intervals, for convenience of notation, let `J and Ì denote the vector of edge lengthsof J and I , respectively. In general, there is no relationship between `J and Ì . However, there is aspecial case of interest. We note that if J is concentric with I and each edge in `J is a fixed multipleof the corresponding edge length in Ì , we can say `J = λ Ì for some constant λ. In this case, we

235

236 CHAPTER 12. LEBESGUE MEASURE

write J = λ I . It then follows that |J | = λk |I|.

We are now ready to define outer measure on <k. Following Definition 11.3.1, we define a suitablecovering family T and premeasure τ . Then, the mapping µ∗ defined in Theorem 11.3.1 will be anouter measure. For ease of exposition, let’s define this here.

Definition 12.1.1 Lebesgue Outer Measure

Let T be the the collection of all open intervals in <k and define the premeasure τ byτ(I) = |I| for all I in T . For any A in X , define

µ∗(A) = inf ∑n

|In| | In ∈ T , A ⊆ ∪n In

We will call a collection (In) whose union contains A a Lebesgue Cover of A.

Then, µ∗ is an outer measure on <k and as such induces a measure through the usual Caratheodorycondition route. It remains to find its properties. The covering family here is not an algebra, so wecan not use Theorem 11.3.3 and Theorem 11.3.4 to conclude

(i): T ⊆M; i.e.M is an OMI-F σ - algebra.

(ii): If A ∈ T , then |A| = µ(A); i.e.M is an OMI-FE σ - algebra.

However, we will be able to alter our original proofs to get these results with just a little work.

Comment 12.1.1 (i): If I is an interval in <k, then (I) covers I itself and so by definition µ∗(I) ≤|I|.

(ii): If x is a singleton set, choose any open interval I that has x as its center. Then, I is a coverof x and so µ∗(x) ≤ |I|. We see the the concentric intervals 1/2n I also are covers ofx and so µ∗(x) ≤ 1/2n for all n. It follows µ∗(x) = 0.

(iii): From (ii), it clear that µ∗(E) = 0 if E is a finite set.

(iv): If E is countable, label its points by (an). Let ε > 0 be given. Then by the Infimum ToleranceLemma, there are intervals In having an as a center so that |In| < ε/2n. Then the intervals(In) cover E and by definition,

µ∗(E) ≤∑n

|In| ≤∑n

ε/2n = ε.

Since ε is arbitrary, we see µast(E) = 0 if E is countable.

We want to see if µ∗(I) = |I|. This is not clear since our covering family is not an algebra. Wenow need a technical lemma.

Lemma 12.1.1 Sums Over Finite Lebesgue Covers Of I Dominate |I|

Let I be any interval of <k and let (I1, . . . , IN ) be any finite Lebesgue cover of I . Then

N∑n=1

|In| ≥ |I|.

12.1. OUTER MEASURE 237

The proof is based on an algorithm that cycles through the covering sets Ii one by one and picksout certain relevant subintervals. We can motivate this by looking at an interval I in<2 whose closureis covered by 3 overlapping intervals I1, I2 and I3. This is shown in Figure 12.1. We do not attemptto indicate the closure of I in this figure nor the fact that the intervals I1 and so forth are open. Wesimply draw boxes and you can easily remove or add edges in your mind to open an interval or closeit.

I

I3

I2I1

An example in <2: cover I1, I2and I3 of I .

Figure 12.1: Motivational Lebesgue Cover

These four intervals all have endpoints on both the x and y axes. If we draw all the possible constantx and constant y lines corresponding to these endpoints, we subdivide the original four intervals intomany smaller intervals as shown in Figure 12.2.In particular, if we looked at interval I1, it is divided into 16 subintervals (J1, i), for 1 ≤ i ≤ 16 asshown in Figure 12.3.

These rectangles are all disjoint and

I1 =

16⋃i=1

J1, i.

although we won’t show it in a figure, I2 and I3 are also sliced up into smaller intervals; using thesame left to right and then downward labeling scheme that we used for I1, we have

• I2 is divided by 4 horizontal and 4 vertical lines into 16 disjoint subintervals, J2,1 to J2,16.Further,

I2 =

16⋃i=1

J2, i.

• I3 is divided by 4 horizontal and 6 vertical lines into 24 disjoint subintervals, J3,1 to J3,24. Wethus know

I3 =

24⋃i=1

J3, i.


I1, I2, I3 and I determine subdi-visions into smaller intervals.

Figure 12.2: Subdivided Lebesgue Cover

Finally, I is also subdivided into subintervals: it is divided by 4 horizontal and 2 vertical lines into 8disjoint subintervals, J1 to J8 and

I =

8⋃i=1

J i.

We also know

|I| =

8∑i=1

|Ji|,

|I1| =

16∑i=1

|J1,i|,

|I2| =

16∑i=1

|J2,i|,

|I3| =

24∑i=1

|J3,i|.

Now look at Figure 12.2 and you see immediately that the intervals Jkj and Jpq are either the same orare disjoint. For example, the subintervals match when interval I2 and I3 overlap. We can concludeeach Ji is disjoint from a Jkj or it equals Jkj for some choice of k and j. Here is the algorithm wewant to use:Step 1: We know I ⊆ I1 ∪ I2 ∪ I3 and J1 = Jn1,q1 where n1 is the smallest index from 1, 2 or 3which works. For this fixed n1, consider the collection

Sn1 = Jn1,1, . . . , Jn1,p(n1)


J1,13

J1,9

J1,5

J1,1

J1,14

J1,10

J1,6

J1,2

J1,15

J1,11

J1,7

J1,3

J1,16

J1,12

J1,8

J1,4

I1

I1 is subdivided into 16 new rect-angles, J1,1 to J1,16.

Figure 12.3: Subdivided I1

where we are using the symbol p(n1) to denote the number of subintervals for In1. Thus, p(1) =

p(2) = 16 and p(3) = 24 in our example. In our example, we find n1 = 1 and

J1 = J1,12

S1 = J1,1, . . . , J1,16.

Look at Figure 12.4 to see what we have done so far.

By referring to Figure 12.2, you can see J1 = J1,12 and J3 = J1,16. Now, let

Tn1≡ T1 = i | ∃k 3 Ji = Jn1,k.

Here T1 = 1, 3. Also, let

Un1 ≡ U1 = k | ∃i 3 Jn1,k = Ji.

We see U1 = 12, 16.

Step 2: Now look at the indices

Vn1 ≡ V1 = 1, 2, 3, . . . , 8 \ T1

= 2, 4, 5, 6, 7, 8.


J7

J5

J3

J1

J8

J6

J4

J2

I

I is subdivided into 8 new rectan-gles, J1 to J8. The shaded part iscovered by I1.

Figure 12.4: The Part Of I Covered by I1

The smallest index in this set is 2. Next, find the smallest index n2 so that

J2 = Jn2,k

for some index k. From Figure 12.2, we see both I2 and I3 intersect I \ I1. The smallest index n2 isthus n2 = 2. The index k that works is 7 and so J2 = J2,7. In figure 12.5, we have now shaded thepart of I not in I1 that lies in I2.

We can see that J2 = J2,7, J4 = J2,11, J5 = J2,14 and J6 = J2,15. Let

Tn2 ≡ T2 = i ∈ V1 | ∃k 3 Ji = Jn2,k.

Here T2 = 2, 4, 5, 6. Also, let

Un2≡ U2 = k | ∃i 3 Jn2,k = Ji.

We see U1 = 7, 11, 14, 15.

Step 3: Now look at the indices

Vn2≡ V2 = 1, 2, 3, . . . , 8 \ (T1 ∪ T2)

= 7, 8.

The smallest index in this set is 7. Next, find the smallest index n3 so that

J7 = Jn3,k

for some index k. From Figure 12.2, we see both I2 and I3 intersect I \ (I1∪ I2). The smallest indexn3 must be 3 and so n3 = 3. The index k that works now is 15 and we have J7 = J3,15. In figure12.6, we have now shaded the part of I not in I1 ∪ I2 that lies in I3.


J7

J5

J3

J1

J8

J6

J4

J2

I

I is subdivided into the 8 newrectangles, J1 to J8. The twoshaded parts are covered by I1(lighter shading) and I2 (darkershading).

Figure 12.5: The Part Of I Covered by I1 and I2

In fact, we have J7 = J3,15 and J8 = J3,16. Thus, we set

Tn3 ≡ T3 = i ∈ V2 | ∃k 3 Ji = Jn3,k= 7, 8.

Also, we let

Un3≡ U3 = k | ∃i 3 Jn3,k = Ji.

We see U1 = 15, 16.

We have now expressed each Ji as some Jn1,k through Jn3,k. We are now ready to finish our argu-ment.

Step 4: We have

1, . . . , 8 = Tn1 ∪ Tn2 ∪ Tn3

= T1 ∪ T2 ∪ T3.

Thus,

∑k∈Un3=U3

|Jn3,k| ≤p(n3)∑k=1

|Jn3,k| =

24∑k=1

|J3,k| ≤ |I3|,

∑k∈Un2

=U2

|Jn2,k| ≤p(n2)∑k=1

|Jn2,k| =

16∑k=1

|J2,k| ≤ |I2|,


J7

J5

J3

J1

J8

J6

J4

J2

I

I is subdivided into the 8 newrectangles, J1 to J8. The threeshaded parts are covered by I1(lighter shading) and I2 (darkershading) and I3 (darkest shading).

Figure 12.6: The Part Of I Covered by I1, I2 and I3

∑k∈Un1

=U1

|Jn1,k| ≤p(n1)∑k=1

|Jn1,k| =

16∑k=1

|J1,k| ≤ |In1 |.

Thus,

|I| =

8∑i=1

|Ji| =

3∑p=1

∑k∈U(np)

|Jnp,k|

≤3∑p=1

|Inp |.

This proves that

|I| ≤3∑i=1

|Ii|.

This is our desired proposition for a particular example set in <2 using three intervals. We are nowready to adapt this algorithm to prove the general result.

Proof 12.1.1We are given intervals I1 to IN in <k whose union covers I . Each interval Ii is the product

(αi1, βi1)× · · · × (αik, βik),

and I is the product(α1, β1)× · · · × (αk, βk).


On the xj axis, the N intervals and the interval I determine a collection of points

(α1j , β1j), xj edge from interval I1;

(α2j , β2j), xj edge from interval I2;

...

(αNj , βNj), xj edge from interval IN ;

(αj , βj), xj edge from interval I.

We do not care if these points are ordered. These xj axis points, for 1 ≤ j ≤ k, “slice” the intervalsI1 through IN and I into smaller intervals just as we did in the example for <2 shown in Figure 12.2.We have

I −→ J1, . . . , Jp

I1 −→ J11, . . . , J1,p(1)

...

IN −→ JN1, . . . , JN,p(1).

Step 1: Look at J1. There is a smallest index n1 so that J − 1 = Jn1,` for some `. Let

Tn1 = i1, . . . , p | ∃ ` 3 Ji = Jn1,`,Un1 = ` | ∃ i 3 Ji = Jn1,`.

This uses up Tn1of the indices 1, . . . , p. You can see this process in Figure 12.4.

Step 2: Let

V1 = 1, . . . , p \ Tn1

and let q be the smallest index from the set V1. For this q, find the smallest index n2 6= n1 so thatJq = Jn2,` for some `. This is the process we are showing in Figure 12.5. We define

Tn2= i ∈ V1 | ∃ ` 3 Ji = Jn2,`,

Un2= ` | ∃ i ∈ V1 3 Ji = Jn2,`.

This uses up more of the smaller subintervals I1 to Ip.

Additional Steps : Let

V2 = 1, . . . , p \ (Tn1∪ Tn2

).

We see V2 is a smaller subset of the original 1, . . . , p than V1. We continue this constructionprocess until we have used up all the indices in 1, . . . , p. This takes say Q steps and we knowQ ≤ p.

Final Step: After the process terminates, we have

|I| =

p∑i=1

|Ji|


=

Q∑p=1

∑` ∈ U(np)

|Jnp,`|

≤Q∑p=1

|Inp | ≤N∑i=1

|Ii|.

this completes the proof.

We can now finally prove that µ∗(I) = |I|. Note that we have to work this hard because our originalcovering family was not an algebra! The final arguments are presented in the next two lemmatta.

Lemma 12.1.2 µ∗(I = |I|

Let I be an open interval in <k. Then µ∗(I) = |I|.

Proof 12.1.2Let (In) be any Lebesgue cover of I . Since I is compact, this cover has a finite subcover, In1

, . . . , InN .Applying Lemma 12.1.1, we see

|I| ≤N∑i=1

|Ini | ≤∑i

|Ii|.

Since (In) is an arbitrary cover of I , we then have |I| is a lower bound for the set

∑n

|In| | (In) is a cover of I.

It follows that

|I| ≤ µ∗(I).

To prove the reverse inequality holds, let U be an open interval concentric with I so that I ⊆ U .Then U is a cover of I and so µ∗(I) ≤ |U |. Hence, for any concentric interval, λI , 1 < λ < 2, wehave µ∗(I) ≤ λk |I|. Since this holds for all λ > 1, we can let λ→ 1 to obtain µ∗(I) ≤ |I|.

Lemma 12.1.3 µ∗(I) = |I|

If I is an open interval of <k, then µ∗(I) = |I|.

Proof 12.1.3We know I is a cover of itself, so it is immediate that µ∗(I) ≤ |I|. To prove the reverse inequality,let λI be concentric with I for any 0 < λ < 1. Then, λI ⊆ I and since µ∗ is an outer measure, it ismonotonic and so

µ∗(λI) ≤ µ∗(I).

But µ∗(λI) = λk |I|. We thus have λk |I| ≤ µ∗(I) for all λ ∈ (0, 1). Letting λ→ 1, we obtain thedesired inequality.

12.2. LOM IS MOM 245

12.2 Lebesgue Outer Measure Is A Metric Outer Measure

We have now shown that if I ∈ T , then |I| = µ∗(I). However, we still do not know that the intervalsI from T are µ∗ measurable. We will do this by showing that Lebesgue outer measure is a metricouter measure. Then, it will follow from Theorem 11.2.1 that the open sets in <k are µ∗ measurable,i.e. are inM. Of course, this implies T ⊆ M as well. Then, since an interval I is measurable, wehave |I| = µ(I). Let’s prove µ∗ is a metric outer measure. We begin with a technical definition.

Definition 12.2.1 The Mδ Form of µ∗

For any set E is <k and any δ > 0, let (In) be a cover of E with each In an interval in <kwith each edge of an In having a length less than δ. Then

Mδ(E) = inf ∑n

|In| | (In).

Next, we need a technical lemma concerning finite Lebesgue covers.

Lemma 12.2.1 Approximate Finite Lebesgue Covers Of I .

Let I be a open interval and let I denote its closure. Let ε and δ be given positive numbers.Then there exists a finite Lebesgue Covering of I , I1, . . . , IN so that each edge of Ii haslength less than δ and

|I1| + · · · + |IN | < |I| + ε

Proof 12.2.1Let

I = Πki=1 (ai, bi)

and divide each component interval (ai, bi) into ni uniform pieces so that (bi − ai)/2 < δ/2. Thisdetermines ni open intervals of the form (aij , bij) for 1 ≤ j ≤ ni with bij − aij < δ/2.

LetN = n1n2 ·nk and let J = (j1, . . . , jk) denote the k - tuple of indices chosen so that 1 ≤ ji ≤ ni.There are N of these indices. Let j indicate any such k - tuple. Then j determines an interval Ijwhere

Ij = Πki=1 (aij , bij), with (bij − aij) < δ/2.

Hence, |Ij | < (δ/2)k. It is also clear that∑|Ij | = |I|.

Now choose concentric open intervals λIj for any λ with 1 < λ < 2. Then since λ > 1, (λIj overall k - tuples j is a Lebesgue cover of I , we have

|λIj | = λk |Ij |


and so ∑|λIj | = λk

∑|Ij |

= λk |I|.

Since λk → 1, for our given ε > 0, there is a η > 0 so that if 1 < λ < 1 + η, we have

λk − 1 <ε

|I| + 1.

In particular, if we pick λ = (1 + η)/2, then

|λIj | <

(1 +

ε

|I| + 1

)|I| < |I| + ε.

Since ε is arbitrary, we see

|λIj | <

(1 +

ε

|I| + 1

)|I| < |I| + ε.

Thus, the finite collection ((1 + η)/2 Ij) is the one we seek as each edge has length ((1 + η)/2 δ/2which is less than δ.

Lemma 12.2.2 Mδ = µ∗

For any subset E of <k, we have Mδ(E) = µ∗(E).

Proof 12.2.2Let’s pick a given δ > 0. The way Mδ is defined then tells us immediately that µ∗(E) ≤ Mδ(E) forany δ > 0 and subset E. It remains to prove the reverse inequality. If µ∗(E) was infinite, we wouldhave µ∗(E) ≥Mδ(E); hence, it is enough to handle the case where µ∗(E) is finite. By the InfimumTolerance Lemma for a given ε > 0, there is a Lebesgue cover (In) of E so that∑

n

|In| < µ∗(E) +ε

2.

By Lemma 12.2.1, there is a finite Lebesgue cover of each (In) which we will denote by (Jnj),1 ≤ j ≤ p(n) so that each interval Jnj has edge length less than δ and satisfies

p(n)∑j=1

|Jnj | < |In| +ε

2n+1.

The combined family of intervals (Jnj for all n and 1 ≤ j ≤ p(n) is clearly a Lebesgue cover of Ealso. Thus, by definition of µ∗, we have

∞∑n=1

p(n)∑j=1

|Jnj | <

∞∑n=1

|In| +

∞∑n=1

ε

2n+1

< µ∗(E) + ε.

12.2. LOM IS MOM 247

Now each edge length of the interval Inj is less than δ and so

Mδ ≤∞∑n=1

p(n)∑j=1

|Jnj |

by definition. We see we have established

Mδ ≤ µ∗(E) + ε

for an arbitrary ε; hence, Mδ ≤ µ∗(E).

We now have enough “ammunition” to prove Lebesgue outer measure is a metric outer measure;i.e. LOM is a MOM!

Theorem 12.2.3 Lebesgue Outer Measure Is a Metric Outer Measure

The Lebesgue Outer Measure, µ∗ is a metric outer measure; i.e., if A and B are two setsin <k with D(A,B) > 0, then µ∗(A ∪ B) = µ∗(A) + µ∗(B).

Proof 12.2.3We always know that µ∗(A ∪B) ≤ µ∗(A) + µ∗(B) for any A and B. Hence, for two sets A and Bwith D(A,B) = δ > 0, it is enough to show µ∗(A) + µ∗(B) ≤ µ∗(A ∪ B). Let ε > 0 be chosen.Since Mδ = µ∗, there is a cover of A ∪B so that the edge length of each In is less than δ/k and

Mδ(A ∪B) = µ∗(A ∪B) ≤∑n

|In| < µ∗(A ∪B) + ε

by an application of the Infimum Tolerance Lemma.

If x and y in A ∪B are both in a given In, then

d(x, y) =

√√√√ k∑i=1

(xi − yi)2 <

√√√√ k∑i=1

(δ

k)2 =

√k2δ2

k2= δ.

However, D(A,B) = δ by assumption. Thus, a given In can not contain points of both A and B.We can therefore separate the family (In) into two collections indexed by U and V , respectively. Ifn ∈ U , then In ∩ A is non empty and if n ∈ V , In ∩ B is non empty. We see Inn∈U is a coverfor A and Inn∈V is a cover for B. Thus, µ∗(A) ≤

∑n∈U |In| and µ∗(B) ≤

∑n∈V |In|. It then

follows that

µ∗(A ∪B) + ε ≥∑n

|In| =∑n∈U

|In| +∑n∈V

|In|

≥ µ∗(A) + µ∗(B).

Since ε is arbitrary, we have shown µ∗(A) + µ∗(B) ≤ µ∗(A ∪ B). This completes the proof thatLebesgue outer measure is a metric outer measure.

This theorem is the final piece we need to fully establish the two conditions

(i): T ⊆M; i.e.M is an OMI-F σ - algebra.

(ii): If I ∈ T , then |I| = µ(I); i.e.M is an OMI-FE σ - algebra.


Comment 12.2.1 We see immediately that since Lebesgue outer measure is a metric outer measure,the σ - algebra of µ∗ measurable subsets contains all the open sets of <k. In particular, any openinterval I is measurable. As mentioned previously, we thus know the Borel σ - algebra of subsets iscontained inM.

By Theorem 11.1.4, we know Lebesgue measure µ is complete.

We can also prove Lebesgue measure µ is regular.

Theorem 12.2.4 Lebesgue Measure Is Regular

For any set E in <k,

µ∗(E) = inf µ(U) | U, E ⊆ U, U is open µ∗(E) = inf µ(F ) | E, E ⊆ F, F is Lebesgue measurable .

Thus, Lebesgue measure is regular.

Proof 12.2.4Since U is open, U is Lebesgue measurable and so µ∗(U) = µ(U). It follows immediately thatµ∗(E) ≤ µ(U) for such U . Hence,

µ∗(E) ≤ inf µ(U) | U, E ⊆ U, U is open .

On the other hand, if ε > 0 is given, the Infimum Tolerance Lemma tells us there is a Lebesgue coverof E, (In), so that

µ∗(E) ≤∑n

|In| < µ∗(E) + ε.

However, this open cover generates an open set G = ∪n In containing E with µ(G) ≤∑n |In|

because µ(In) = |In|. We conclude, using the definition of µ∗ that

µ(G) ≤∑n

|In| < µ∗(E) + ε.

Hence, we must have

inf µ(U) | U, E ⊆ U, U is open ≤ µ∗(E) + ε.

Since ε is arbitrary, the result follows.

Since each open U is measurable, we then know

µ∗(E) = inf µ(U) | U, E ⊆ U, U is open ≥ inf µ(F ) | E, E ⊆ F, F ∈ M

by the first argument. To obtain the reverse inequality, note that since µ∗(F ) = µ(F ) for all measur-able F , monotonicity of µ∗ says µ∗(E) ≤ µ∗(F ) for all measurable F . We conclude

µ∗(E) ≤ inf µ(F ) | E, E ⊆ F, F ∈ M.

12.3. APPROXIMATION RESULTS 249

Now recall the definition of a regular measure from Definition 11.3.3. Using the Infimum ToleranceLemma again, there is are measurable sets (Fn) so that E ⊆ Fn for all n and

µ∗(E) ≤ µ(Fn) < µ∗(E) +1

n.

Then, ∩nFn is also measurable and so by our equivalent form of µ∗, we have µ∗(E) ≤ µ(∩n Fn).However, ∩n Fn ⊆ Fn always and hence,

µ∗(E) ≤ µ(∩n Fn) ≤ µ(Fn) < µ∗(E) +1

n.

We conclude for all n, µ∗(E) ≤ µ(∩n Fn) < µ∗(E) + 1n . Letting n go to infinity, we find

µ∗(E) = µ(∩n Fn) which shows µ is regular.

12.3 Approximation Results

We now present some approximation results for Lebesgue measurable sets and some applications tothe Lp spaces. We begin with the following result.

12.3.1 Approximating Measurable Sets

Theorem 12.3.1 Measurability and Approximation Conditions

A set E in <k is Lebesgue measurable if and only if for all ε > 0, there is a pair of sets(F,G) so that F is closed, G is open with F ⊆ E ⊆ G and µ(G \ F ) < ε. Moreover, ifE is bounded, we can choose the open set G so that G is compact.

Proof 12.3.1First, we prove that if the condition holds for all ε > 0, thenE must be measurable. For each positiveinteger n, there are therefore closed sets Fn and open sets Gn so that

Fn ⊆ E ⊆ Gn

with µ(Gn \ Fn) < 1n . Let F = ∪nFn and G = ∩nGn. Then, G and F are measurable and

G \ F ⊆ Gn \ Fn

with µ(G \ F ) ≤ µ(Gn \ Fn) < 1n . Hence, as n goes to∞, we see µ(G \ F ) = 0. Since Lebesgue

measure is complete and E \F ⊆ G\F , we also know E \F is measurable with measure 0. Finally,since E = F ∪ (E \ F ) and each piece is measurable, E is measurable.

To prove the other direction is a bit more complicated. We now start with the assumption that E ismeasurable. First, let’s assume E is a bounded set. Since E is bounded, there is a bounded openinterval I so that µ(E) ≤ |I| with E ⊆ I . Further, for any ε that is positive, by Theorem 12.2.4,there is an open set H so that E ⊆ H and µ(H) < µ(E) + ε

2 . But then the set G = H ∩ I is alsoopen and since it contains E, we must have

µ(E) ≤ µ(G) ≤ µ(H) < µ(E) +ε

2.

It is also clear G is compact.


Now choose a compact set A so that E ⊆ A (we can do this because E is bounded). Next, since theset A \ E is measurable, by Theorem 12.2.4 we can find an open set B so that A \ E ⊆ B and

µ(B) < µ(A \ E) +ε

2.

Define the closed set F by F = A\B. Then, an easy calculation shows F ⊆ E. We thus have shown

µ(F ) = µ(A) − µ(A ∩B) ≥ µ(A) − µ(B)

> µ(A) − µ(A \ E) − ε

2

= µ(E) − ε

2.

We conclude µ(G \ F ) = µ(G)− µ(F ) < ε. This shows the result holds if E is bounded.To show the result is valid if E is not bounded, let Sn = x : ||x|| ≤ n be the closed ball of radiusn in <n. Define the sets En as follows:

E1 = E ∩ S1

E2 = E ∩ (S2 \ S1)

E3 = E ∩ (S3 \ S2)

...

En = E ∩ (Sn \ Sn−1)

...

Then E = ∪nEn and each En is bounded and measurable. Hence, for each n, the result for abounded measurable set applies. Thus, there are closed sets Fn and open sets Gn with Fn ⊆ En ⊆Gn and

µ(Gn \ Fn) <ε

2n.

Let F = ∪Fn and G = ∪Gn. We see

G \ F ⊆ ∪n(Gn \ Fn)

andµ(G \ F ) ≤

∑n

µ(Gn \ Fn) <∑n

ε

2n< ε.

Since F ⊆ E ⊆ G, all that is left to prove is that F is closed.To do this, let xi be a sequence in F which converges to x. Since the sequence converges (xi) is

bounded, all the xi live in some SN . But Fn ⊆ E ∩ (Sn ⊆ Sn) for all n ≥ 2 and hence

FN+1 ⊆ SN+1 \ SNFN+2 ⊆ SN+2 \ SN+1 ⊆ SN+2 \ SN

and so forth. Thus, Fn ⊆ Sn \ SN ⊆ SCN for all n > N . Hence, the points xi must all live in∪Nn=1 Fn which is a bounded set. But then the limit point x must also be in this set which tells us x isin F . Thus, F is closed and we have proven the desired result.


We easily prove Theorem 12.3.1 in a more general setting. First, let’s introduce a common definition.

Definition 12.3.1 Borel Measure

Let X be a metric space and S a sigma-algebra of subsets of X . We say the measure µ onS is a Borel Measure if the Borel sets in X are µ measurable.

Theorem 12.3.2 Finite Measure Borel Sets Can Be Approximated By Closed Sets

Let X be a metric space, µ a Borel measure on X , ε > 0 and B a Borel Set with µ(B) <∞. Then B contains a closed set F with µ(B \ F ) < ε.

Proof 12.3.2First, let’s assume µ(X) <∞. Let F be the defined as follows:

F = E ⊆ X and ∀ γ > 0, ∃ closed set K ⊆ E with µ(E \K) < γ.

It is easy to see that closed sets with finite measure are in F . Let ε > 0 be chosen. Now supposesets E1 to EN are in F for some positive integer N . Then, there are closed sets K1 to KN so thatKi ⊆ Ei and µ(Ei \Ki) <

ε2i . Then

µ(∩Ni=1 Ei \ ∩Ni=1 Ki) ≤ µ(∩Ni=1 (Ei \Ki)

<

N∑i=1

ε

2i= 1− 1

2N< ε.

Hence, the closed set ∩Ni=1 Ki is contained in the set ∩Ni=1 Ei and satisfies the γ condition. Thus,F is closed under finite intersections. From this argument, it is also clear that F is closed undercountable intersections using minor changes in the reasoning as a countable intersection of closedsets is closed. Now, let’s look at a finite union. We again suppose sets E1 to EN are in F for somepositive integer N . Then, there are closed sets K1 to KN so that Ki ⊆ Ei and µ(Ei \ Ki) <

ε2i .

Then,

µ(∪Ni=1 Ei \ ∪Ni=1 Ki) ≤ µ(∪Ni=1 (Ei \Ki)

<

N∑i=1

ε

2i= 1− 1

2N< ε.

Since the finite union of closed sets is closed, we see the closed set ∪Ni=1 Ki is contained in theset ∪Ni=1 Ei and satisfies the γ condition also. Hence, F is closed under finite unions. To handlecountable unions, define the sequence of sets

Fi = ∪∞i=1 Ei \ ∪Ni=1 Ki.

Then, . . . , FN ⊆ FN−1 ⊆ . . . ⊆ F1 and µ(F1) < ∞ because µ(X) < ∞. Hence, we can invokeLemma 9.1.2 to conclude

limNµ(∪∞i=1 Ei \ ∪Ni=1 Ki) = µ(∪∞i=1 Ei \ ∪∞i=1 Ki)


≤ µ(∪∞i=1 (Ei \Ki)

<

N∑i=1

ε

2i= 1− 1

2N< ε.

Thus, for large enough choice of N , we must have

µ(∪∞i=1 Ei \ ∪Ni=1 Ki) < ε.

The set ∪Ni=1Ki is a closed subset of ∪∞i=1Ei and satisfies the γ condition. Hence, F is closed undercountable unions. Since F contains the closed subsets, it now contains the open sets as well. Weconclude F contains the Borel sets.If µ(X) =∞, given a set B with µ(B) <∞, note SB = E ∩B : E ∈ S is a sigma-algebra forX and µB defined as the restriction of µ to this sigma-algebra is finite. Apply the argument for thefinite measure case to µB to get the desired result.

Theorem 12.3.3 Finite Measure Borel Sets Can Be Approximated By Open Sets

LetX be a metric space, µ as Borel measure onX , ε > 0 andB a Borel set. If µ(X) <∞or more generally, if B is contained in a countable union of open sets Ui, each of finitemeasure, then B is contained in an open set G with µ(G \B) < ε.

Proof 12.3.3We use Theorem 12.3.2 here. Since the µ(X) is finite, µ(BC) is finite. So there is a closed set K inBC so that µ(BC \K) < ε. Let the open set G = KC . Then µ(G \B) < ε and we are done.In the more general situation, choose a closed set Ki contained in each Ui \B so that

µ

((Ui \B) \Ki

)<

ε

2i.

Next, note that B ∩Ui ⊆ Ui \Ki which is an open set. Define G = ∪∞i=1 (Ui \Ki). Then G is open,contains B and µ(G \B) < ε.

12.3.2 Approximating Measurable Functions

We can use Theorem 12.3.1 in many ways. One way is to construct continuous functions on suitabledomains which approximate summable functions. Let’s start with the characteristic function of themeasurable set E, IE .

Theorem 12.3.4 Continuous Approximation Of A Characteristic Function

Let E in <k be Lebesgue measurable. Then given ε > 0, there is a continuous function φEso that ||IE − φE ||1 < ε. Moreover, if E is bounded, the support of φE is compact.

Proof 12.3.4Using Theorem 12.3.1, given ε > 0, we see there is a closed set F and an open set G with F ⊆ E ⊆G with µ(FC ∩G) < ε

2 . For any subset C in <k, it is easy to see the distance function

d(x,C) = infd(x, y) : y ∈ C


is continuous on <k, where d denotes the standard Euclidean metric. It follows that the function fdefined on <k by

f(x) =d(x,GC)

d(x,GC) + d(x, F )

is continuous and satisfies φE is 1 on F and 0 onGC and 0 ≤ f(x) ≤ 1 always. Since φE is nonzeroon GC , it follows that φE has compact support if E is bounded.

Finally, note

∫|φE − IE | dµ =

∫F

|φE − IE | dµ +

∫FC∩E

|φE − IE | dµ

+

∫EC∩G

|φE − IE | dµ +

∫GC|φE − IE | dµ.

However, since E ⊆ G, GC ⊆ EC which tells us both IE and φE are 0 on GC . Hence, the lastintegral vanishes. Also, note FC ∩ E ⊆ FC ∩G and EC ∩G ⊆ FC ∩G. Thus,∫FC∩E

|φE − IE | dµ +

∫EC∩G

|φE − IE | dµ ≤ µ(FC ∩ E) + µ(EC ∩ E) ≤ 2µ(FC ∩G) < ε.

This allows us to conclude∫|φE − IE | dµ < ε which is the desired result.

Next, we see we can approximate simple functions arbitrarily close in the L1 “norm”.

Theorem 12.3.5 Continuous Approximation Of A Simple Function

Let µ denote Lebesgue measure in <k. Let φ : <k− > < be a simple function. Then givenany ε > 0, there is a continuous function g so that

∫|φ − g| dµ < ε. Moreover, if the

support of φ is bounded, the support of g is compact.

Proof 12.3.5The simple function φ has the standard representation φ =

∑ni=1 aiIEi where the numbers ai are

distinct and nonzero and the sets Ei are measurable and disjoint. Let A = max1≤i≤n |ai|. ByTheorem 12.3.4, there are continuous functions gi so that

∫(IEi−gi)dµ < ε

A n . Let g =∑ni=1 aigi.

Then

∫|g − φ| dµ =

n∑i=1

|ai||IEi − gi| dµ

≤ A

n∑i=1

ε

A n< ε.

It is clear that g has compact support if ∪ni=1Ei is bounded.

Now, we can approximate a summable function f with a continuous function as well.

Theorem 12.3.6 Approximation Of A Summable Function With A Continuous Function

Let f be summable on <k with respect to Lebesgue measure. Then given ε > 0, there is acontinuous function of compact support g so that

∫|f − g| dµ < ε.


Proof 12.3.6First, define the functions fn by

fn = f I[−n,n].

Then, for all n, fn is dominated by the summable function f and we also know fn → f . By theDominated Convergence Theorem, we then have

∫fn dµ→

∫f dµ. Hence, given ε > 0, there is an

N so that if n > N , then

∫|f − fn| dµ =

∫[−n,n]C

|f | dµ <ε

4.

Choose a particular p > N . Then since fp is summable, we know there are sequences of simplefunctions, (φj) and (ψj) so that φj ↑ f+

p and ψj ↑ f−p with fp = f+p − f−p . We know also that∫

φj dµ ↑∫f+p dµ and

∫ψj dµ ↑

∫f−p dµ. These simple functions are constructed using the

technique given in Theorem 8.8.2. Since fp has compact support, it follows that for each and eachj these simple functions have compact support also. We see there is an M and a simple functionζ = φM − ψM so that

∫(fp − ζ) dµ < ε

4 . Using Theorem 12.3.5, there is a continuous function ofcompact support g so that

∫(ζ − g) dµ < ε

4 . Combining, we see

∫|f − g| dµ ≤

∫|f − fp| dµ +

∫|fp − ζ| dµ +

∫|ζ − g| < ε.

This result allows us to show that L1 is separable.

Theorem 12.3.7 L1 Is Separable

Let [a, b] be a finite interval. The space ([a, b],M, µ) where µ is Lebesgue measure on[a, b] andM is the sigma-algebra of Lebesgue measurable subsets is separable.

Proof 12.3.7Let f be a summable function. By Theorem 12.3.6 there is a continuous function g so that

∫ ba|f −

g|dµ < ε4 . Further, there is a polynomial p on [a, b] by the Weierstrass Approximation Theorem 5.1.2

so that supx∈[a,b] |g(x)− p(x)| < ε4(b−a) . Further, there is a polynomial q with rational coefficients

so that supx∈[a,b] |p(x)− q(x)| < ε4(b−a) . Combining, we have

∫|f − q| dµ ≤

∫|f − g| dµ +

∫|g − p| dµ +

∫|p− q| dµ

≤ ε

4+ (b− a) ||g − p||∞ + (b− a) ||p− q||∞

≤ ε

4+

ε

4+

ε

4< ε.

since the set of all polynomials on [a, b] with rational coefficients is countable, we see we have shownL1 is separable here.

Comment 12.3.1 It is straightforward to extend these results to any 1 ≤ p <∞. We can also extendthe separability result to a bounded subset Ω of <k but we will not do that here.

12.4. NON MEASURABLE SETS 255

Comment 12.3.2 We see Theorem 12.3.6 tells us the continuous functions with compact support aredense in L1 on <k.

12.4 The Existence Of Non Lebesgue Measurable SetsWe now show that there are subsets of <n which are not Lebesgue measurable. We begin by showingLebesgue measure is translation invariant: this means if E is measurable, so it t+E for any t in <nand µ(E) = µ(E + t).

Theorem 12.4.1 Lebesgue Measure Is Translation Invariant

Let E in <n be Lebesgue measurable and let t in <n be arbitrary. Then, µ∗(E) = µ∗(t+E), t+ E is also measurable and hence, µ(E) = µ(T + E).

Proof 12.4.1We will provide a sketch. The proof is left as an exercise.

Step 1: First, to show µ∗(E) = µ∗(t+ E) for all t, we use a standard Lebesgue Cover argument.

Step 2:

we show if E is measurable, so is t + E. This is done by showing the set t + E satisfies theCaratheodory condition.

• Prove T ∩ (t+B) = (T − t) ∩B + t for all t and sets B and T .

• Prove (t+B)C = t+BC .

• Finally, prove t + E satisfies the Caratheodory condition which essentially finishes theproof.

The translation invariance of Lebesgue measure is important in the construction of a non Lebesguemeasurable set. We establish this via two preliminary results which we put into Lemmas and thenthe final theorem.

Lemma 12.4.2 Non Measurable Set Lemma 1

Let θ be in (0, 1). Let E be a measurable set in <n with µ(E) > 0. Then, there is an openinterval I so that µ(E ∩ I) > θµ(I).

Proof 12.4.2First do the case µ(E) is finite. Start with a Lebesgue cover of E using the supremum tolerancelemma for positive epsilon δ. If you think about this carefully, by choosing δ right this can be phrasedas

µ(E) >1

1 + δ

∑n

|In|.

Now pick δ so that 11+δ = θ. Then we know µ(E) > θ

∑n |In|. The result then follows.

If the measure ofE is not finite and for all intervals J we had µ(E∩J) = 0, we would find µ(E) = 0as well which is not possible. So there must be one interval J with µ(E ∩ J) positive. Now use thefirst part to get the result.


Lemma 12.4.3 Non Measurable Set Lemma 2

Given E in <n, we define the set of all differences x− y for x and y in E to the differenceset of E which we denote byEd. IfE is measurable in<n and if µ(E) > 0, theEd containsa neighborhood of 0.

Proof 12.4.3This one is a bit trickier. First choose λ so that

1− 1

2n+1< λ < 1.

Then by Lemma 12.4.2, we can find an open interval I so that λµ(I) < µ(E∩I . Let δ be the smallestedge length of the interval I and let J be the open interval whose ith edge is (− δ2 ,

δ2 ).

• Show if x is from J , then I + x contains the center of I .

• For any x from J , look at I and I + x on their ith edge. For convenience, let the point x fromJ satisfy 0 ≤ xi <

di2 . Then, show the common part of I and I + x here has length larger

than di2 . Thus, µ(I ∩ (I + x)) > ( 1

2n µ(I).

• Then show for x in J , µ(I ∩ (I + x)) < 2λµ(I).

• For any x from J , let A = E ∩ I and B = E ∩ I + x. both µ(A) and µ(B) are larger thanλµ(I). If A and B were disjoint, this would imply µ(A ∪ B) > 2λµ(I). However, we haveshown µ(A ∪B) < 2λµ(I) which is a contradiction. Hence A and B can not be disjoint.

The above tells us E ∩ I and (E ∩ I) + x have a common point for each x in J . It then follows wecan write x as u− v for some u and v from E implying x is in Ed. This completes the proof.

We can now show there is a set in <n which is not Lebesgue Measurable. We already know thesigma-algebra of Lebesgue measurable subsets, L , of <n strictly contains the Borel sigma algebra,B. Theorem 12.4.4 tells us that L is strictly contained in power set of <n. From this, we can thenprove every measurable set of positive measure contains a non measurable set.

Theorem 12.4.4 Non Lebesgue Measurable Set

There is a non Lebesgue measurable set in <n. Further, any set of positive measure con-tains a non measurable set.

Proof 12.4.4Let z be a fixed irrational number and let the set M = Q. It is easy to show Mn is countable andclosed under addition and subtraction and dense in <n.Then, to prove the result:

• Prove the relation∼ on <n×<n defined by x ∼ y if x−y is inMn is an equivalence relation.

• Let µ∗(E) be defined by

µ∗(E) = sup µ(F ) : F ⊆ E, F measurable .

Prove if E is measurable, then µ(E) = µ∗(E) = µ∗(E).

• Let M be the collection of equivalence classes determined by ∼ and use the Axiom Of Choice(look this up!) to pick a unique point from each equivalence class to form the set S. Recall, we

12.5. METRIC SPACES 257

know that µ∗(S) = sup µ(E) : E ⊆ S, E measurable. If µ(E) > 0, use Lemma 12.4.3 tonote Ed contains a neighborhood of 0. Since M is dense, Ed ∩M contains a point x which isnot 0. Prove this point x = u − v with u, v from E implying u ∼ v with x 6= 0 which is notpossible by the way S was constructed. This implies µ∗(E) = 0. Thus, by the previous result,if S were measurable, µ(S) = 0.

• Let (xn) be an enumeration ofM . Then, show<n = ∪n(S+xn) and prove that S measurableimplies ∞ = 0 giving a contradiction. We must therefore conclude that S can not be ameasurable set.

• Show if E is measurable with positive measure, then E contains a non measurable set.

Now go back and reread Section 13.4 where we used these results!

12.4.1 ExercisesExercise 12.4.1 Prove Lemma 12.4.2 in detail.

Exercise 12.4.2 Prove Lemma 12.4.3 in detail.

Exercise 12.4.3 Prove Theorem 12.4.4 in detail.

12.5 Metric Spaces Of Finite Measure SetsWe can do many things with measurable sets. This material is a connection of sorts to the standarddiscussions of linear analysis.

Theorem 12.5.1 The Metric Space of Finite Measurable Sets

Let M be the collection of all Lebesgue measurable sets in <n with finite measure. Recall,the symmetric difference of sets A and B is

A∆B =

(A ∩BC

)∪(AC ∩B

).

Then

1. The relation ∼ on M ×M defined by

A ∼ B if µ(A∆B) = 0

is an equivalence relation.

2. Let M denote the set of all equivalence classes with respect to this relation and let[A] be a typical equivalence class with representative A. Prove M is a metric spacewith metric D defined by

D([A], [B]) := µ(A∆B).

Proof 12.5.1A sketch of the proof is given below:


• First prove if A and B are in M ,

|µ(A)− µ(B)| ≤ µ(A∆B).

• Then, prove if A ∼ E and B ∼ F , then µ(A∆B) = µ(E∆F ).Hint: Show

|µ(A∆B)− µ(E∆F )| ≤ µ(A∆E) + µ(B∆F ).

• Prove D is a metric which requires you prove D is well-defined by showing its value is inde-pendent of the choice of equivalence class representatives.

Theorem 12.5.2 The Metric Space of All Finite Measure Sets Is Complete

The metric space (M , D) is complete.

Proof 12.5.2We provide a detailed sketch. Let (En) be a Cauchy sequence.

1. Prove there is a subsequence (Enm) of (En) so that

µ(Enm∆En) <1

2mfor all n > nm,

µ(Enk∆Enm) <1

2mfor all nk > nm.

2. For expositional convenience, let Gm = Enm in what follows. Let H = lim supGm andG = lim inf Gm.

• Prove µ(H \Gm)→ 0 and µ(Gm \G)→ 0.

• Prove µ(H∆Gm)→ 0. This argument uses the disjoint decomposition idea from Lemma9.1.5.Step 1: Since

µ(HC ∩Gj) = limm∩n≥m(GCn ∩Gj)

overestimate the right hand side by µ(GCn∩Gj) which is also bounded above by µ(Gn∆Gj)which is as small as we want. This shows µ(HC ∩Gj)→ 0.Step 2: First note

H ∩GCj = ∩m ∪n≥m Gn ∩GCjwhich shows H ∩ GCj is a decreasing limit of sets. Now look at ∪n≥mGn ∩ GCj whichis contained in GCj always. However, since the measure of Gj is finite, the measureof its complement is not, so this upper bound is not very useful. For convenience, letBm = ∪n≥mGn. To understand how to use the properties of our sequence Gn, werewrite the union Bm like this:

Um = Gm

Um+1 = Gm+1 \ (Gm)

Um+2 = Gm+2 \ (Gm ∪Gm+1)

...

12.5. METRIC SPACES 259

Um+n = Gm+n \ (Gm ∪Gm+1 ∪ · · · ∪Gm+n−1)

...

From the above, we see for all m > j

µ(Bm ∩Gj) =

∞∑n=0

µ(Um+n ∩GCj ).

However, we can overestimate a bit using

µ(Um ∩GCj ) <1

2m

µ(Um+1 ∩GCj ) = µ(Gm+1 ∩GCm ∩GCj )

≤ µ(Gm+1 ∩GCm)

<1

2m.

µ(Um+2 ∩GCj ) = µ(Gm+2 ∩GCm ∩GCm+1 ∩GCj )

≤ µ(Gm+2 ∩GCm+1)

<1

2m+1.

...

µ(Um+n ∩GCj ) = µ(Gm+n ∩GCm ∩ · · · ∩GCm+n−1 ∩GCj )

≤ µ(Gm+n ∩GCm+n−1)

<1

2m+n−1.

Thus,

µ(Bm ∩Gj) <1

2m+

∞∑n=1

1

2m+n−1

=1

2m+

1

2m−1.

We conclude

µ(H \Gj) = limm

µ(Bm ∩GCj ) ≤ limn

(1

2m+

1

2m−1) = 0.

This shows the desired result.

3. Prove µ(H∆Em)→ 0 also which proves completeness.

Comment 12.5.1 We can also prove µ(G∆H) = 0 so that G ∼ H .


12.5.1 ExercisesExercise 12.5.1 Prove Theorem 12.5.1.


Chapter 13

Cantor Set Experiments

We now begin a series of personal investigations into the construction of an important subset of [0, 1]called the Cantor Set. We follow a great series of homework exercise outlined, without solutions, ina really hard but extraordinarily useful classical analysis text by Stromberg, (Stromberg (11) 1981).

13.1 The Generalized Cantor SetLet (an) for n ≥ 0 be a fixed sequence of real numbers which satisfy

a0 = 1, 0 < 2an < an−1 (13.1)

Define the sequence (dn) by

dn = an−1 − 2an

Note each dn > 0. We can use the sequence (an) to define a collection of intervals Jn,k and In,k asfollows.

(0) J0,1 = [0, 1] which has length a0.

(1) J1,1 = [0, a1] and J1,2 = [1 − a1, 1]. You can see each of these intervals has length a1. Welet W1,1 = J1,1 ∪ J1,2 and I1,1 = J0,1 −W1,1 where the minus symbol used here representsset difference. This step creates an open interval of [0, 1] which has length d1 > 0. LetP1 = J1,1 ∪ J1,2. This is a closed set.

(2) Set J2,1 = [0, a2], J2,2 = [a1 − a2, a1], J2,3 = [1− a1, 1 + a2 − a1], and J2,4 = [1− a2, 1].These 4 closed subintervals have length a2. It is not so mysterious how we set up the J2,k

intervals. Step (1) created a closed interval [0, a1], an open interval (a1, 1 − a1) and anotherclosed interval [1 − a1, 1]. The first closed subinterval is what we have called J1,1. Divideit into three parts; the first part will be a closed interval that starts at the beginning of J1,1

and has length a2 and the third part will be closed interval of length a2 that ends at the lastpoint of J1,1. When these two closed intervals are subtracted from J1,1, an open interval willremain. The length of J1,1 is a1. So the open interval must have length a1−2a2 = d2. A littlethought tells us that the first interval must be [0, a2] (which we have named J2,1 ) and the thirdinterval must be [a1 − a2, a1] (which we have named J2,2). To get the intervals J2,3 and J2,4,we divide J1,2 into the same type of three subintervals as we did for J1,1. The first and thirdmust have length a2 which will give an open interval in the inside of length d2. This will giveJ2,3 = [1− a1, 1− a1 + a2] and j2,4 = [1− a2, 1].

261

262 CHAPTER 13. CANTOR SETS

Then let W2,1 = J2,1 ∪ J2,2, and W2,2 = J2,3 ∪ J2,4. Then create new intervals by lettingI2, 1 = J1,1 − W2,1 and I2, 2 = J1,2 − W2,2. We have now created 4 open subintervalsof length d2. Let P2 = J2,1 ∪ J2,2 ∪ J2,3 ∪ J2,4. We can write this more succinctly asP2 = ∪J2,k|1 <= k <= 22. Again, notice that P2 is a closed set that consists of 4 closedsubintervals of length a2.

Let’s look even more closely at the details. A careful examination of the process above withpen and paper in hand gives the following table that characterizes the left hand endpoint ofeach of the intervals J2,k.

J2,1 0J2,2 a2 + d2

J2,3 2a2 + d2 + d1

J2,4 3a2 + 2d2 + d1

Since we know the left hand endpoint and the length is always a2, this fully characterizes thesubintervals J2,k. Also, as a check, the last endpoint 3a2 + 2d2 + d1 plus one more a2 shouldadd up to 1. We find

4a2 + 2d2 + d1 = 4a2 + 2(a1 − 2a2) + (a0 − 2a1)

= a0 = 1.

(3) Step (2) has created 4 closed subintervals J2,k of length a2 and 2 new open intervals I2,i oflength d2. There is also the first open interval I1,1 of length d1 which was abstracted from[0, 1]. Now we repeat the process described in Step (2) on each closed subinterval J2,k. Wedo not need to use the auxiliary sets W3,i now as we can go straight into the subdivision algo-rithm. We divide each of these intervals into 3 pieces. The first and third will be of length a3.This leaves an open interval of length d3 between them. We label the new closed subintervalsso created by J3,k where k now ranges from 1 to 8. The new intervals have left hand endpoints

J3,1 0J3,2 a3 + d3

J3,3 2a3 + d3 + d2

J3,4 3a3 + 2d3 + d2

J3,5 4a3 + 2d3 + d2 + d1

J3,6 5a3 + 3d3 + d2 + d1

J3,7 6a3 + 3d3 + 2d2 + d1

J3,8 7a3 + 4d3 + 2d2 + d1

Each of these subintervals have length a3 and a simple calculation shows (7a3 + 4d3 + 2d2 +d1) + a3 = 1 as desired. There are now 4 more open intervals I3,i giving a total of 6 opensubintervals arranged as follows:

Parent LengthI1,1 J0,1 d1

I2,1 J1,1 d2

I2,2 J1,2 d2

I3,1 J2,1 d3

I3,2 J2,2 d3

I3,3 J2,3 d3

I3,4 J2,4 d4

13.2. REPRESENTATION 263

We define P3 = ∪J3,k|1 <= k <= 23 and note that P1 ∩ P2 ∩ P3 = P3.

We can, of course, continue this process recursively. Thus, after Step n, we have constructed 2n

closed subintervals Jn,k each of length an. The union of these subintervals is labeled Pn and istherefore defined by Pn = ∪Jn,k|1 <= k <= 2n. The left hand endpoints of Jn,k can be writtenin a compact and illuminating form, but we will delay working that out until later. Now, we caneasily see the form of the left hand endpoints for the first few intervals:

Jn,1 0Jn,2 an + dnJn,3 2an + dn + dn−1

Jn,4 3an + 2dn + dn−1

Definition 13.1.1 The Generalized Cantor Set

Let (an), N ≥ 0 satisfy Equation 13.1. We call such a sequence a Cantor Set GeneratingSequence and we define the Cantor Set generated by (an) to be the set P = ∩∞n=1Pn,where the sets Pn are defined recursively via the discussion in this section. We will denotethe generalized Cantor Set generated by the Cantor Sequence (an) by Ca.

Comment 13.1.1 The Cantor Set generated by the sequence (1/3n), n ≥ 0 is very famous and iscalled the Middle Thirds set because we are always removing the middle third of each interval in theconstruction process. We will denote the Middle Third Cantor set by C.

Exercise 13.1.1 Write out the explicit endpoints of all these intervals up to and including Step 4.Illustrate this process with clearly drawn tables and graphs.

Exercise 13.1.2 Write out explicitly P1, P2, P3 and P4. Illustrate this process with clearly drawntables and graphs.

Exercise 13.1.3 Do the above two steps for the choice an = 3−n for n >= 0. Illustrate this processwith clearly drawn tables and graphs.

Exercise 13.1.4 Do the above two steps for the choice an = 5−n for n >= 0. Illustrate this processwith clearly drawn tables and graphs.

Exercise 13.1.5 As mentioned, the above construction process above can clearly be handled viainduction. Prove the following:

(a) Pn−1 − Pn = ∪In,k | 1 <= k <= 2n−1

(b) Let P = ∩∞n=0 Pn. Then P0 − P = ∪∞n=1

(Pn−1 − Pn

)

Exercise 13.1.6 Prove any Cantor set is in the Borel σ-algebra B.

13.2 Representing The Generalized Cantor SetWe are now in a position to prove additional properties about the Cantor Set Ca for a Cantorgenerating sequence (an). Associate with (an) the sequence (rn) whose entries are defined byrn = an−1 − an. Let S denote the set of all sequences of real numbers whose values are either 0 or1; i.e. S = x = (xn) | xn = 0 or xn = 1. Now define the mapping f : S → Ca by


f(x) =

∞∑n=1

xnrn (13.2)

Theorem 13.2.1 Representing The Cantor Set

1. f is well - defined.

2. f(x) is an element of Ca.

3. f is 1− 1 from S to Ca.

4. f is onto Ca.

Proof 13.2.1You will prove this Theorem by establishing a series of results.

Exercise 13.2.1 For any Cantor generating sequence (an), we have limn an = 0.

Exercise 13.2.2 Show

∞∑j=n+1

rj = limm

m∑j=n+1

rj = limm

(an − am) = an

Exercise 13.2.3 rn >∑∞j=n+1 rj .

Exercise 13.2.4 For n >= 1 and any finite sequence (x1, x2, ..., xn) of 0’s and 1’s, define the closedinterval

J(x1, ..., xn) = [

n∑j=1

xjrj , an +

n∑j=1

xjrj ]

Show

1. Show J(0) = [0, a1] = J1,1.

2. Show J(1) = [r1, a1 + r1] = [1− a1, 1] = J1,2.

3. Now use induction on n to show that the intervals J(x1, ..., xn) are exactly the 2n intervalsJn,k for 1 <= k <= 2n that we described in the previous section.

Hint 13.2.1 i.e. assume true for n−1. Then we can assume that there is a unique (x1, ..., xn−1)choice so that Jn−1,k = J(x1, ..., xn−1).

Recall how the J’s are constructed. At Step n− 1, the interval Jn−1,k is used to create 2 moreintervals on level n by removing a piece. The 2 intervals left both have length an and we woulddenote them by Jn,2k−1 and Jn,2k. Now use the definition of the closed intervals J(x1, ..., xn)to show that (remember our x1, ..., xn−1 are fixed)

J(x1, ..., xn−1, 0) = Jn,2k−1

J(x1, ..., xn−1, 1) = Jn,2k

This will complete the induction.

13.3. CANTOR FUNCTION 265

Exercise 13.2.5 Let x be in S. Show that f(x) is in J(x1, ..., xn) for each n.Sketch Of Argument: We know that each J(x1, ..., xn) = Jn,k for some k. Let this k be writ-ten k(x, n) to help us remember that it depends on the x and the n. Also remember that 1 <=k(x, n) <= 2n. So f(x) is in Jn,k(x,n) which is contained in Pn Hence, f(x) is in Pn for all nwhich shows f(x) is in Ca. This shows f maps S into Ca.

Exercise 13.2.6 Now let x and y be distinct in S. Choose an index j so that xj is different from yj .Show this implies that f(x) and f(y) then belong to different closed intervals on the jth level. Thisimplies f(x) is not the same as f(y) and so f is 1− 1 on S.

Exercise 13.2.7 Show f is surjective. To do this, let z be in Ca. Since z is in P1, either z is in J(0)or z is in J(1). Choose x1 for that z is in J(x1). Then assuming x1, ..., xn−1 have been chosen, wehave z is in J(x1, ..., xn−1). Now z is in

Pn ∩ J(x1, ..., xn−1) = J(x1, ..., xn−1, 0) ∪ J(x1, ..., xn−1, 1).

This tells us how to choose xn.

Hence, by induction, we can find a sequence (xn) in S so that z is in intersection over n ofJ(x1, ..., xn). But by our earlier arguments, f(x) is in the same intersection!

Finally, each of these closed intervals has length an which we know goes to 0 in the limit on n.So z and f(x) are both in a decreasing sequence of sets whose lengths go to 0. Hence z and f(x)must be the same. (This uses what is called the Cantor Intersection Theorem).

We can also prove a result about the internal structure of the generalized Cantor set: it can notcontain any open intervals.

Exercise 13.2.8 Prove Ca contains no open intervals.

In addition, we have the following result:

Exercise 13.2.9 The limit of 2nan always exists and is in [0, 1].

13.3 The Cantor Function

We now prove additional interesting results that arise from the use of generalized Cantor sets via aseries of exercises that you complete. As usual, let (an) be a Cantor Set generating sequence. Usingthe function f defined in the previous section, let’s define the mapping φ by

φ((xn)) =

∞∑j=1

xj (1/2j)

Hence, φ : S → [0, 1]. and φ f : S → [0, 1]. Let the mapping Ψ = φ f−1. Note Ψ : Ca → [0, 1].

Exercise 13.3.1 φ maps S one to one and onto [0, 1] with a suitable restriction on the base 2 repre-sentation of a number in [0, 1].

Exercise 13.3.2 x < y in Ca implies Ψ(x) ≤ Ψ(y).


Exercise 13.3.3 Ψ(x) = Ψ(y) if and only if (x, y) is one of the intervals removed in the Cantor setconstruction process, i.e.

(x, y) =

(n−1∑j=1

xjrj + an,

n−1∑j=1

xjrj + rn

)

Exercise 13.3.4 In the case where Ψ(x) = Ψ(y) extend the mapping Ψ to [0, 1]−Ca by

Ψ(t) = Ψ(x) = Ψ(y), x < t < y.

Finally, define Ψ(0) = 0 and Ψ(1) = 1. Prove Ψ : [0, 1] → [0, 1] is a non decreasing continuousmap of [0, 1] onto [0, 1] and is constant on each component interval of [0, 1]−Ca where componentinterval means the In,k sets we constructed in the Cantor set construction process.

Comment 13.3.1 If Ca is the Cantor set constructed from the sequence (1/3n), we call Ψ theLebesgue Singular Function.

Now, let C be a Cantor set constructed from the generating sequence (an) where lim 2nan = 0.Let Ψ be the mapping discussed above for this C. Define the mapping g : [0, 1] → [0, 1] byg(x) = (Ψ(x) + x)/2.

Exercise 13.3.5 Prove g is strictly increasing and continuous from [0, 1] onto [0, 1].

Exercise 13.3.6 Prove that

g(

∞∑j=1

xj rj) =

∞∑j=1

xj r′j

where rj ′ = (1/2j + rj)/2.

Exercise 13.3.7 Prove C′ = g(C) is also a generalized Cantor set.

Comment 13.3.2 Note that the sequence a′j = (1/2)(1/2j+aj) is also a Cantor generating sequencethat gives the desired rj ′ for the previous exercise.

Exercise 13.3.8 Compute the outermeasure of the Cantor set generated by an when lim 2nan = 0and also the outermeasure of the Cantor set C′ = g(C).

Exercise 13.3.9 Compute the Lebesgue measure of the Cantor set generated by an when lim 2nan =0 and also the Lebesgue measure of the Cantor set C′ = g(C).

Exercise 13.3.10 If a subset of < has content 0, it also has Lebesgue measure 0.

Next, we show this function g is of great importance in developing a better understanding of mea-sures.

13.4 Interesting Consequences

We now argue the Borel sigma - algebra is strictly contained in the Lebesgue sigma - algebra byusing the special functions we have constructed here. If C is a Cantor set constructed from thegenerating sequence (an) where lim 2nan = 0, the Lebesgue measure of C is 0. Further, if Ψ isthe mapping we defined earlier associated with C we can define the mapping g : [0, 1] → [0, 1]by g(x) = (Ψ(x) + x)/2. The mapping g is quite nice: it is 1 − 1, onto, strictly increasing andcontinuous. We also showed in the exercises in Section 13.3 that g(C) is another Cantor set withlim 2na′n = 1/2, where (a′n) is the generating sequence for g(C). We also know

13.4. CONSEQUENCES 267

µB(C) = µL(C) = 0

µB(g(C)) = µL(g(C)) = 1/2.

A nonconstructive argument using the Axiom of Choice then allows us to show any Lebesgue mea-surable set with positive Lebesgue measure must contain a subset which is not in the Lebesgue sigma- algebra. So since µL(g(C)) = 1/2, there is a set F ⊆ g(C) which is not isL. Thus, g−1(F ) ⊆ Cwhich has Lebesgue measure 0. Lebesgue measure is a measure which has the property that everysubset of a set of measure 0 must be in the Lebesgue sigma - algebra. Then, using the monotonicityof µL, we have µL(g−1(F )) is also 0. From the above remarks, we can infer something remarkable.

Let the mapping h be defined to be g−1. Then h is also continuous and hence it is measurablewith respect to the Borel sigma-algebra. Note since B ⊆ L, this tells us immediately that h is alsomeasurable with respect to the Lebesgue sigma - algebra. Thus, h−1(U) is in the Borel sigma -algebra for all Borel sets U . But we know h−1 = g, so this tells us g(U) is in the Borel sigma-algebra if U is a Borel set. Hence, if we chose U = g−1(F ), then g(U) = F would have to be aBorel set if U is a Borel set. However, we know that F is not inL and so it is also not a Borel set. Wecan only conclude that g−1(F ) can not be a Borel set. However, g−1(F ) is in the Lebesgue sigma- algebra. Thus, there are Lebesgue measurable sets which are not Borel! Thus, the Borel sigma -algebra is strictly contained in the Lebesgue sigma - algebra!

We can use this example to construct another remarkable thing. Using all the notations fromabove, note the indicator function of CC , the complement of C, is defined by

ICC (x) =

1 x ∈ CC

0 x ∈ C.

We see f = ICC is Borel measurable. Next, define a new mapping like this:

φ(x) =

1 x ∈ CC

2 x ∈ C \ g−1(F )3 x ∈ g−1(F ).

Note that φ = f a.e. with respect to Borel measure. However, φ is not Borel measurable becauseφ−1(3) is the set g−1(F ) which is not a Borel set.

We conclude that in this case, even though the two functions were equal a.e. with respect toBorel measure, only one was measurable! The reason this happens is that even though C has Borelmeasure 0, there are subsets of C which are not Borel sets! We discussed this example earlier inChapter 9 as well.

Chapter 14

Lebesgue Stieljes Measure

We now show how to construct a very important class of measures using right continuous monotonicfunctions on <. We know that if λ on < is a measure, then for any point b

limx→b

λ

((x, b]

)= λ

((a, b]

).

In fact, if we define the function h by h(x) = λ([a, x]), then we see that h must be continuous fromthe right at each a; i.e. limx→a+ h(x) = h(a+) = h(a). On the other hand, using a monotonicsequence of sets that decrease, we see

limx→b−

λ

((a, x]

)= h(b−).

Hence, it is not required that h must be continuous from the left at each b. Now if g is any monotoneincreasing function on [a, b], g is of bounded variation and so it always has right and left hand limitsat any point. From the discussion above, we see we can define outer measures in a more generalfashion. For Lebesgue Outer Measure, we use the premeasure τ((a, b]) = b − a which correspondsto the continuous, strictly monotone increasing choice g(x) = x. We can also define a premeasureτg by τg((a, b]) = g(b)− g(a). This will generate a measure in the usual way. Call this measure λg .Then

λg

([a, b]

)= λg

(a)

+ λg

((a, b]

).

But since

λg

(a)

= limε→0+

λg

((a− ε, a+ ε)

)= g(a+)− g(a−),

we see for all x > a,

λg

([a, x]

)= g(a+)− g(a−) + g(x)− g(a).

269

270 CHAPTER 14. LEBESGUE STIELJES MEASURE

Letting x→ a+, we find

λg

(a)

= limx→a+

(g(a+)− g(a−) + g(x)− g(a)

).

Simplifying, we have

limx→a+

g(x) = g(a).

which tells us the monotonic non decreasing function g we use to construct a new measure must beright continuous.

14.1 Lebesgue - Stieljes Outer Measure and MeasureFrom the discussion above, to construct new measures, we choose any g which is non -decreasingfunction on < and continuous from the right. Moreover, the unbounded limits are well - definedlimx→−∞ g(x) and limx→∞ g(x). These last two limits could be −∞ and ∞ respectively. Then,define the mapping τg on U by

τg(∅) = 0,

τg

((a, b]

)= g(b) − g(a),

τg

((−∞, b]

)= g(b) − lim

x→−∞g(x),

τg

((a,∞)

)= lim

x→∞g(x) − g(a),

τg

((−∞,∞)

)= lim

x→∞g(x) − lim

x→−∞g(x).

This defines τg on the collection of sets U consisting of the empty set, intervals of the form (a, b]for finite numbers a and b and unbounded intervals of the form (−∞, b] and (a,∞). Let A be thealgebra generated by finite unions of sets from U . Note A contains <.

Let’s extend the mapping τg to be additive on A. If E1, E2, . . . , En is a finite collection of disjointsets in A, we extend the definition of τg to this finite disjoint unions as follows:

τg

( n⋃i−1

Ei

)=

n∑i=1

τg(Ei). (14.1)

Lemma 14.1.1 Extending τg To Additive Is Well - Defined

The extension of τg from U to the algebra A is well - defined; hence, τg is additive on A.

Proof 14.1.1For (a, b] ∈ A, write

(a, b] =

n⋃i=1

(ai, bi],

14.1. LEBESGUE-STIELJES 271

for any positive integer n with a1 = a, bn = b and the in between points satisfy ai+1 = bi for alli. Of course, there are many such decompositions of (a, b] we could choose. Also, these are the onlydecompositions we can have. If we use the unbounded sets, we can not recapture (a, b] using a finitenumber of unions! Then, using Equation 14.1, we have

τg((a, b]) =

n∑i=1

τg((ai, bi])

=

n∑i=1

g(bi) − g(ai).

But since ai+1 = bi, this sum collapses to

τg((a, b]) = g(b) − g(a).

This was the original definition of τg on the element (a, b] in U . We conclude the value of τg onelements of the form (a, b] is independent of the choice of decomposition of it into a finite union ofsets from U .

For an unbounded interval of the form (a,∞), any finite disjoint decomposition can have only oneinterval of the form (b,∞) giving (a,∞) = (a, b] ∪ (b,∞), with the piece (a, b] written as any finitedisjoint union (a, b] = ∪ni=1 (ai, bi] as before. The same arguments as used above then show τg iswell - defined on this type of element of U also. We handle the sets (−∞, b] is a similar fashion.

Next, if we look at any arbitrary A in A, then A can be written as a finite union of membersA1, . . . , Ap of U . Each of these elements Ai can then be written using a finite disjoint decompositioninto intervals (aij , bij ], 1 ≤ j ≤ p(i) as we have done above. Thus,

A = ∪mi=1 ∪p(i)j=1 (aij , bij ]

where we abuse notation, for convenience, by noting it is possible a11 = −∞ and bm p(m) =∞. Wesimply interpret a set of the form (a,∞] as (a,∞). We then combine these intervals and relabel asnecessary to write A as a finite disjoint union

A = ∪Ni=1 (ai, bi]

with bi ≤ ai+1 and again it is possible that a1 = −∞ and bN =∞. We therefore know that

τg(A) = ∪Ni=1 τg((ai, bi]).

Now assume A has been decomposed into another finite disjoint union, A = ∪Mj=1Bj , each Bj ∈ A.Let

Cj = i |(ai, bi] ⊆ Bj.

Note a given interval (ai, bi] can not be in two different sets Bj and Bk because they are assumeddisjoint. Hence, we have

Bj = ∪i∈Cj (ai, bi]

and

τg(Bj) =∑i∈Cj

τg((ai, bi]).


Thus,

M∑j=1

τg(Bj) =

M∑j=1

∑i∈Cj

τg((ai, bi])

=

N∑i=1

τg((ai, bi]).

This shows that our extension for τg is independent of the choice of finite decomposition and so theextension of τg is a well - defined additive map on A

We can now apply Theorem 11.3.3 to conclude that since the covering family A is an algebra andτg is additive on A , the σ - algebra,Mg , generated by τg contains A and the induced measure, µg ,is regular. Next, we want to know that µg(A) = τg(A) for all A in A. To do this, we will prove theextension τg is actually a pseudo-measure. Thus, we will be able to invoke Theorem 11.3.4 to get thedesired result.

Lemma 14.1.2 Lebesgue - Stieljes Premeasure Is a Pseudo-Measure

The mapping τg is a pseudo-measure on A.

Proof 14.1.2We need to show that if (Tn) is a sequence of disjoint sets from A whose union ∪nTn is also in A,then

τg( ∪n Tn) =∑n

τg(Tn).

First, notice that if there was an index n0 so that τg(Tn0) = ∞, then letting B = ∪n Tn \ Tn0

, wecan write ∪nTn as the finite disjoint union B ∪ Tn0

and hence

τg( ∪n Tn) = τg(B) + τg(Tn0) = ∞.

Since the right hand side sums to∞ in this case also, we see there is equality for the two expressions.Therefore, we can restrict our attention to the case where all the individual Tn sets have finite τg(Tn)values. This means no elements of the form (−∞, b] or (a,∞) can be part of any decomposition ofthe sets Tn. Hence, we can assume each Tn can be written as a finite union of intervals of the form(a, b]. It follows then that it suffices to prove the result for a single interval of the form (a, b].

Since τg is additive on finite unions, if C ⊆ D, we have

τg(D) = τg(C) + τg(D \ C) ≥ τg(C).

Now assume we can write the interval (a, b) as follows:

(a, b] = ∪∞n=1 (ai, bi]

with the sets (ai, bi] disjoint. For any n, we have

(a, b] = ∪nk=1 (ak, bk] ∪ ∪∞k=n+1 (ak, bk].


Therefore

τg

((a, b]

)= τg

(∪nk=1 (ak, bk]

)+ τg

(∪∞k=n+1 (ak, bk]

).

The finite additivity on disjoint intervals then gives us

τg

(∪nk=1 (ak, bk]

)=

n∑k=1

τg

((ak, bk]

)= g(b1) − g(a1) + g(b2) − g(a2) + . . . + g(bn) − g(an).

We know g is non decreasing, thus g(b1)−g(a2) ≤ 0, g(b2)−g(a3) ≤ 0, and so forth until we reachg(bn−1)− g(an) ≤ 0. Dropping these terms, we find

τg

(∪nk=1 (ak, bk]

)≤ g(bn) − g(a1) ≤ g(b) − g(a).

Thus, these partial sums are bounded above and so the series of non negative terms∑n τg((ak, bk])

converges. This tells us that

τg

(∪∞k=1 (ak, bk]

)≤ τg

((a, b]

).

To obtain the reverse inequality, let ε > 0 be given. Then, since the series above converges, theremust be a positive integer N so that if n ≥ N ,

∞∑k=n+1

τg

((ak, bk]

)< ε

We conclude that

τg

((a, b]

)=

n∑k=1

τg

((ak, bk]

)+ τg

(∪∞k=n+1 (ak, bk]

)

≥n∑k=1

τg

((ak, bk]

)+ τg

(∪Kk=n+1 (ak, bk]

)

=

n∑k=1

τg

((ak, bk]

)+

K∑k=n+1

τg

((ak, bk]

).

We know that

limK

K∑k=n+1

τg

((ak, bk]

)= 0.

Thus, letting K →∞, we find for all n > N , that

τg

((a, b]

)≥

n∑k=1

τg

((ak, bk]

).


However, the sequence of partial sums above converges. We have then the inequality

τg

((a, b]

)≥

∞∑k=1

τg

((ak, bk]

).

Combining the two inequalities, we have that our extension τg is a pseudo-measure.

Comment 14.1.1 It is worthwhile to summarize what we have accomplished at this point. We knownow that the premeasure τg defined by the non decreasing and right continuous map g on the algebraof sets, A, generated by the collection U consisting of the empty set, finite intervals like (a, b] andunbounded intervals of the form (−∞, b] and (a,∞) when defined to be additive on A generates aninteresting outer measure µ∗b . We have also proven that the extension τg becomes a pseudo-measureon A. Thus,

(i): The sets A in A are in the σ - algebra of sets that satisfy the Caratheodory condition using µ∗gwhich we denote byMg . We denote the resulting measure by µg . This is because τg is additiveon the algebra A by Theorem 11.3.3.

(ii): We know µg is regular by Theorem 11.3.3 and complete by Theorem 11.1.4.

(iii): We know that µg(A) = τg(A) for all A in A since τg is a pseudo-measure by Theorem 11.3.4.

(iv): Since any open set can be written as a countable disjoint union of open intervals, this meansany open set is inMg becauseMg contains open intervals as they are inA and the σ - algebraMg is closed under countable disjoint unions. This also tells us that the Borel σ - algebra iscontained inMg .

(v): Since open sets are µ∗g measurable, by Theorem 11.2.2, it follows that µ∗g is a metric outermeasure.

Comment 14.1.2 The measures µg induced by the outer measures µ∗g are called Lebesgue - Stieljesmeasures . Since open sets are measurable here, these measures are also called Borel measures .

Comment 14.1.3 So for a given non decreasing right continuous g, we can construct a Lebesgue -Stieljes measure satisfying

µg

((a, b]

)= g(b) − g(a).

So what about the open interval (a, b)? We know that

(a, b) =⋃n

(a, b − 1

n].

Then

µg

((a, b)

)= lim

ng

(b − 1

n

)− g(a)

= g(b−) − g(a).

What about the singleton b? We know

b =⋂n

(b − 1

n, b

]


and so

µg

(b)

= limn

g(b) − g

(b − 1

n

)= g(b) − g(b−).

Note this tells us that the Lebesgue - Stieljes measure of a singleton need not be 0. However, at anypoint b where g is continuous, this measure will be zero. Since our g can have at most a countablenumber of discontinuities, we see there are only a countable number of singleton sets whose measureis non - zero.

We can also prove approximation lemmas for Lebesgue-Stieljes measures also. A typical one is givenbelow.

Theorem 14.1.3 Approximating Sets With A Lebesgue-Stieljes Outer Measure

LetX be a metric space, µ∗g be a Lebesgue-Stieljes outer measure,Mg the induced sigma-algebra of measurable sets and µg the induced measure. Then given any E contained inX , there is a Gδ set G (i.e. a countable intersection of open sets)and a Fσ set (i.e. acountable union of closed sets) F so that F ⊆ E ⊆ G and µg(F ) = µ∗g(E) = µg(G).

Proof 14.1.3If µ∗g(E) = ∞, we can simply choose G = X and be done. Hence, we can assume µ∗g(E) < ∞.Therefore, we know µg is regular and complete and so there is a set F with µ∗g(E) = µg(T ) andEm = ∪n Emn , F = ∩m Em where for each m, the family of sets (Emn ) is in the algebra A used toconstruct the outer measure, satisfies A ⊆ ∪n Emn and

∑n

τg(Emn ) < µ∗(A) +

1

m.

It is straightforward to see we can rewrite each Emn as a countable union of sets of the form (amn , bmn ]

and thus we have Emn = ∪n (amn , bmn ]. For any r > 0, we see Emn = ∪n (amn , b

mn + r), a countable

union of open sets of finite µg measure. Hence, applying Theorem 12.3.3, for any j there is an openset Gj so that µg(Gj \ F ) < 1

j . Hence, if G = ∩jGj , a Gδ set, we have F ⊆ G and

µg(∩jGj \ F ) ≤ µg

( N⋂j=1

Gj \ F)

= µg

( N⋂j=1

(Gj \ F )

)≤ µg(GN \ F ) <

1

N.

As N →∞, we see µg(G \ F ) = 0 and µ∗g(E) = µg(F ) = µg(G).

Now if µg(E) is bounded, we have µ∗g(E) = µg(T ) where T = ∩m ∪n Emn ) as before. Thus, T is aBorel set for which µg(T ) <∞. We can therefore apply Theorem theorem:measapproxtwo, since µgis a Borel measure and < is a metric space, to obtain a closed set Fn ⊆ T so that µg(T \ Fn) < 1

nfor each positive integer n. Let F = ∪nFn, a Fσ set. Then, for all n, we have


µg(T \ F ) = µg(∩nT \ Fn)

≤ µg(T \ Fn) <1

n.

This immediately implies that µg(T \ F ) = 0. Thus, in this case, µ∗g(E) = µg(T ) = µg(F ).

In the case that µ∗g(E) is infinite, we let Ej = E ∩ (j, j + 1] for all integers n. Then E = ∪jEjand the sets Ej are disjoint. Hence, there are Borel sets Tj with bounded µ∗g values satisfyingµ∗g(Ej) = µf (Tj) where Tj = ∩m ∪n Ej,mn ) where the component sets Ej,mn are defined as before.The argument in the first part now applies. There are Fσ sets Fj so that µ∗g(Ej) = µg(Tj) = µg(Fj).Let F = ∪jFj . Then

µg(E \ F ) = µg(∪jEj \ F ) = µg(∪j(Ej \ F )

≤ µg(∪jEj \ Fj) =∑j

µg(Ej \ Fj) = 0.

This proves the result.

14.2 Properties Of Lebesgue-Stieljes Measures

It is clear Lebesgue measure on <, as developed in Chapter 12 should coincide with the Lebesgue-Stieljes measure generated by the function g(x) = x. In Theorem 14.2.1 below, we see that sinceLebesgue measure satisfies Conditions (1), (2) and (3), the function g constructed in the proof isexactly this function, g(x) = x. Hence, the Lebesgue-Stieljes measure for g(x) = x is just Lebesguemeasure.

It is also clear that if g is of bounded variation and right continuous, g can be written as a differenceof two non decreasing functions u and v which are right continuous. The function u determines aLebesgue-Stieljes measure µu; the function v, the Lebesgue-Stieljes measure µv and finally g definesthe charge µg = µu − µv . Then we see

∫fdµg =

∫fdµu −

∫fdµv .

We now summarize our discussions and a give converse result.

Theorem 14.2.1 Characterizations of Lebesgue-Stieljes Measures

14.2. PROPERTIES 277

Let g be non decreasing and right continuous on <. let µ∗g be the outer measure associatedwith g as discussed above. LetMg be the associated set of µ∗g measurable functions andµg be the generated measure. Then,

1. µ∗g is a metric outer measure and so all Borel sets are µ∗g measurable.

2. If A is a bounded Borel set, the µg(A) is finite.

3. Each setA ⊆ < has a measurable cover U which is of typeGδ; i.e. U is a countableintersection of open sets. This means A ⊆ U and µ∗g(A) = µg(U).

4. For every half open interval (a, b], µg((a, b]) = g(b)− g(a).

Conversely, if µ∗ is an outer measure on < with resulting measure space (<,M, µ). Ifconditions (1), (2) and (3) above are satisfied by µ∗ and µ, then there is a non decreasingright continuous function g on < so that µ∗g(A) = µ∗(A) for all A ⊆ <. In particular,µg(A) = µ(A) for all A ∈M.

Proof 14.2.1The first half of the theorem has been proved in our extensive previous comments and using Theorem14.1.3. It remains to prove the converse. Define g on < by

g(x) =

µ((0, x]), if x > 00, if x = 0−µ((x, 0]) if x < 0.

It is clear that g is non decreasing on <. Is g right continuous? Let (hn) be any sequence of positivenumbers so that hn → 0. Assume x > 0. Then, we know

(0, x] =

∞⋂n=1

(0, x+ hn].

By Condition (2) on the measure µ, we see µ((0, x+ h1]) is finite. Thus, by Lemma 9.1.2, we have

µ((0, x]) = limn→∞

µ((0, x+ hn])

or using our definition of g, g(x) = limn g(x + hn). We conclude g is right continuous for positivex. The arguments for x < 0 and x = 0 are similar.

Next, we show µ∗g = µ∗. By definition of g, µg((a, b]) = µ((a, b]) for all half-open intervals (a, b].Now both µ and µg are countably additive. Since every open interval is a countable disjoint unionof half-open intervals, it follows that µg(I) = µ(I) for any open interval I . Since any open set G isa countable union of open intervals, we then see µg(G) = µ(G) for all open sets G. Further, if H isa countable intersection of open sets, we can write H = ∩nGn where each Gn is open. Let

H1 = G1

H2 = G1 ∩G2

...


Hn = ∩nj=1Gj

Then (Hn) is a decreasing sequence of open sets because finite intersections of open sets are stillopen. If we assume H is bounded, then using Lemma 9.1.2 again, we have µ(H) = limn µ(Hn).But we already have shown that µg(Hn) = µ(Hn) since Hn is an open set. We conclude thereforethat µg(H) = limn µg(Hn) = limn µ(Hn) = µ(H). Thus, µ and µg match on bounded sets of typeGδ .

Now let A be any bounded subset of <. Since µ satisfies Condition (3), there is a Gδ set H1 so thatµ∗(A) = µ(H1) with A ⊆ H . Since g is non decreasing and right continuous, we know there is alsoa Gδ set H2 so that A ⊆ H2 and µ∗g(A) = µg(H2). Let H = H1 ∩H2. Then A ⊆ H . Since Borelsets are inMg andM, we also know µ∗(H) = µ(H) and µ∗g(H) = µg(H). It then follows that

µ∗g(H) ≥ µ∗g(A) = µg(H2) ≥ µg(H) = µ∗g(H)

µ∗(H) ≥ µ∗(A) = µ(H1) ≥ µ(H) = µ∗(H)

we see µ∗g(A) = µg(H) and µ∗(A) = µ(H). But H is of type Gδ and so the values µ(H) andµg(H) must match. We conclude

µ∗g(A) = µ(H) = µ∗(A)

for all bounded subsets A. To handle unbounded sets A, note An = A ∩ [−n, n] is bounded andA = ∪nAn. It is clear An increases monotonically to A. For each An, there is a Gδ set Hn so thatµ∗g(An) = µg(Hn) = µ(Hn) = µ∗(An).

Claim 1 If ν∗ is an regular outer measure, then for any sequence of sets (An),

ν∗(lim inf An) ≤ lim inf ν∗(An).

Proof Since ν∗ is regular, there is a measurable setF so that ν∗(lim inf An) = ν(F ) with lim inf An) ⊆F . Further, there are measurable sets Fn with An ⊆ Fn with ν∗(An) = ν(Fn). Hence,

ν(F ) = ν∗(lim inf An) ≤ ν∗(lim inf Fn).

Recall

lim inf Fn =⋃m

⋂n≥m

Fn.

Hence, we see that for all N , we have ν(∪Nm=1 ∩n≥m Fn) ≤ supm infn≥m ν(Fn). This implies

limNν

(∪Nm=1 ∩n≥m Fn

)≤ lim inf ν(Fn).

.To finish we note that since the set convergence is monotonic upward to lim inf Fn, we have ν(lim inf Fn) ≤lim inf ν(Fn). Combining inequalities, we find

14.3. HOMEWORK 279

ν(F ) = ν∗(lim inf An) ≤ ν∗(lim inf Fn)

≤ lim inf ν(Fn) = lim]infν∗(An)

which completes the proof.

Claim 2: If (An) is a monotonically increasing sequence with limit A and ν∗ is a regular outermeasure then ν∗(An)→ ν∗(A).

Proof Since the sequence is monotonic increasing, it is clear the limit exists and is bounded above byν∗(A). On the other hand, we know from Claim 1 that ν∗(lim inf An) = ν∗(A) ≤ lim inf ν∗(An).But since the limit exists, the lim inf matches the limit and we are done.

Using Claim 2 applied to µ∗g and µ∗ on the monotonically increasing sets An, we have µ∗g(An) →µ∗g(A) and µ∗(An) → µ∗(A). Since µ∗g(An) = µ∗(An), we have µ∗g(A) = µ∗(A). This shows theµ∗g = µ∗ on <.

14.3 HomeworkExercise 14.3.1 Let h be our Cantor function

h(x) = (x + Ψ(x))/2.

We know τh defines a Lebesgue-Stieljes measure and since the resulting sigma-algebra contains theBorel sets, it is a Borel-Stieljes measure. Determine if τh is absolutely continuous with respect to theBorel measure on < (Borel measure is just Lebesgue measure restricted to B).

Part VI

Abstract Measure Theory Two

281

Chapter 15

Modes Of Convergence

There are many ways a sequence of functions in a measure space can converge. In this chapter, wewill explore some of them and the relationships between them.

There are several types of convergence here:

(i): Convergence pointwise,

(ii): Convergence uniformly,

(iii): Convergence almost uniformly,

(iv): Convergence in measure,

(v): Convergence in Lp norm for 1 ≤ p <∞,

(vi): Convergence in L∞ norm.

We will explore each in turn. We have already discussed the p norm convergence in Chapter 10so there is no need to go over those ideas again. However, some of the other types of convergencein the list above are probably not familiar to you. Pointwise and pointwise a.e. convergence havecertainly been mentioned before, but let’s make a formal definition so it is easy to compare it to othertypes of convergence later.

Definition 15.0.1 Convergence Pointwise and Pointwise a.e.

Let (X,S) be a measurable space. Let (fn) be a sequence of extended real valued mea-surable functions: i.e. (fn) ⊆ M(X,S). Let f : X → < be a function. Then, we say fnconverges pointwise to f on X if limn fn(x) = f(x) for all x in X . Note that this type ofconvergence does not involve a measure although it does use the standard metric, || on <.We can write this as

fn → f [ptws].

If there is a measure µ on S, we can also say the sequence converges almost everywhere ifµ(x | fn(x) 6→ f(x)) = 0. We would write this as

fn → f [ptws a.e.].

Next, you have probably already seen uniform convergence in the context of advanced calculus.We can define it nicely in a measure space also.

283

284 CHAPTER 15. CONVERGENCE MODES

Definition 15.0.2 Convergence Uniformly

Let (X,S) be a measurable space. Let (fn) be a sequence of real valued measurablefunctions: i.e. (fn) ⊆M(X,S). Let f : X → < be a function. Then, we say fn convergesuniformly to f onX if for any ε > 0, there is a positive integerN (depending on the choiceof ε so that if n > N , then | fn(x)− f(x) |< ε for all x in X . We can write this as

fn → f [unif ].

However, if we are in a measure space, we can relax the idea of uniform convergence of the wholespace by taking advantage of the underlying measure.

Definition 15.0.3 Almost Uniform Convergence

Let (X,Sµ) be a measure space. Let (fn) ⊆M(X,S, µ) be a sequence of functions whichare finite a.e. Let f : X → < be a function. We say fn converges almost uniformly to f onX if for any ε > 0, there is a measurable set E such that µ(EC) < ε and (fn) convergesuniformly to f on E. We write this as

fn → f [a.u.].

Finally, we can talk about a brand new idea: convergence using only measure itself.

Definition 15.0.4 Convergence In Measure

Let (X,Sµ) be a measure space. Let (fn) ⊆ M(X,S, µ) be a sequence of functionswhich are finite a.e. Let f : X → < be a function. Let E be a measurable set. We sayfn converges in measure to f on E if for any pair (ε, δ) of positive numbers, there exists apositive integer N (depending on ε and δ) so that if n > N , then

µ(x | | fn(x)− f(x) | ≥ δ) < ε.

We write this asfn → f [meas on E].

If E is all of X , we would just write

fn → f [meas].

15.1 Extracting Subsequences

In some cases, when a sequence of functions converges in one way, it is possible to prove that thereis at least one subsequence that converges in a different manner. We will now make this idea precise.

Definition 15.1.1 Cauchy Sequences In Measure

15.1. EXTRACTING SUBSEQUENCES 285

Let (X,S, µ) be a measure space and (fn) be a sequence of extended real valued measur-able functions. We say (fn) is Cauchy in Measure if for all α > 0 and ε > 0, there is apositive integer N so that

µ

(|fn(x) − fm(x)| ≥ α

)< ε, ∀ n, m > N.

We can prove a kind of completeness result next.

Theorem 15.1.1 Cauchy In Measure Implies A Convergent Subsequence

Let (X,S, µ) be a measure space and (fn) be a sequence of extended real valued mea-surable functions which is Cauchy in Measure. Then there is a subsequence (f1

n) and anextended real valued measurable function f such that f1

n → f [a.e.], f1n → f [a.u.] and

f1n → f [meas].

Proof 15.1.1For each pair of indices n and m, there is a measurable set Enm on which the definition of thedifference fn − fm is not defined. Hence, the set

E =⋃n

⋃m

Enm

is measurable and on EC , all differences are well defined. We do not know the sets Enm havemeasure 0 here as the members of the sequence do not have to be summable or essentially bounded.

Now, let’s get started with the proof.(Step 1): let α1 = 1/2 and ε1 = 1/2 also. Then, (fn) Cauchy in Measure implies

∃N1 3 n, m > N1 ⇒ µ

(|fn(x) − fm(x)| ≥ 1/2

)< 1/2.

Letg1 = fN1+1.

(Step 2): let α2 = 1/22 and ε1 = 1/22 also. Then, (fn) Cauchy in Measure again implies thereis an N2 > N1 so that

n, m > N2 ⇒ µ

(|fn(x) − fm(x)| ≥ 1/4

)< 1/4.

Letg2 = fN2+1.

It is then clear by our construction that

µ

(|g2(x) − g1(x)| ≥ 1/2

)< 1/2.

(Step 3): let α3 = 1/23 and ε1 = 1/23 also. Then, (fn) Cauchy in Measure again implies there isan N3 > N2 so that

n, m > N3 ⇒ µ

(|fn(x) − fm(x)| ≥ 1/8

)< 1/8.


Letg3 = fN3+1.

It follows by construction that

µ

(|g3(x) − g2(x)| ≥ 1/4

)< 1/4.

Continuing this process by induction, we find a subsequence (gn) of the original sequence (fn)so that for all k ≥ 1,

µ

(|gk+1(x) − gk(x)| ≥ 1/2k

)< 1/2k.

Define the sets

Ej =

(|gj+1(x) − gj(x)| ≥ 1/2j

)and

Fk =

∞⋃j=k

Ej .

Note if x ∈ FCk ,

|gj+1(x) − gj(x)| < 1/2j

for any index j ≥ k. Each set Fk is then measurable and they form an increasing sequence. Let’s geta bound on µ(Fk). First, if A and B are measurable sets, then

µ

(A ∪ B

)= µ

(A ∪ BC

): + µ

(A ∩ B

)+ µ

(AC ∪ B

)

But adding in µ(A ∩ B

)simply makes the sum larger. We see

µ

(A ∪ B

)≤ µ

(A ∪ BC

): + µ

(A ∩ B

)+ µ

(A ∩ B

)+ µ

(AC ∪ B

)= µ(A) + µ(B).

This result then extends easily to finite unions. Thus, if (An) is a sequence of measurable sets, thenby the sub additive result above,

µ

( n⋃i=1

Ai

)≤

n∑i=1

µ(Ai).

Hence, the sets ∪ni=1 Ai form an increasing sequence and we clearly have

µ

( ∞⋃i=1

Ai

)= lim

nµ

( n⋃i=1

Ai

)≤

∞∑i=1

µ(Ai).


We can apply this idea to the increasing sequence (Fk) to obtain

µ(Fk) ≤∞∑j=k

µ(Ej)

<

∞∑j=k

1/2j = 1/2k−1.

Now, for any i > j, we have

|gi(x) − gj(x)| ≤i−1∑`=j

|g`+1 − g`|.

Choosing the indices i and j so that i > j ≥ k, we then find if x 6∈ Fk, that

|g`+1(x) − g`(x)| < 1/2`.

Hence, for these indices,

|gi(x) − gj(x)| ≤i−1∑`=j

|g`+1 − g`|

<

i−1∑`=j

1/2` =

∞∑`=j

1/2` = 1/2j−1.

We conclude that if x ∈ FCk and i > j ≥ k we have

|gi(x) − gj(x)| ≤ 1/2j − 1. (∗)

Now let F = ∩k Fk. The F is measurable and µ(F ) = limk µ(Fk) = 0. Let x be in FC . By DeMorgan’s Laws, x ∈ ∪k FCK which implies x is in some FCk . Call this k∗. Then given ε > 0, chooseJ so that 1/2J−1 < ε. Then, by Equation ∗, if i > j ≥ J ≥ k∗,

|gi(x) − gj(x)| ≤ 1/2j−1 < 1/2J−1 < ε.

Thus, the sequence gk(x) is a Cauchy sequence of real numbers for each x in FC . Hence, limk gk(x)exists for such x. Defining f by

f(x) =

limk gk(x), x ∈ FC0, x ∈ F,

we see f is measurable and it is the pointwise limit a.e. of the subsequence (gk). This completes theproof of the first claim. To see that (gk) converges in measure to f , look again at Equation ∗:

|gi(x) − gj(x)| ≤ 1/2j − 1, ∀ i > j ≥ k, ∀ x ∈ FCk .

Now let i→∞ and use the continuity of the absolute value function to obtain

|f(x) − gj(x)| ≤ 1/2j − 1, ∀ j ≥ k, ∀ x ∈ FCk . (∗∗)

Equation ∗∗ says that (gk) converges to f uniformly on FCk . Further, recall µ(Fk) < 1/2k−1.


Note given any δ > 0, there is an integer k∗ so that 1/2k∗−1 < δ and gk converges uniformly on

FCk∗ . We therefore conclude that (gk) converges almost uniformly to f as well.To show the last claim, given an arbitrary α > 0 and ε > 0, choose a positive integer k∗ so that

µ(F ∗k ) < 1/2k∗−1 < min(α, ε).

Then, by Equation ∗∗, we have(|f(x) − gj(x)| ≥ α

)⊆

(|f(x) − gj(x)| > 1/2k

∗−1

).

Then, again by Equation ∗∗, we have

⊆(|f(x) − gj(x)| > 1/2k

∗−1

)⊆ Fk∗ .

Combining, we have

µ

(|f(x) − gj(x)| ≥ α

)≤ µ

(Fk∗

)< 1/2k

∗−1 < ε.

This shows that (gk) converges to f in measure.

The result above allows us to prove that Cauchy in Measure implies there is a function which theCauchy sequence actually converges to.

Theorem 15.1.2 Cauchy In Measure Implies Completeness

Let (X,S, µ) be a measure space and (fn) be a sequence of extended real valued mea-surable functions which is Cauchy in Measure. Then there is an extended real valuedmeasurable function f such that fn → f [meas] and the limit function f is determineduniquely a.e.

Proof 15.1.2By Theorem 15.1.1, there is a subsequence (f1

n) and a real valued function measurable function fso that f1

n → f [meas]. Let α > 0 be given. If |f(x) − fn(x)| ≥ α, then given any f1n in the

subsequence, we have

α ≤ |f(x)− fn(x)| ≤ |f(x)− f1n(x)| + |fn(x)− f1

n(x)|.

Note, just as in the previous proof, there is a measurable set E where all additions and subtractionsof functions are well-defined. Now, let β = |f(x)− f1

n(x)| and γ = |fn(x)− f1n(x)|. The equation

above thus says

β + γ ≤ α

Since β and γ are non negative and both are less than or equal to α, we can think about this inequalityin a different way. If there was equality

β∗ + γ∗ = α

with both β∗ and γ∗ not zero, then we could let t = β∗/α and we could say β∗ = t α and γ∗ =(1 − t) α as γ∗ = α − β∗. Now imagine β and γ being larger α. Then, β and γ would have to


where n1 denotes the index of the function f1n. Further, since (fn) is Cauchy in measure, there is a

positive integer N2 so that

µ

(|fn(x)− f1

n(x)| ≥ α/2

)< ε/2, ∀ n, n1 > N2.

So if n1 is larger than N = max(N1, N2), we have

µ

(|f(x)− fn(x)| ≥ α/2

)< ε, ∀ n > N.

This shows fn → f [meas] as desired.

To show the uniqueness a.e. of f , assume there is another function g so that fn → g [meas].Then, by arguments similar to ones we have already used, we find

x | |f(x)− g(x)| ≥ α ⊆ x | |fn(x)− f(x)| ≥ 1/2 α.

Then, mutatis mutandi, we obtain

µ

(x | |f(x)− g(x)| ≥ α

)≤ µ

(x | |fn(x)− f(x)| ≥ 1/2 α

)+ µ

(x | |fn(x)− g(x)| ≥ 1/2 α

)< ε.

Since ε > 0 is arbitrary, we see for any α > 0,

µ

(x | |f(x)− g(x)| ≥ α

)= 0.

However, we know

µ

(x | |f(x)− g(x)| > 0

)=

⋃n

(x | |f(x)− g(x)| ≥ 1/n

),

which immediately tells us that

µ

(x | |f(x)− g(x)| > 0

)= 0.

This says f = g a.e. and we are done.

Theorem 15.1.3 p-Norm Convergence Implies Convergence in Measure

Assume 1 ≤ p < ∞. Let (fn) be a sequence in Lp(X,S, µ) and let f ∈ Lp(X,S, µ) sothat fn → f [p− norm]. Then fn → f [meas] which is Cauchy in Measure.

Proof 15.1.3Let α > 0 be given and let

En(alpha) = x | |fn(x) − f(x)| ≥ α.


Then, given ε > 0, there is a positive integer N so that∫|fn − f |p dµ < αp ε, ∀ n > N.

Thus, ∫En(α)

|fn − f |p dµ ≤∫|fn − f |p dµ < αp ε, ∀ n > N.

But on En(α), the integrand in the first term is bigger than or equal to αp. We obtain

αp µ(En(α)) < αp ε, ∀ n > N.

Canceling the αp term, we have µ(En(α)) < ε, for all n > N . This implies fn → f [meas].

Comment 15.1.1 Let’s assess what we have learned so far. We have shown

(i):

fn → f [p− norm] ⇒ fn → f [meas]

by Theorem 15.1.3.

(ii): It is a straightforward exercise to show

fn → f [meas] ⇒ (fn) Cauchy In Measure .

Then,

(fn) Cauchy In Measure ⇒ ∃ (f1n) ⊆ (fn) 3 f1

n → f [a.e.]

by Theorem 15.1.1. Note, we proved the existence of such a subsequence already in the proofof the completeness of Lp as discussed in Theorem 10.1.10.

(iii): Finally, we can also apply Theorem 15.1.1 to infer

fn → f [meas] ⇒ ∃ (f1n) ⊆ (fn) 3 f1

n → f [a.u.]

Theorem 15.1.4 Almost Uniform Convergence Implies Convergence In Measure

Let (X,S) be a measurable space. Let (fn) be a sequence of real valued measurablefunctions: i.e. (fn) ⊆M(X,S). Let f : X → < be measurable. Then

fn → f [a.u.] ⇒ fn → f [meas].

Proof 15.1.4If fn converges to f a.u., given arbitrary ε > 0, there is a measurable set Eε so that µ(Eε) < ε andfn converges uniformly on ECε . Now let α > 0 be chosen. Then, there is a positive integer Nα sothat

|fn(x) − f(x)| < ε, ∀ n > Nα, ∀ x ∈ ECε .


Hence, if n > Nα and x satisfies |fn(x) − f(x)| ≥ α, we must have that x ∈ Eε. We conclude(|fn(x) − f(x)| ≥ α

)⊆ Eε, ∀ n > Nα.

This implies immediately that

µ

(|fn(x) − f(x)| ≥ α

)≤ µ(Eε) < ε, ∀ n > Nα.

This proves fn → f [meas].

Comment 15.1.2 We have now shown

fn → f [p− norm] ⇒ fn → f [meas]

by Theorem 15.1.3. This then implies by Theorem 15.1.1

∃ (f1n) ⊆ (fn) 3 f1

n → f [a.u.]

15.2 Egoroff’s TheoremA famous theorem tells us how pointwise a.e. convergence can be phrased “almost” like uniformconvergence. This is Egoroff’s Theorem.

Theorem 15.2.1 Egoroff’s Theorem

Let (X,S, µ) be a measure space with µ(X) < ∞. Let f be an extended real valuedfunction which is measurable. Also, let (fn) be a sequence of functions in M(X,S) suchthat fn → f [a.e.]. Then, fn → f [a.u.] and fn → f [meas].

Proof 15.2.1From previous arguments, the way we handle converge a.e. is now quite familiar. Also, we know howwe deal with the measurable set on which addition of the function fn are not well defined. Hence,we may assume without loss of generality that the convergence here is on all of X and that additionis defined on all of X . With that said, let

Enk =

∞⋃k=n

(|fk(x)− f(x)| ≥ 1/m

).

Note that each Enk is measurable and En+1,k ⊆ Enk so that this is an decreasing sequence of setsin the index n. Given x in X , we have fn → f(x). Hence, for ε = 1/m, there is a positive integerN(x, ε) so that

|fn(x) − f(x)| < ε = 1/m, ∀ n > N(x, ε).

Thus, (|fn(x) − f(x)| ≥ 1/m

)= ∅, ∀ n > N(x, ε). (∗)

Now consider Fm =⋂∞n=1 Enm. If x ∈ Fm, then x is in Enm for all n. In particular, letting

n∗ = N(x, ε) + 1, we have x ∈ En∗ m. Looking at how we defined En∗ m, we see this implies that

15.2. EGOROFF’S THEOREM 293

there is a positive integer k′ > n∗, so that |f ′k(x) − f(x)| ≥ 1/m. However, by Equation ∗, thisset is empty. This contradiction means our original assumption that Fm was non empty is wrong.Hence, Fm = ∅. Now, since µ(X) <∞, µ(E1m is finite also. Hence, by Lemma 9.1.2,

0 = µ(Fm) = limn

µ(En m.

This implies that given δ > 0, there is a positive integer Nm so that

µ(En m < δ/2m, ∀m > Nm,

since limm µ(Enm = 0. For each integer m, choose a positive integer nm > Nm. We can arrangefor these integers to be increasing; i.e., nm < nm+1. Then,

µ

(Enm m

)< δ/2m

and letting

Eδ =

∞⋃m=1

Enm m,

we have

µ(Eδ) ≤∞∑m=1

δ/2m < δ.

Finally, note

ECδ =

( ∞⋃m=1

Enm m

)C=

∞⋂m=1

ECnm m.

Next, note

ECnm m =

( ∞⋃k=nm

(|fk(x) − f(x)| ≥ 1/m

))C

=

∞⋂k=nm

(|fk(x) − f(x)| ≥ 1/m

)C=

∞⋂k=nm

(|fk(x) − f(x)| < 1/m

).

Thus, since x ∈ ECδ means x is in ECnm m for all m, the above says |fk(x) − f(x)| < 1/m for allk > nm. Therefore, given ε > 0, pick a positive integer M so that 1/M < ε. Then, for all x in ECδ ,we have

|fk(x) − f(x)| < 1/M < ε, ∀ k ≥ nM .

This says fn converges uniformly to f onECδ with µ(Eδ) < δ. Hence, we have shown fn → f [a.u.]

Finally, if fn → f [a.u.], by Theorem 15.1.4, we have fn → f [meas] also.

Next, let’s see what we can do with domination by a p-summable function.


Theorem 15.2.2 Pointwise a.e. Convergence Plus Domination Implies p-Norm Convergence

Let 1 ≤ p < ∞ and (X,S, µ) be a measure space. Let f be an extended real valuedfunction which is measurable. Also, let (fn) be a sequence of functions in Lp(X,S) suchthat fn → f [a.e.]. Assume there is a dominating function g which is p-summable; i.e.|fn(x)| ≤ g(x) a.e. Then, if fn → f [a.e.], f is p-summable and fn → f [p− norm].

Proof 15.2.2Since |fn(x)| ≤ g(x) a.e., we have immediately that |f | ≤ g a.e. since fn → f [a.e.]. Thus,|f |p ≤ gp and we know f is in Lp(X,S). Since all the functions here are p-summable, the set whereall additions is not defined has measure zero. So, we can assume without loss of generality that thisset has been incorporated into the set on which convergence fails. Hence, we can say

|fn(x)− f(x)| ≤ |fn(x)| + |f(x)| ≤ 2 g(x), a.e.

So,

|fn(x)− f(x)|p ≤ 2p |g(x)|p, a.e.

By assumption, g is p-summable, so we have 2p gp is in L1(X,S). Applying Lebesgue’s DominatedConvergence Theorem, we find

limn

∫|fn(x)− f(x)|p dµ =

∫limn|fn(x)− f(x)|p dµ = 0.

Thus, fn → f [p− norm].

15.3 Vitali’s Theorem

This important theorem is one that gives us more technical tools to characterize p-norm convergencefor a sequence of functions. We need a certain amount of technical infrastructure to pull this off; sobear with us as we establish a series of lemmatta.

Lemma 15.3.1 p-Summable Functions Have p-Norm Arbitrarily Small Off a Set

Let 1 ≤ p < ∞ and (X,S, µ) be a measure space. Let f be in Lp(X,S). Then givenε > 0, there is a measurable set Eε so that µ(Eε) < ∞ and if F ⊆ ECε is measurable,then || f IF ||p < ε.

Proof 15.3.1Let En = (|fn(x)| ≥ 1/n). Note En ∈ S and the sequence (En) is increasing and ∪nEn = X . Letfn = fIEn . It is straightforward to verify that fn ↑ f as fn ≤ fn+1 for all n. Further, |fn|p ≤ |f |p;hence, by the Dominated Convergence Theorem,

limn

∫|fn|p dµ =

∫limn|fn|p dµ =

∫|f |p dµ < ∞.

The definition of fn and En then implies

µ(En)/np ≤∫En

|f |p dµ ≤∫|f |p dµ < ∞.

15.3. VITALI’S THEOREM 295

This tells us µ(En) <∞ for all n.

Now choose ε > 0 arbitrarily. Then there is a positive integer N so that∫|f |p dµ −

∫|fn|p dµ < εp, ∀ n > N.

Thus, since fn = fIEn , we can say∫En

|f |p dµ +

∫ECn

|f |p dµ −∫En

|f |p dµ < εp, ∀ n > N.

or ∫ECn

|f |p dµ < εp, ∀ n > N.

So choose Eε = EN+1 and we have ∫ECε

|f |p dµ < εp.

which implies the desired result.

Lemma 15.3.2 p-Summable Inequality

Let 1 ≤ p < ∞ and (X,S, µ) be a measure space. Let (fn) be a sequence of functionsin Lp(X,S). Define βn on S by

βn(E) = || fn IE ||p, ∀ E.

Then,

|βn(E) − βm(E)| ≤ || fn − fm ||p, ∀ E, ∀ n, m.

Proof 15.3.2By the backwards triangle inequality, for any measurable E,

|| fn − fm ||p ≥ | || fn IE ||p − || fm IE ||p | = |βn(E) − βm(E)|.

Lemma 15.3.3 p-Summable Cauchy Sequence Condition I

Let 1 ≤ p < ∞ and (X,S, µ) be a measure space. Let (fn) be a Cauchy Sequence inLp(X,S). Define βn on S as done in Lemma 15.3.2. Then, there is a positive integer Nand a measurable set Eε of finite measure, so that if F is a measurable subset of Eε, thenβn(E) < ε for all n > N .

Proof 15.3.3Since (fn) is a Cauchy sequence in p-norm, there is a function f in Lp(X,S) so that fn → f [p −


norm]. By Lemma 15.3.1, given ε > 0, there is a measurable set Eε with finite measure so that∫ECε

|f |p dµ < (ε/2)p.

Now given a measurable F contained inECε , recalling the meaning of βn(F ) as described in Lemma15.3.2, we can write

βn(F ) ≤ || fn ||p ≤ || fn IECε ||p≤ || (fn − f) IECε ||p + || f IECε ||p< ε/2 + || (fn − f) IECε ||p .

Since fn → f [p− norm], there is a positive integer N so that if n > N ,

|| (fn − f) IECε ||p < ε/2.

This shows βn(F ) < ε when n > N as desired.

Lemma 15.3.4 Continuity Of The Integral

Let (X,S, µ) be a measure space and f be a summable function. Then for all ε > 0 thereis a δ > 0, so that

|∫E

f dµ| < ε, ∀ E ∈ S, with µ(E) < δ.

Proof 15.3.4Define the measure γ on S by γ(E) =

∫E|f | dµ. Note, by Comment 9.4.3, we know that γ is

absolutely continuous with respect to µ. Now assume the proposition is false. Then, there is anε > 0 so that for all choices of δ > 0, we have a measurable set Eδ for which µ(Eδ) < δ and|∫Eδ

f dµ|/geqε. In particular, for the sequence δn = 1/2n, we have a sequence of sets En withµ(En) < 1/2n and

|∫En

f dµ|/geqε.

Let

Gn =

∞⋃k=n

Ek, G =

∞⋂n=1

Gn.

Then,

µ(G) ≤ µ(Gn) ≤∞∑k=n

µ(Ek) <

∞∑k=n

1/2k = 1/2n−1.

This implies mu(G) = 0 and thus, since γ is absolutely continuous with respect to µ, γ(G) = 0also. We also know the sequence Gn is decreasing and so γ(Gn)→ γ(G) = 0. Finally, since

γ(Gn) ≥ γ(En) ≥ |∫E

f dµ| ≥ ε,

15.3. VITALI’S THEOREM 297

we have γ(G) = limn γ(Gn) ≥ ε as well. This is impossible. Hence, our assumption that theproposition is false is wrong.

Lemma 15.3.5 p-Summable Cauchy Sequence Condition II

Let 1 ≤ p < ∞ and (X,S, µ) be a measure space. Let (fn) be a Cauchy Sequence inLp(X,S). Define βn on S as done in Lemma 15.3.2. Then, given ε > 0, there is a δ > 0and a positive integer N so that if n > N , then

βn(E) < ε, ∀ E ∈ S, with µ(E) < δ.

Proof 15.3.5Since Lp(X,S, µ) is complete, there is a p-summable function f so that fn → f [p− norm]. Then,by Lemma 15.3.4, given an ε > 0, there is a δ > 0, so that∫

E

|f |p dµ < (ε/2)p, if µ(E) < δ.

Hence, using the convenience mapping βn(E) previously defined in Lemma 15.3.2, we see

βn(E) = || fn IE ||p = || (f − fn) IE ||p + || f IE ||p≤ || (f − fn) IE ||p + ε/2

when µ(E) < δ. Finally, since fn → f [p − norm], there is a positive integer N so that if n > N ,then || (f − fn) IE ||p< ε/2. Combining, we have βn(E) < ε if n > N for µ(E) < δ.

Theorem 15.3.6 Vitali Convergence Theorem

Let 1 ≤ p < ∞ and (X,S, µ) be a measure space. Let (fn) be a sequence of functionsin Lp(X,S). Then, fn → f [p−norm] if and only if the following three conditions hold.

(i): fn → f [meas],(ii): For all ε > 0, there exists N and a set Eε ∈ S with µ(Eε) < ∞ so that if F is ameasurable set in ECε , then

∫F|fn|p dµ < εp, for all n > N .

(iii): For all ε > 0, there is a δ > 0 and an N so that if E is measurable with µ(E) < δ,then

∫E|fn|p dµ < εp, for all n > N .

Proof 15.3.6

⇒: If fn → f [p−norm], then by Theorem 15.1.3, fn → f [meas] which shows (i) holds. Then,since fn → f [p − norm], (fn) a Cauchy sequence. Thus, by Lemma 15.3.1, condition (ii) holds.Finally, since (fn) is a Cauchy sequence, by Lemma 15.3.5, condition (iii) holds.⇒: Now assume conditions (i), (ii) and (iii) hold. Let ε > 0 be given. From condition (ii), we seethere is a measurable set Eε of finite measure and a positive integer N1 so that∫

ECε

|fn|p dµ < (ε/4)p

if n > N1. Thus, for indices n and m larger than N1, we have

|| fn − fm ||p = || (fn − fm) IEε + (fn − fm) IECε ||p


≤ || (fn − fm) IEε ||p + || (fn − fm) IECε ||p≤ || (fn − fm) IEε ||p + || fn IECε ||p + || fm IECε ||p< || (fn − fm) IEε ||p + ε/2.

We conclude

|| fn − fm ||p < || fm IEε ||p + ε/2, ∀ n, m > N1. (∗)

Now let β = µ(Eε and set

α =ε

4 β1/p

and define the sets Hnm by

Hnm = x | |fn(x)− fm(x)| ≥ α.

Apply condition (ii) for our given ε now. Thus, there is a δ(ε) and a positive integer N2 so that∫E

|fn|p dµ < (ε/8)p, n > N − 2, when µ(E) < δ(ε). (∗∗)

Since fn → f [meas], (fn) is Cauchy in measure. Hence, there is a positive integer N3 so that

µ(Hnm < δ(ε), ∀ n, m > N3. (∗ ∗ ∗)

Finally, using the Minkowski Inequality, we have

|| (fn − fm) IEε ||p = || (fn − fm) IEε\Hnm + (fn − fm) IHnm ||p≤ || (fn − fm) IEε\Hnm ||p + || fn IHnm ||p + || fm IHnm ||p

Now let N = maxN1, N2, N3. Then, if n and m exceed N , we have Equation ∗, Equation ∗∗ andEquation ∗ ∗ ∗ all hold. This implies

|| (fn − fm) IEε ||p ≤(αp µ(Eε \Hnm)

)1/p

+ ε/8 + ε/8

& ≤ α

(µ(Eε)

)1/p

+ ε/4

= α β1/p + ε/4 = ε/(4 β1/p) β1/p + ε/4

= ε/2.

From Equation ∗, we have for these indices n and m,

|| fn − fm ||p < || fm IEε ||p + ε/2

< ε.

Thus, (fn) is a Cauchy sequence in p-norm. Since Lp(X,S, µ) is complete, there is a function g sothat fn → g [p− norm]. So by Theorem 15.1.3, fn → g [meas]. It is then straightforward to showthat f = g a.e. This tells us f and g belong to the same equivalence class of Lp(X,S, µ).

15.4. SUMMARY 299

15.4 Summary

We can summarize the results of this chapter as follows. If we know the measure is finite, we can sayquite a bit.

Theorem 15.4.1 Convergence Relationships On Finite Measure Space One

Let (X,S, µ) be a measure space with µ(X) <∞. Let f and (fn) be in M(X,S). Then,

fn → f [p− norm]fn → f [unif ]fn → f [a.u.]fn → f [a.e.]

⇒ fn → f [meas].

Certain types of convergence give us pointwise convergence.

Theorem 15.4.2 Convergence Relationships On Finite Measure Space Two

Let (X,S, µ) be a measure space with µ(X) <∞. Let f and (fn) be in M(X,S). Then,fn → f [unif ]fn → f [a.u.]

⇒ fn → f [a.e.].

Uniform convergence on a finite measure space implies p norm convergence.

Theorem 15.4.3 Convergence Relationships On Finite Measure Space Three


fn → f [unif ] ⇒fn → f [a.u.]fn → f [p− norm]

The final result says that pointwise convergence is “almost” uniform convergence.

Theorem 15.4.4 Convergence Relationships On Finite Measure Space Four


fn → f [a.e.]. ⇒ fn → f [a.u.].

If the measure of X is infinite, we have many one way implications.

Theorem 15.4.5 Convergence Relationships On General Measurable Space


Let (X,S, µ) be a measure space. Let f and (fn) be in M(X,S). Then,(i):

fn → f [p− norm]fn → f [unif ]fn → f [a.u.]


(ii):

fn → f [unif ]fn → f [a.u.]

⇒ fn → f [a.e.].

(iii):

fn → f [unif ] ⇒ fn → f [a.u.].

Next, if we can dominate the sequence by an Lp function, we can say even more.

Theorem 15.4.6 Convergence Relationships With p-Domination One

Let (X,S, µ) be a measure space. Let f and (fn) be in M(X,S). Assume there is ag ∈ Lp so that |fn| ≤ g. Then, we know the following implications:

fn → f [p− norm]fn → f [unif ]fn → f [a.u.]fn → f [a.e.]


Theorem 15.4.7 Convergence Relationships With p-Domination Two


fn → f [unif ]fn → f [a.u.]

⇒ fn → f [a.e.].

Theorem 15.4.8 Convergence Relationships With p-Domination Three


fn → f [unif ] ⇒fn → f [a.u.]fn → f [p− norm]

Theorem 15.4.9 Convergence Relationships With p-Domination Four

15.5. HOMEWORK 301


fn → f [a.e.]. ⇒ fn → f [a.u.].

Theorem 15.4.10 Convergence Relationships With p-Domination Five


fn → f [a.e.]fn → f [a.u.]

⇒ fn → f [p− norm].

Theorem 15.4.11 Convergence Relationships With p-Domination Six


fn → f [meas]. ⇒ fn → f [p− norm].

There are circumstances where we can be sure we can extract a subsequence that converges insome fashion.

Theorem 15.4.12 Convergent Subsequences Exist

Let (X,S, µ) be a measure space. It doesn’t matter whether or not µ(X) is finite. Let fand (fn) be in M(X,S). Then, we know the following implications:

(i):

fn → f [p− norm]fn → f [meas]

⇒ ∃ subsequence f1

n → f [a.u.].

(ii):

fn → f [p− norm]fn → f [meas]

⇒ ∃ subsequence f1

n → f [a.e.].

Further, the same implications hold if we know there is a g ∈ Lp so that |fn| ≤ g.

15.5 Homework

Exercise 15.5.1 Characterize convergence in measure when the measure in counting measure.

Exercise 15.5.2 Let (X,Sµ) be a measure space. Let (fn), (gn) ⊆ M(X,S, µ) be sequences offunctions which are finite a.e. Let f, g : X → < be functions. Prove if fn → f [meas on E] andgn → g[meas on E], then (fn + gn)→ (f + g)[meas on E].


Exercise 15.5.3 Let (X,Sµ) be a measure space with µ(X) < ∞. Let (fn), (gn) ⊆ M(X,S, µ)be sequences of functions which are finite a.e. Let f, g : X → < be functions. Prove if fn →f [meas on E] and gn → g[meas on E], then (fn gn)→ (f g)[meas on E]. Hint: first consider thecase that fn → 0[meas on E] and gn → 0[meas on E].

Exercise 15.5.4 Let (X,S, µ) be a measure space. Let (fn) ⊆ M(X,S, µ) be a sequence offunctions which are finite a.e. Let f : X → < be a function. Prove if fn → f [a.u.], thenfn → f [ptws a.e.] and fn → f [meas].

Exercise 15.5.5 Let (X,S, µ) be a finite measure space. for any pair of measurable functions f andg, define

d(f, g) =

∫| f − g |

1+ | f − g |dµ.

(i): Prove M(X,S, µ) is a semi-metric space.

(ii): Prove if (fn) is a sequence of measurable functions and f is another measurable function,then fn → f [meas] if and only if d(fn, f)→ 0.

Hint: You don’t need any high power theorems here. First, let φ(t) = t/(1 + t) so that d(f, g) =∫φ(|f − g|dµ. Then try this:

(⇒): We assume fn → f [meas]. Then, given any pair of positive numbers (δ, ε), we have there isan N so that if n > N , we have

µ(|fn(x)− f(x)| ≥ δ) < ε/2.

Let Eδ denote the set above. Now for such n > N , note

d(fn, f) =

∫Eδ

φ(|fn − f |) dµ +

∫ECδ

φ(|fn − f |) dµ.

Since φ is increasing, we see that on ECδ , φ(|fn(x) − f(x)|) < φ(δ). Thus, you should be able toshow that if n > N , we have

d(fn, f) < µ(Eδ) + φ(δ) µ(X) = ε/2 + φ(δ) µ(X).

Then a suitable choice of δ does the job.(⇐): If we know d(fn, f) goes to zero, break the integral up the same way into a piece on Eδ andECδ . This tells us right away that given ε > 0, there is an N so that n > N implies

φ(δ) µ(Eδ) < ε.

This gives us the result with a little manipulation.

Exercise 15.5.6 Let (<,M, µ) denote the measure space consisting of the Lebesgue measurable setsM and Lebesgue measure µ. Let the sequence (fn) of measurable functions be defined by

fn = n I[1/n,2/n].

Prove fn → 0 on all <, fn → 0 [meas] but fn 6→ 0 [p− norm] for 1 ≤ p <∞.

Chapter 16

Decomposition Of Measures

We now examine the structure of a charge λ on a σ - algebra S. For convenience, let’s recall that acharge is a mapping on S to < which assigns the value 0 to ∅ and which is countably additive. Weneed some beginning definitional material before we go further.

16.1 The Jordan Decomposition Of A Charge

Definition 16.1.1 Positive and Negative Sets For a Charge

Let λ be a charge on (X,S). We say P ∈ S is a positive set with respect to λ if

λ(E ∩ P ) ≥ 0, ∀ E ∈ S.

Further, we say N ∈ S is a negative set with respect to λ if

λ(E ∩ N) ≤ 0, ∀ E ∈ S.

Finally, M ∈ S is a null set with respect to λ is

λ(E ∩ M) = 0, ∀ E ∈ S.

Definition 16.1.2 The Positive and Negative Parts Of a Charge

Let λ be a charge on (X,S). Define the mapping λ+ on S by

λ+(E) = sup λ(A) |A ∈ S, A ⊆ E.

Also, define the mapping λ− on S by

λ−(E) = − inf λ(A) |A ∈ S, A ⊆ E.

Theorem 16.1.1 The Jordan Decomposition Of A Finite Charge

Let λ be a finite valued charge on (X,S). Then, λ+ and λ− are finite measures on S andλ = λ+ − λ−. The pair (λ+, λ−) is called the Jordan Decomposition of λ.

303

304 CHAPTER 16. DECOMPOSING MEASURES

Proof 16.1.1Let’s look at λ+ first. Given any measurable E, since ∅ is contained in E, by the definition of λ+,we must have λ+(E) ≥ λ(∅) = 0. Hence, λ+ is non negative.

Next, if A and B are measurable and disjoint, By definition of λ+, for C1 ⊆ A C2 ⊆ B, we musthave

λ+(A ∪ B) ≥ λ(C1 ∪ C2)

= λ(C1) + λ(C2).

This says λ+(A ∪ B) − λ(C2) is an upper bound for the set of numbers λ(C1). Hence, bydefinition of λ+(A), we have

λ+(A ∪ B) ≥ λ(C1 ∪ C2)

= λ+(A) + λ(C2).

A similar argument then shows that λ(C2) is bounded above by λ+(A ∪ B) − λ+(A). Thus, wehave

λ+(A ∪ B) ≥ λ(C1 ∪ C2)

= λ+(A) + λ+(B).

On the other hand, if C ⊆ A ∪B, then we have

λ(C) = λ(C ∩A ∪ C ∩B)

≤ λ+(A) + λ+(B).

This immediately implies that

λ+(A ∪ B) ≤ λ+(A) + λ+(B).

Thus, it is clear λ+ is additive on finite disjoint unions.

We now address the question of the finiteness of λ+. To see λ+ is finite, assume that it is not. Sothere is some set E with λ+(E) = ∞. Hence, by definition, there is a measurable set A1 so thatλ(A1) > 1. Thus, by additivity of λ+, we have

λ+(A1) + λ+(E \A1) = λ+(E) = ∞.

Thus, at least one of of λ+(A1) and λ+(E \A1) is also∞. Pick one such a set and call it B1. Thus,λ+(B1) =∞. Let’s do one more step. Since λ+(B1) =∞, there is a measurable set A2 inside it sothat λ(A2) > 2. Then,

λ+(A2) + λ+(B1 \A2) = λ+(B1) = ∞.

Thus, at least one of of λ+(A2) and λ+(B1 \ A2) is also ∞. Pick one such a set and call it B2.Thus, we have λ+(B2) = ∞. You should be able to see how we construct the two sequences (An)and (Bn). When we are done, we know An ⊆ Bn−1, λ(An) > n and λ(Bn) =∞ for all n.

Now, if for an infinite number of indices nk, Bnk = Bnk−1 \Ank , what happens? It is easiest to seewith an example. Suppose B5 = B4 \ A5 and B8 = B7 \ A8. By the way we construct these sets,we see A6 does not intersect A5. Hence, A7 ∩ A5 = ∅ also. Finally, we have A8 ∩ A5 = ∅ too.Hence, extrapolating from this simple example, we can infer that the sequence (Ank is disjoint. By

16.1. JORDAN DECOMPOSITION 305

the countable additivity of λ, we then have

λ

(⋃k

Ank

)=

∑k

λ(Ank >∑k

nk = ∞.

But λ is finite on all members of S. This is therefore a contradiction.

Another possibility is that there is an indexN so that if n > N , the choice is always that ofBn = An.In this case, we have

E ⊇ AN+1 ⊇ AN+2 . . .

Since λ is finite and additive,

λ(AN+j−1 \AN+j) = λ(AN+j−1) − λ(AN+j)

for j > 2 since all the λ values are finite. We now follow the construction given in the proof of thesecond part of Lemma 9.1.2 to finish our argument. Construct the sequence of sets (En) by

E1 = ∅E2 = AN+1 \AN+2

E3 = AN+1 \AN+3

......

...

En = AN+1 \AN+n−1.

Then (En) is an increasing sequence of sets which are disjoint and so λ(∪n En) = limn λ(En).Since λ(AN+1) is finite, we then know that λ(En) = λ(AN+1) − λ(AN+n). Hence, λ(∪n En) =λ(AN+1)− limn λ(AN+n). Next, note by De Morgan’s Laws,

λ

(∪n En

)= λ

(⋃n

AN+1 ∩ACN+n

)= λ

(AN+1

⋂∪nACN+n

)= λ

(AN+1

⋂(∩nAN+n

)C)

= λ

(AN+1 \

(∩nAN+n

)).

Thus, since λ(AN+1) is finite and ∩nAN+n ⊆ AN+1, it follows that

λ(∪nEn) = λ(AN+1)− λ(∩nAN+n).

Combining these results, we have

λ(AN+1)− limnλ(AN+n) = λ(AN+1)− λ(∩nAN+n).

Canceling λ(AN+1) from both sides, we find

λ(∩nAN+n) = limnλ(AN+n) ≥ lim

nN + n = ∞.


We again find a set ∩nAN+n with λ value ∞ inside E. However, λ is always finite. Thus, in thiscase also, we arrive at a contradiction.We conclude at this point that if λ+(E) = ∞, we force λ to become infinite for some subsets. Sincethat is not possible, we have shown λ+ is finite. Since λ− = (−λ)+, we have established that λ− isfinite also. Next, given the relationship between λ+ and λ−, it is enough to prove λ+ is a measure tocomplete this proof.It is enough to prove that λ+ is countably additive. Let (En) be a countable sequence of measurablesets and let E be their union. If A ⊆ E, then A = ∪n A ∩ En and so

λ(A) =∑n

λ(A ∩ En)

≤∑n

λ+(En),

by the definition of λ+. Since this holds for all such subsets A, we conclude∑n λ+(En) is an

upper bound for the collection of all such λ(A). Hence, by the definition of a supremum, we haveλ+(E) ≤

∑n λ

+(En).To show the reverse, note λ+(E) is finite by the arguments in the first part of this proof. Now, pickε > 0. Then, by the Supremum Tolerance Lemma, there is a sequence (An) of measurable sets, eachAn ⊆ En so that

λ+(En) − ε/2n < λ(An) ≤ λ+(En).

Let A = ∪nAn. Then A ⊆ E and so we have λ(A) ≤ λ+(E). Hence,∑n

λ+(En) <∑n

(λ(An) + ε/2n

)<

∑n

λ(An) + ε

since the second term is a standard geometric series. Next, since (An) is a disjoint sequence, thecountable additivity of λ gives ∑

n

λ+(En) < λ

(∪nAn

)+ ε.

But A = ∪nAn and since this holds for all ε > 0, we can conclude∑n

λ+(En) < λ(A) ≤ λ+(E).

Combining these inequalities, we see λ+ is countably additive and hence is a measure.

Comment 16.1.1 If we had allowed the charge in Definition 9.0.2 to be extended real valued; i.e.take on the values of∞ and −∞, what would happen? First, note by applying the arguments in thefirst part of the proof above, we can say if λ+(E) = ∞, λ(E) = ∞ and similarly, if λ−(E) = ∞,λ(E) = −∞. Conversely, note by definition of λ+, if λ(E) = ∞, then λ+(E) = ∞ also and ifλ(E) = −∞, then λ−(E) =∞. So if λ+(E) =∞, what about λ−(E)? If λ−(E) =∞, that wouldforce λ(E) = −∞ contradicting the value it already has. Hence λ−(E) is finite. Next, given anymeasurable set F , what about λ−(F )? There are several cases. First, if F ⊆ E, then

λ−(E)& = λ−(F ) + λ−(F \ E).

16.2. HAHN DECOMPOSITION 307

Since λ− ≥ 0, if λ−(F ) = ∞, we get λ−(E), which is finite, is also infinite. Hence, this can nothappen. Second, if F and E are disjoint, with λ−(F ) =∞, we find

λ−(E ∪ F ) = λ−(E) + λ−(F ).

The right hand side is∞ and so since λ−(E ∪F ) is infinite, this forces λ(E ∪F ) = −∞. But sinceλ is additive on disjoint sets, this leads to the undefined expression

λ(E ∪ F ) = λ(E) + λ(F )

−∞ = (∞) + (−∞).

This is not possible because by assumption, λ takes on a well defined value in < for all measurablesubsets. Thus, we conclude if there is a measurable set E so that λ+(E) is infinite, then λ− will befinite everywhere. The converse is also true: if λ−(E) is infinite, then λ+ will be finite everywhere.Thus, we can conclude if λ is extended real valued, only one of λ+ or λ− can take on∞ values.

Theorem 16.1.2 The Jordan Decomposition Of A Charge

Let λ be an extended valued charge on (X,S). Then, λ+ and λ− are measures on S andλ = λ+ − λ−. The pair (λ+, λ−) is called the Jordan Decomposition of λ.

Proof 16.1.2By Comment 16.1.1, we can assume the charge λ takes on only ∞ on some measurable sets. Thefirst part of the proof of Theorem 16.1.1 is the same. It is clear both λ+ and λ− are nonnegative andadditive on finite unions of disjoint sets. The argument λ− is countably additive is the same as theone in the proof of Theorem 16.1.1 because we have assumed λ− is finite. There are two cases toshow λ+ is countably additive. These are

1. λ+(∪nEn) < ∞. In this case, we can use the argument of Theorem 16.1.1 to show thatλ+(∪nEn) ≥

∑n λ

+(En) which shows countable additivity in this case.

2. λ+(∪nEn) =∞. Here, the argument in the previous theorem shows there is a set A ⊆ ∪nEnso that λ(A) = ∞. Hence, in this case, there is at least one N so that λ(A ∩ EN ) = ∞.However, this implies λ+(En) = ∞ and so

∑n λ

+(En) = ∞ also. Hence, the two sidesmatch and again we have established countable additivity.

16.2 The Hahn Decomposition Associated With A ChargeNow we show that any charge λ has associated Positive and Negative sets.

Theorem 16.2.1 The Hahn Decomposition Associated With A Charge

Let λ be a charge on (X,S). Then, there is a positive set P and a negative set N so thatX = P ∪N and P ∩N = ∅. The pair (P,N) is called a Hahn Decomposition associatedwith the charge λ.

Proof 16.2.1We may assume that λ+ is finite; hence, by the Supremum Tolerance Lemma there are measurablesets An so that


λ(An) > λ+(X) − 1

2n(α)

for all n. Let A = lim supAn. Then, AC = lim inf An. Recall

lim supAn =

∞⋂m=1

∞⋃n=m

An

lim inf An =

∞⋃m=1

∞⋂n=m

ACn .

Let Bm = ∩∞n=mAn. Since Bm ⊆ An for m ≥ n, we see λ+(Bm) ≤ λ+(Am). Hence, we have

lim inf λ+(Bm) ≤ lim inf λ+(An).

But the sequence of setsBm increases monotonically to ∪mBm and so we know by Lemma 9.1.2 thatlimλ+(Bm) = λ+(∪mBm). Since lim inf λ+(Bm) = limλ+(Bm) here, we obtain

λ+(∪mBm) = λ+(lim inf λ+(An) ≤ lim infmλ+(An).

.Now, from Equation α, it then follows since λ+ is finite, that

λ+(ACn ) = λ+(X)− λ+(An)

≤ λ+(X)− λ(An)

≤ 1

2n.

Thus,

0 ≤ λ+(AC) ≤ lim inf λ+(ACn ) = 0

and we have shown that λ+(AC) = 0. It remains to show λ−(A) = 0. Note

0 ≤ λ−(An) = λ+(An)− λ(An)

≤ λ+(X)− λ(An) ≤ 1

2n.

Hence, for all m,

0 ≤ λ−(A) ≤ λ−( ∞⋃n=m

An

)

≤∞∑n=m

λ−(An) ≤∞∑n=m

1

2n

16.3. VARIATION 309

<1

2m−1.

This clearly implies that λ−(A) = 0 and we have the desired decomposition.

We can use the Hahn Decomposition to characterize λ+ and λ− is a new way.

Lemma 16.2.2 The Hahn Decomposition Characterization of a Charge

Let (A,B) be a Hahn Decomposition for the charge λ on (X,S). Then, ifE is measurable,λ+(E) = λ(E ∩ A) and λ−(E) = −λ(E ∩ B).

Proof 16.2.2Let D be a measurable subset of E ∩A. Then λ(D) ≥ 0 by the definition of the positive set A. Sinceλ is countably additive, we then have

λ

(E ∩ A

)= λ

((E ∩ A) ∩ D

)+ λ

((E ∩ A) ∩ DC

)= λ

(D

)+ λ

((E ∩ A) ∩ DC

)

But the second set is contained in E ∩ A and so its λ measure is non negative. Hence, we canoverestimate the left hand side as

λ

(E ∩ A

)≥ λ

(D

)≥ 0.

Since this is true for all subsets D, the definition of λ+ implies λ+(E ∩A) ≤ λ(E ∩A). Now,

λ+(E) = λ+(E ∩A) + λ+(E ∩B).

If F is a measurable subset of By the definition of E ∩ B, then λ(F ) ≤ 0 and so sup λ(F ) ≤ 0.This tells us λ+(E ∩ B) = 0. Thus, we have established that λ+(E) = λ+(E ∩ A). and soλ+(E) ≤ λ(E ∩A).

The reverse inequality is easier. SinceE∩A is a measurable subset ofE, the definition of λ+ impliesλ(E ∩A) ≤ λ+(E). Combining these results, we have λ+(E) = λ(E ∩A) as desired.

A similar argument shows that λ−(E) = −λ(E ∩B).

16.3 The Variation Of A Charge

A charge λ has associated with it a concept that is very similar to that of the variation of a function.We now define the variation of a charge.

Definition 16.3.1 The Variation of a Charge


Let (X,S) be a measure space and λ be a charge on S. For a measurable set E, a meshin E is a finite collection of disjoint measurable sets inside E, E1, . . . , En for somepositive integer n. Define the mapping Vλ by

Vλ(E) = sup ∑i

|λ(Ei)| | Ei is a mesh in E

where we interpret the sum as being over the finite number of sets in the given mesh. Wesay Vλ(E) is the total variation of λ on E and Vλ is the total variation of λ.

Theorem 16.3.1 The Variation of a Charge is a Measure

Let (X,S) be a measure space and λ be a finite charge on S. Then Vλ is a measure on S.

Proof 16.3.1Given a measurable set E, the Jordan Decomposition of λ implies that for a mesh E1, . . . , En inE, |λ(Ei)| ≤ λ+(Ei) + λ−(Ei). Hence, since λ+ and λ− are measures and countably additive, wehave ∑

i

|λ(Ei)| ≤∑i

λ+(Ei) +∑i

λ−(Ei)

≤ λ+(E) + λ−(E) < ∞

since λ+ and λ− are both finite. We conclude Vλ is a finite mapping.

Since the only mesh in ∅ is ∅ itself, we see Vλ(∅) = 0. It remains to show countable additivity. Let(En) be a countable disjoint family in S and let E be their union. Let A1, . . . , Ap be a mesh in E.Then each Ai is inside E and they are pairwise disjoint. Let Ai n = Ai ∩ En. Note Ai is the unionof the sets Ai n. Then it is easy to see A1n, . . . , Apn is a mesh in En. For convenience, call thismesh Mn. Then

p∑i=1

|λ(Ai)| =

p∑i=1

∑n

|λ(Ai n| =∑n

( p∑i=1

|λ(Ai n|).

The term in parenthesis is the sum over the mesh Mn of En. By definition, this is bounded above byVλ(En). Thus, we must have

p∑i=1

|λ(Ai)| ≤∑n

Vλ(Ai).

To get the other inequality, we apply the Supremum Tolerance Lemma to the definition of Vλ(En) tofind meshes

M εn = Aε1 n, . . . , Aεpn n,

where pn is a positive integer, so that

Vλ(En) <

pn∑i=1

|λ(Aεi n)| + ε/2n.

16.3. VARIATION 311

It follows that the union of a finite number of these meshes is a mesh of E. For each positive integerN , let

MN =

N⋃i=1

M εi

denote this mesh. Then,

N∑n=1

Vλ(En) <

N∑n=1

( pn∑i=1

|λ(Aεi n)| + ε/2n).

The first double sum corresponds to summing over a mesh of E and so by definition, we have

N∑n=1

Vλ(En) < Vλ(E) +

N∑n=1

ε/2n ≤ Vλ(E) +

∞∑n=1

ε/2n = Vλ(E) + ε.

Since N is arbitrary, we see the sequence of partial sums on the left hand side converges to a finitelimit. Thus,

∞∑n=1

Vλ(En) ≤ Vλ(E) + ε.

Since ε is arbitrary, the other desired inequality follows.

Theorem 16.3.2 Vλ = λ+ + λ−

Let (X,S) be a measure space and λ be a finite charge on S. Then Vλ = λ+ + λ−.

Proof 16.3.2Choose a measurable set E and let ε > 0 be chosen. Then, by the Supremum Tolerance Lemma,there is a mesh M ε = Aε1, . . . , Aεp so that

Vλ(E) − ε <∑i

|λ(Aεi)| ≤ Vλ(E).

Let F be the set of indices i in the mesh above where λ(Aεi) ≥ 0 and G be the other indices whereλ(Aεi) < 0. Let F be the union over the indices in F and G be the union over the indices in G. Notewe have

Vλ(E) − ε <∑i

|λ(Aεi)|

=∑F

|λ(Aεi)| +∑F

|λ(Aεi)|.

Now in F ,

|λ(Aεi)| = λ+(Aεi)− λ−(Aεi)

≤ λ+(Aεi),

and in G,

|λ(Aεi)| = λ−(Aεi)− λ+(Aεi)


≤ λ−(Aεi).

Thus, we can say

Vλ(E) − ε ≤∑F

λ+(Aεi) +∑G

λ−(Aεi)

= λ+(F) + λ−(G)

≤ λ+(E) + λ−(E).

Thus, for all ε > 0, we have

Vλ(E) ≤ λ+(E) + λ−(E) + ε.

This implies

Vλ(E) ≤ λ+(E) + λ−(E).

To prove the reverse, note ifA ⊆ E forE ∈ S, thenA itself is a mesh (a pretty simple one, of course)and so |λ(A)| ≤ Vλ(A). Further, λ(E) = λ(A) + λ(E \A). Thus, we have

2 λ(A) ≤ λ(A) + |λ(A)| ≤ λ(E) − λ(E \A) + |λ(A)|≤ λ(E) + |λ(E \A)| + |λ(A)|

But the collection A,E \A is a mesh for E and so

2 λ(A) ≤ λ(E) + Vλ(E).

Next, using the definition of λ+, we find

2 λ+(E) ≤ λ(E) + Vλ(E).

Finally, using the Jordan Decomposition of λ, we obtain

2 λ+(E) ≤ λ+(E) − λ−(E) + Vλ(E).

This immediately leads to λ+(E)− λ−(E) ≤ Vλ(E).

16.4 Absolute Continuity Of ChargesNow we are ready to look at absolute continuity in the context of charges.

Definition 16.4.1 Absolute Continuity Of Charges

Let (X,S, µ) be a measurable space and let λ be a charge on S. Then λ is said to beabsolutely continuous with respect to µ if whenever E is a measurable set with µ(E) = 0,then λ(E) = 0 also. We write this as λ µ. The set of all charges that are absolutelycontinuous with respect to µ is denoted by AC[µ].

There is an intimate relationship between the absolute continuity of Vλ, λ, λ+ and λ−; essentially,one implies all the others.

Theorem 16.4.1 Equivalent Absolute Continuity Conditions For Charges

16.4. ABSOLUTE CONTINUITY 313

Let (X,S, µ) be a measurable space. Then for the statements(1): λ+ and λ− are in AC[µ],(2): Vλ is in AC[µ], and(3): λ is in AC[µ], we have (1) ⇔ (2) ⇔ (3).

Proof 16.4.1

(1)→ (2): if µ(E) = 0, then λ+(E) and λ−(E) are also zero by assumption. Applying the JordanDecomposition of λ, we see λ(E) = 0 too. Hence, λ is in AC[µ].(2)→ (3): if µ(E) = 0, then Vλ(E) = 0. But, by Theorem 16.3.2, we have both λ+(E) and λ−(E)are zero. Then, applying the Jordan Decomposition again, we have λ(E) = 0. This tells us λ isabsolutely continuous with respect to µ.(3) → (1): Let (A,B) be a Hahn Decomposition of X due to λ. If µ(E) = 0, then λ(E) = 0by assumption. Thus, λ(E ∩ A) = λ(E ∩ B) = 0 as well. By Lemma 16.2.2, we then have thatλ+(E) = λ−(E) = 0 showing that (1) holds.

There is another characterization of absolute continuity that is useful.

Lemma 16.4.2 ε− δ Version Of Absolute Continuity Of a Charge

Let λ be a charge on S. Then

λ µ ⇔ ∀ ε > 0, ∃ δ > 0 3 |λ(E)| < ε for measurable E with µ(E) < δ.

Proof 16.4.2

(⇒): If λ is absolutely continuous with respect to µ, then note by Theorem 16.4.1 Vλ is also inAC[µ]. We will prove this result by contradiction. Assume the desired implication does not hold forVλ. Then, there is a positive ε so that for all n, there is a measurable set En with µ(En) < 1/2n andVλ(En) ≥ ε.Let Gn =

⋃∞k=n Ek and G =

⋂n Gn. Then,

µ(G) ≤ µ(Gn) ≤∞∑k=n

Ek <

∞∑k=n

1/2k = 1/2n−1.

Since this holds for all n, this implies µ(G) = 0. Since Vλ is in AC[µ], we then have Vλ(G) = 0.But

Vλ(G) = limn

Vλ(Gn) ≥ ε.

This contradiction implies that our assumption that the right hand side did not hold must be false.Hence, the condition holds for Vλ. It is easy to see that since Vλ = λ+ + λ−, that the conditionholds for them also. This then implies the condition holds for λ = λ+ − λ−.(⇐): We assume the condition on the right hand side holds. Now let (A,B) be a Hahn Decomposi-tion for X with respect to λ. In particular, if µ(E) = 0, then µ(E ∩A) = 0 also. The condition thenimplies λ(E ∩A) < ε. However, the choice of ε is arbitrary which then implies |λ(E ∩A)| = 0. Butthe absolute values are unnecessary as λ is non negative onA. We conclude λ+(E) = λ(E∩A) = 0.A similar argument then shows λ−(E) = −λ(E ∩ B) = 0. This tells us λ(E) = 0 by the JordanDecomposition.


Lemma 16.4.3 The Absolute Continuity Of The Integral

Let (X,S, µ) be a measure space and f be a summable function. Define the map λ byλ(E) =

∫Ef dµ for all measurable E. Then, λ is a charge with

λ+(E) =

∫E

f+ dµ, λ−(E) = −∫E

f− dµ.

Moreover, if Pf = x |f(x) ≥ 0 and Nf = PCf , then (Pf , Nf ) is a Hahn Decompositionfor X with respect to λ. Finally, since λ µ, we know for all positive ε, there is a positive

δ, so that if E is a measurable set with µ(E) < δ, then∣∣∣∣∫E f dµ

∣∣∣∣ < ε.

Proof 16.4.3It is easy to see that ν1 =

∫Ef+ dµ and ν2 =

∫Ef− dµ define measures and that λ = ν1 − ν2.

Hence, λ is a charge which is absolutely continuous with respect to µ. It is also easy to see that(Pf , Nf ) is a Hahn Decomposition for λ. Now if B is measurable and contained in the measurableset E, we have

λ(B) =

∫B∩Pf

f+ dµ −∫B∩Nf

f+ dµ

≤∫B∩Pf

f+ dµ

≤∫E∩Pf

f+ dµ.

Next, note that∫E∩Pf f

+dµ =∫Ef+dµ because the portion ofE that lies inNf does not contribute

to the value of the integral. Thus, for any B ⊆ E, we have

λ(B) ≤∫E

f+ dµ = ν1(E).

The definition of λ+ then implies two things: first, the inequality above tells us λ+(E) ≤ ν1(E) andsecond, since E∩Pf is a subset of E, we know λ(E∩Pf ) ≤ λ+(E). However, λ(E∩Pf ) = ν1(E)and hence, ν1(E) ≤ λ+(E) also. Combining, we have λ+(E) = ν1(E).

A similar argument shows that λ−(E) = ν2(E).

The last statement of the proposition follows immediately from Lemma 16.4.2.

16.5 The Radon - Nikodym Theorem

From our work above, culminating in Lemma 16.4.3, we know that integrals of summable functionsdefine charges which are absolutely continuous with respect to the measure we are using for theintegration. The converse of this is that if a measure is absolutely continuous, we can find a summablefunction so that the measure can be found by integration. That is if λ µ, there exists f summableso that λ(E) =

∫f dµ. This result is called the Radon - Nikodym theorem and as you might expect,

its proof requires some complicated technicalities to be addressed. Hence, we begin with a lemma.

Lemma 16.5.1 Radon - Nikodym Technical Lemma

16.5. RADON-NIKODYM 315

Let (X,S, µ) be a measurable space with µ(X) finite. Let λ be a measure which is finitewith λ(X) > 0 and λ µ. Then there is a positive ε and a measurable set A withµ(A) > 0 so that

ε µ(E ∩A) ≤ λ(E ∩A), ∀ E ∈ S.

Proof 16.5.1Pick a fixed ε > 0 and assume the set A exists. Let ν = λ− ε µ. Then, ν is a finite charge also. Note,our assumption tells us that

ν(B) = λ(B) − ε µ(B) ≥ 0,

for all measurable subsets B of A. Hence, by the definition of ν−, we must have that −ν−(A) ≥ 0or ν−(A) ≤ 0. But ν− is always non negative. Combining, we have ν−(A) = 0. This gives ussome clues as to how we can find the desired A. Note if (A,B) is a Hahn Decomposition for ν, thenwe have this desired inequality, ν−(A) = 0. So, we need to find a positive value of ε∗ so that when(A,B) is a Hahn Decomposition of

ν∗(A) = λ(A) − ε∗ µ(A),

we find ν∗(A) > 0.

To do this, for ε = 1/n, let (An, Bn) be a Hahn Decomposition for νn = λ−(1/n)µ. LetG = ∪nAnand H = ∩n Bn. We also know An ∪ Bn = X and An ∩ Bn = ∅ for all n. Further,

HC =

(⋂n

Bn

)C=⋃n

BCn =⋃n

An = G.

We conclude X = G ∪ H; it is easy to see G ∩ H = ∅. Now, H ⊆ Bn for all n, so νn(H) =−ν−n (H) ≤ 0 as Bn is a negative set. Hence, we can say

λ(H) − (1/n) µ(H) ≤ 0

which implies λ(H) ≤ (1/n) µ(H) for all n. Since λ is a measure, we then have

0 ≤ λ(H) ≤ µ(H)/n

which implies by the arbitrariness of n that λ(H) = 0. Hence,

λ(X) = λ(G) + λ(H) = λ(G).

Thus, λ(G) > 0 as λ(X) > 0. Since λ µ, it then follows that µ(G) > 0 also. SinceG = ∪nAn, itmust be true that there is at least one n with µ(An) > 0. Call this index N . Then, νN (E ∩AN ) ≥ 0as AN is a positive set for νN . This implies

λ(E ∩ AN ) − µ(E ∩ AN )

N≥ 0,

which is the result we seek using A = AN and ε = 1/N .

Theorem 16.5.2 The Radon - Nikodym Theorem: Finite Charge Case


Let (X,S, µ) be a measurable space with µ σ - finite. Let λ be a charge with λ µ. Then,there is a summable function f so that

λ(E) =

∫E

f dµ

for all measurable E. Moreover, if g is another summable function which satisfies thisequality, then f = g µ a.e. The summable function f is called the Radon - Nikodymderivative of λ with respect to µ and is often denoted by the usual derivative symbol:f = dλ

dµ . Hence, this equality is often written

λ(E) =

∫E

dλ

dµdµ

Proof 16.5.2We will do this in three steps.Step 1: We assume µ(X) is finite and λ is a finite measure.Step 2: We assume µ is σ - finite and λ is a finite measure.Step 3: We assume µ is σ - finite and λ is a finite charge.

As is usual, the proof of Step 1 is the hardest.Proof Step 1: Let

F = f : X → < | f ≥ 0, f summable and∫E

f dµ ≤ λ(E), ∀ E ∈ S.

Note since f = 0X is in F , F is nonempty. From the definition of F , we see∫Xf dµ ≤ λ(X) <∞

for all f in F . Hence,

c = supf∈F

∫X

f dµ < ∞.

We will find a particular f ∈ F so that c =∫Xf dµ. Let (fn) ⊆ F be a minimizing sequence:

i.e.∫Xfn dµ → c. We will assume without loss of generality that each fn is finite everywhere as

the set of points where all are infinite is a set of measure zero. Now, there are details that shouldbe addressed in that statement, but we have gone through those sort of manipulations many timesbefore. As an exercise, you should go through them again on scratch paper for yourself. With thatsaid, we will define a new sequence of finite functions (gn) by

gn = f1 ∨ f2 ∨ . . . ∨ fn

= max f1, . . . , fn.

This is a pointwise operation and it is clear that (gn) is an increasing sequence of non negativefunctions. Since f1 and f2 are summable, let A be the set of points where f1 > f2. Then,∫

X

f1 ∨ f2 dµ =

∫A

f1 dµ +

∫AC

f2 dµ

≤∫X

f1 dµ +

∫X

f2 dµ.

This tells us f1 ∨ f2 is summable also. A simple induction argument then tells us gn is summable


for all n.

Is gn ∈ F? Let E be measurable. Define the measurable sets (En) by

E1 = x | gn(x) = f1(x) ∩ E,

E2 = x | gn(x) = f2(x) ∩ (E \ E1),

...

En = x | gn(x) = fn(x) ∩ (E \ ∪n−1i=1 Ei).

Then, it is clear E = ∪i Ei, each Ei is disjoint from the others and gn(x) = fi(x) on Ei. Thus,since each fi is in F , we have∫

E

gn dµ =

n∑i=1

∫Ei

fi dµ

≤n∑i=1

λ(Ei) = λ(∪ni=1 Ei)

= λ(E).

We conclude each gn is in F for all n. Next, if g = sup gn, then gn ↑ g and∫E

gn dµ ≤ λ(E) ≤ λ(X)

for all n. Now apply the Monotone Convergence Theorem to see g is summable and∫E

gn dµ→∫E

g dµ ≤ λ(E).

Let’s define f by

f(x) =

g(x) g(x) < ∞,0 g(x) = ∞.

Since g is summable, the set of points where it takes on the value ∞ is a set of measure 0. Thus,f = g µ a.e. and f is measurable. It is easy to see f is in F .

Moreover, since fn ≤ gn, we have

c = limn

∫X

fn dµ

≤ limn

∫X

gn dµ ≤ c

because gn ∈ F . Thus,

c = limn

∫X

gn dµ =

∫X

g dµ.

This immediately tells us that∫Xf dµ = c with f ∈ F .


Next, define m : S → < by

m(E) = λ(E) −∫E

f dµ,

for all measurable E. It is straightforward to show m is difference of two measures and hence is afinite charge. Also, since f is in F , we see m is non negative and thus is a measure. In addition,since λ µ and the measure defined by

∫Ef dµ is also absolutely continuous with respect to µ, we

have that m µ too. Now if m(X) = 0, this would imply, since m(E) ≤ m(X), that

0 ≤ λ(E) −∫E

f dµ ≤ m(X) = 0.

But this says λ(E) =∫Ef dµ for all measurable E which is the result we seek.

Hence, it suffices to show m(X) = 0. We will do this by contradiction. Assume m(X) > 0. Nowapply Lemma 16.5.1 to conclude there is a positive ε and measurable set A so that µ(A) > 0 and

ε µ(E ∩A) ≤ m(E ∩A), (∗)

for all measurable E. Define a new function h using Equation ∗ by h = f + ε IA. Then for a givenmeasurable E, we have ∫

E

h dµ =

∫E

f dµ + ε µ(E ∩A)

≤∫E

f dµ + m(E ∩A)

by Equation ∗. Now replace m by its definition to find∫E

h dµ ≤∫E

f dµ + λ(E ∩A) −∫E∩A

f dµ

=

∫E∩AC

f dµ + λ(E ∩A).

Finally, use the fact that f is in F to conclude∫E

h dµ ≤ λ(E ∩AC) + λ(E ∩A) = λ(E).

This shows that h is in F . However,∫X

h dµ =

∫X

f dµ + ε µ(A) > c!

which is our contradiction. This completes the proof of Step 1.

Proof Step 2: Now µ is σ finite. This means there is a countable sequence of disjoint measurablesets (Xn) with µ(Xn) finite for each n and we can write X = ∪n Xn. Let Sn be the σ - algebra ofsubsets of Xn given by S ∩Xn. By Step 1, there are summable non negative functions fn so that

λ(F ) =

∫F

fn d µ,


for each F in Sn. Now define f by f(x) = fn(x) when x ∈ Xn. This is a well - defined function andit is easy to see f is measurable. We also know by Theorem 9.4.5 that µf defined by µf (E) =

∫Ef dµ

is a measure. Now if E is measurable, then E = ∪n E ∩Xn, E = ∪n E ∩Xn and

µf (E) =

∫E

f d µ =

∫∪n E∩Xn

f d µ = limn

∫∪ni=1 E∩Xn

f d µ

Then, for any n,∫∪ni=1 E∩Xi

f d µ =

n∑i=1

∫E∩Xi

f d µ =

n∑i=1

∫E∩Xi

fn d µ

=

n∑i=1

λ(E ∩Xi) = λ(∪ni=1 E ∩Xi) ≤ λ(E),

which is a finite number. Hence, the series of non negative terms∑n

∫E∩Xn f d µ converges to a

finite number and∫E

f d µ =∑n

∫E∩Xn

fn d µ = λ(∪n E ∩Xn) = λ(E).

This establishes the result for Step 2.

Proof Step 3: Here, we have µ is σ - finite and λ is a finite charge. By the Jordan Decomposition ofλ, we can write

λ(E) = λ+(E) − λ−(E),

for all measurable E. Now apply Step 2 to find non negative summable functions f+ and f− so that

λ+(E) =

∫E

f+ d µ,

λ−(E) =

∫E

f− d µ.

Let f = f+ − f− and we are done with the proof of Step 3.

Finally, it is clear from the proof above, that the Radon - Nikodym derivative of λ with respect to µ,is unique up to redefinition on a set of µ measure 0.

We can also prove the Radon- Nikodym theorem for the case that λ is a sigma-finite charge. Ofcourse, this includes the case where λ is a charge with values in <. Recall by Comment 16.1.1 thatin this case, the charge λ can only take on∞ or −∞ values as it must be additive.

Theorem 16.5.3 The Radon - Nikodym Theorem: Sigma-Finite Charge Case


Let (X,S, µ) be a measurable space with µ σ - finite. Let λ be a sigma-finite measure withλ µ. Then, there is a function f so that

λ(E) =

∫E

f dµ

for all measurable E. Moreover, if g is another function which satisfies this equality, thenf = g µ a.e. The function f is called the Radon - Nikodym derivative of λ with respectto µ and is often denoted by the usual derivative symbol: f = dλ

dµ . Hence, this equality isoften written

λ(E) =

∫E

dλ

dµdµ

Proof 16.5.3We can still do this in three steps.Step 1: We assume µ and λ are finite measures.Step 2: We assume µ and λ are sigma-finite measures.Step 3: We assume µ is σ - finite and λ is a sigma-finite charge.

The proof of Step 1 is identical to the one presented in Theorem 16.5.2. The proof of Step 2 is alsoquite similar. The only difference is that we choose a disjoint countable collection of sets Xn forwhich both µ and λ are finite.

Proof Step 2: Since µ and λ are σ finite, there is a countable sequence of disjoint measurable sets(Xn) with µ(Xn) and λ(Xn) finite for each n and X = ∪nXn. Let Sn be the σ - algebra of subsetsof Xn given by S ∩Xn. By Step 1, there are summable non negative functions fn so that

λ(F ) =

∫F

fn d µ,

for each F in Sn. Now define f by f(x) = fn(x) when x ∈ Xn. This is a well - defined function andit is easy to see f is measurable. If E is measurable, then E = ∪n E ∩Xn, E = ∪n E ∩Xn and∫

E

f d µ =

∫∪n E∩Xn

f d µ.

Then, for any n,∫∪ni=1 E∩Xi

f d µ =

n∑i=1

∫E∩Xi

f d µ =

n∑i=1

∫E∩Xi

fn d µ

=

n∑i=1

λ(E ∩Xi) = λ(∪ni=1 E ∩Xi) ≤ λ(E).

Now, since λ is sigma-finite, it is possible for this value to be ∞. However, whether it is a finitenumber or infinite in value, the partial sums defined on the left hand side are clearly bounded aboveby this number. Hence, the series of non negative terms

∑n

∫E∩Xn f d µ converges and∫

E

f d µ =∑n

∫∪n E∩Xn

fn d µ = λ(E ∩Xn) = λ(E).

16.6. LEBESGUE DECOMPOSITION 321

We can no longer say the function f is summable, of course. This establishes the result for Step 2.

Proof Step 3: Here, we have µ is σ - finite and λ is a sigma-finite charge. From Theorem 16.1.2, wemay assume λ has a Jordan Decomposition λ = λ+ − λ− with λ+ a finite measure. Thus, we canwrite

λ(E) = λ+(E) − λ−(E),

for all measurable E. Now apply Step 2 to find non negative functions f+ and f− so that

λ+(E) =

∫E

f+ d µ,

λ−(E) =

∫E

f− d µ.

Since λ+ is assumed finite, it is clear f+ is actually summable and so the addition defined above iswell defined in all cases. We let f = f+ − f− and we are done with the proof of Step 3.

16.6 The Lebesgue Decomposition of a Measure

Definition 16.6.1 Singular Measures

Let (X,S, µ) be a measure space and let λ be a charge on S. Assume there is a decompo-sition of X into disjoint measurable subsets U and V (X = U ∪ V and U ∩ V = ∅) sothat µ(U) = 0 and λ(E ∩ V ) = 0 for all measurable subsets E of V . In this case, we sayλ is perpendicular to µ and write λ ⊥ µ.

Comment 16.6.1 If λ ⊥ µ, let (U, V ) be a decomposition of X associated with the singular measureλ. We then know that µ(U) = 0 and λ(E ∩ V ) = 0 for all measurable E. Note, if E is measurable,then

E =

(E ∩ U

)∪(E ∩ V

).

Thus,

λ(E) = λ

(E ∩ U

)+ λ

(E ∩ V

)= λ

(E ∩ U

).

Further,

µ(E) = µ

(E ∩ U

)+ µ

(E ∩ V

)= µ

(E ∩ V

).

Comment 16.6.2 If λ ⊥ µ with λ 6= 0, then there is a measurable set E so that λ(E ∩ U) 6= 0. Butfor this same set µ(E ∩ U) = 0 as E ∩A is a subset of U . Thus, λ 6 µ.

Comment 16.6.3 If λ ⊥ µ and λ µ, then for any measurable set E, we have λ(E) = λ(E ∩ U).But, since µ(E ∩ U) = 0, we must have λ(E ∩ U) = 0 because λ µ. Thus, λ = 0.

Comment 16.6.4 It is easy to prove that λ ⊥ µ implies Vλ ⊥ µ, λ+ ⊥ µ and λ− ⊥ µ. Also, ifλ+ ⊥ µ and λ− ⊥ µ, this implies λ ⊥ µ.


Theorem 16.6.1 Lebesgue Decomposition Theorem

Let (X,S, µ) be a σ - finite measure space. Let λ be a finite charge on S. Then, there aretwo unique finite measures, λac µ and λp ⊥ µ such that λ = λac + λp.

Proof 16.6.1We will prove this result in four steps.Step 1: λ and µ are finite measures.

Step 2: µ is a σ - finite measure and λ is a finite measure.

Step 3: µ is a σ - finite measure and λ is a finite charge.

Step 4: The decomposition is unique.

Proof Step 1: As is usual, this is the most difficult step. We can see, in this case, that λ + µ is ameasure. Note that (λ+ µ)(E) = 0 implies that λ(E) is 0 too. Hence, λ (λ+ µ). By the Radon- Nikodym Theorem, there is then a non negative λ+ µ summable f so that for any measurable E,

λ(E) =

∫E

f d (λ + µ).

Hence, f is µ and λ summable as well and

λ(E) =

∫E

f d λ +

∫E

f d µ).

Let

A1 = x | f(x) = 1,A2 = x | f(x) > 1, and

B = x | f(x) < 1.

Also, for each n, let

En = x | f(x) ≥ 1 + 1/n.

Then, we see immediately A2 = ∪n En and X = A ∪B. Now, we also have

λ(En) =

∫En

f d (λ + µ)

≥ (1 + 1/n)

(λ(En) + µ(En)

).

This implies λ(En) ≥ (1 + 1/n) λ(En) which tells us λ(En) ≤ 0. But since λ is a measure, thisforces λ(En) = 0. From the same inequality, we also have λ(En) ≥ λ(En) + µ(En). which forcesµ(En) = 0 too.Next, note the sequence of sets (En) increases to A2 and so

limn

µ(En) = µ(A2),

limn

λ(En) = λ(A2).

16.6. LEBESGUE DECOMPOSITION 323

Since µ(En) = λ(En) = 0 for all n, we conclude µ(A2) = λ(A2) = 0.

Also,

λ(A1) =

∫A1

f d (λ + µ)

=

∫A1

1 d (λ + µ)

= µ(A1) + λ(A1),

which implies µ(A1) = 0. Let A = A1 ∪ A2. Then, the above remarks imply µ(A) = 0. Wenow suspect that A and B will gives us the decomposition of X which will allow us to construct themeasures λac µ and λp ⊥ µ. Define λac and λp by

λac = λ(E ∩B),

λp = λ(E ∩A).

Then,

λ(E) = λ(E ∩A) + λ(E ∩B)

= λac(E) + λp(E),

showing us the we have found a decomposition of λ into two measures.Is λac µ? Let µ(E) = 0. Then µ(E ∩B) = 0 as well. Now, we know

λ(E ∩B) =

∫E∩B

f d (λ + µ)

=

∫E∩B

f d λ +

∫E∩B

f d µ.

However, the second integral must be zero since µ(E ∩B) = 0. Thus, we have

λ(E ∩B) =

∫E∩B

f d λ.

We also have λ(E ∩B) =∫E∩B 1 dλ and so∫

E∩B1 d λ =

∫E∩B

f d λ.

Thus, ∫E∩B

(1 − f

)d λ = 0.

But on E∩B, 1−f > 0; hence, we must have λ(E∩B) = 0. This means λac(E∩B) = 0 implyingλac µ.

Is λp ⊥ µ? Note, for any measurable E, we have

λp(E ∩B) = λ

((E ∩B) ∩A

)= λ(∅) = 0.


Thus, λp ⊥ µ. In fact, we have shown

λ(E) =

∫E∩B

f d λ + λp(E).

Proof Step 2:Note that once we find a decompositionX = A∪B withA andB measurable and disjoint satisfyingµ(A) = 0 and λ(E ∩B) = 0 if µ(E) = 0, then we can use the technique in the proof of Step 1. Welet λac(E) = λ(E ∩B) and λp(E) = λ(E ∩A). This furnishes the decomposition we seek. Hence,we must find a suitable A and B.

The measure µ is now σ - finite. Hence, there is a sequence of disjoint measurable sets Xn withµ(Xn) <∞ and X = ∪nXn. Let Sn denote the σ - algebra of subsets S ∩Xn. By Step 1, there is adecompositionXn = An∪Bn of disjoint and measurable sets so that µ(An) = 0 and λ(E∩Bn) = 0if µ(E) = 0. Since the sets Xn are mutually disjoint, we know the sequences (An) and (Bn) aredisjoint also. Let A = ∪n An and B = AC and note AC = ∩n Bn. Then, since µ is a measure, wehave

µ(∪ni=1 Ai) =

n∑i=1

µ(Ai) = 0

for all n. Hence,

µ(A) = limn

µ(∪ni=1 Ai) = 0.

Next, if µ(E) = 0, then µ(E ∩Bn) = 0 for all n by the properties of the decomposition (An, Bn) ofXn. Since

E ∩ B = ∩n(E ∩Bn

),

and λ(E ∩B1) is finite, we have

λ(E ∩ B) = limn

λ(E ∩Bn).

However, each λ(E∩Bn) is zero because µ(E) = 0 by assumption. Thus, we conclude λ(E∩B) =0. We then have the A and B we need to construct the decomposition.

Proof Step 3: The mapping λ is now a finite charge. Let λ = λ+−λ− be the Jordan Decompositionof the charge λ. Applying Step 2, we see there are pairs of measurable sets (A1, B1) and (A2, B2)so that

X = A1 ∪ B1, A1 ∩ B1 = ∅, µ(A1) = 0, µ(E) = 0 ⇒ λ+(E ∩ B1) = 0,

and

X = A2 ∪ B2, A2 ∩ B2 = ∅, µ(A2) = 0, µ(E) = 0 ⇒ λ−(E ∩ B2) = 0.

Let A = A1 ∪ A2 and B = B1 ∩ B2. Note BC = A. It is clear then that µ(A) = 0. Finally, ifµ(E) = 0, then λ+(E ∩B1) = 0 and λ−(E ∩B2) = 0. This tells us

λ(E ∩ B) = λ+(E ∩ B) − λ−(E ∩ B)

= λ+(E ∩ B1 ∩ B2) − λ−(E ∩ B1 ∩ B2)

16.7. HOMEWORK 325

= λ+

((E ∩ B1) ∩ B2

)− λ−

((E ∩ B2) ∩ B1

).

Both of the terms on the right hand side are then zero because we are computing measures of subsetsof a set of measure 0. We conclude λ(E ∩B) = 0. The decomposition is then

λac(E) = λ(E ∩B) =

(λ+ − λ−

)(E ∩ B),

λp(E) = λ(E ∩A) =

(λ+ − λ−

)(E ∩ A).

Proof Step 4: To see this decomposition is unique, assume λ = λ1 + λ2 and λac + λp are twoLebesgue decompositions of λ. Then, λac−λ1 = λ2−λp. But since λ1 and λac are both absolutelycontinuous with respect to µ, it follows that λac − λ1 µ also. Further, since both λ2 and λp aresingular with respect to µ, we see λ2 − λp ⊥ µ. However, λac − λ1 = λ2 − λp by assumption andso λac − λ1 µ and λac − λ1 ⊥ µ. By Comment 16.6.3, this tells us λac = λ1. This then impliesλ2 = λp.

16.7 HomeworkExercise 16.7.1 Let (X,S) be a measurable space and λ is a charge on S. Prove if P1 and P2 arepositive sets for λ, then P1 ∪ P2 is also a positive set for λ.

Exercise 16.7.2 Let g1(x) = 2x, g2(x) = I[0,∞), g3(x) = x I[0,∞) and g4(x) = arctan(x). All ofthese functions generate Borel - Stieljes measures on <.

(i): Determine which are absolutely continuous with respect to Borel measure. Then, if absolutelycontinuous with respect to Borel measure, find their Radon - Nikodym derivative.

(ii): Which of these measures are singular with respect to Borel measure?

Exercise 16.7.3 Let λ and µ be σ - finite measures on S, a σ - algebra of subsets of a set X . Assumeλ is absolutely continuous with respect to µ. If g ∈M+(X,S), prove that∫

g dλ =

∫g f dµ

where f = dλ/dµ is the Radon - Nikodym derivative of λ with respect to µ.

Exercise 16.7.4 Let λ, ν and µ be σ - finite measures on S , a σ - algebra of subsets of a set X . Usethe previous exercise to show that if ν λ and λ µ, then

dν

dµ=

dν

dλ

dλ

dµ, µ a.e.

Further, if λ1 and λ2 are absolutely continuous with respect to µ, then

d(λ1 + λ2)/dµ = dλ1/dµ + dλ2/dµ µ a.e.

Exercise 16.7.5 Prove the results of Comment 16.6.4.

Chapter 17

Connections To Riemann Integration

Theorem 17.0.1 Every Riemann Integrable Function Is Lebesgue Integrable and The Two In-tegrals Coincide

Let f [a, b] → < be a Riemann integrable function. If we let inf fdµ =∫

[a,b]fdµ rep-

resent the Lebesgue integral of f with µ denoting Lebesgue measure on the real line and∫ baf(x)dx be the Riemann integral of f on [a, b], then

∫[a,b]

fdµ =

∫ b

a

f(x)dx.

Proof 17.0.1For each positive integer n, let πn denote the uniform partition of [a, b] which divides the intervalinto pieces of length 1

2n . Hence, πn = a + i b−a2n : 0 ≤ i ≤ 2n. Let mi and Mi be defined asusual in Chapter 4. Define the simple functions φn and ψn as follows:

φn(x) =

2n∑i=1

miI[xi,xi+1)

ψn(x) =

2n∑i=1

MiI[xi,xi+1).

Then φn ≤ φn+1 ≤ f and ψn ≥ ψn+1 ≥ f for all n Further, these monotonic limits define mea-surable functions g and h so that φn ↑ g and ψn ↓ h pointwise on [a, b]. Finally, it follows thatg ≤ f ≤ h on [a, b].

Next, note that

∫φn dµ =

2n∑i=1

miµ([xi, xi+1))

=

2n∑i=1

mi(xi+1 − xi)

327

328 CHAPTER 17. CONNECTIONS TO RIEMANN INTEGRATION

= L(f, πn)∫ψn dµ =

2n∑i=1

Miµ([xi, xi+1))

=

2n∑i=1

Mi(xi+1 − xi)

= U(f, πn)

Since f is Riemann integrable by assumption, we know L(f, πn) →∫ baf(x)dx and U(f, πn) →∫ b

af(x)dx. Hence, limn

∫φndµ ≤

∫ baf(x)dx. It follows from Levi’s Theorem 9.5.4 that g is

summable and∫φn dµ ↑

∫gdµ. We can apply Levi’s Theorem again to −psin to conclude h is

summable and∫ψn dµ ↓

∫hdµ.

However, we also know that

∫(h− g) dµ = lim

n

∫(psin − φn) dµ

= limn

(U(f, πn)− L(f, πn)

)= 0.

We conclude h = g a.e. Since g ≤ f ≤ h, this tells us f = g = h a.e. Since Lebesgue measure iscomplete and g and h are measurable, we now know f is measurable and summable with

∫f dµ = lim

n

∫φn dµ = lim

nL(f, πn) =

∫ b

a

f(x)dx.

We can prove a connection between Riemann-Stieljes integrals and Lebesgue-Stieljes integrals also.

Theorem 17.0.2 Riemann-Stieljes Integrable Functions Are Lebesgue-Stieljes Integrable andThe Two Integrals Coincide: One

Let f : [a, b] → < and g : [a, b] → < be bounded functions. Assume g is monotoneincreasing and continuous from the right on [a, b]. Assume further that f is Riemann-Stieljes integrable with respect to the integrator g on [a, b]. Let µg denote the Lebesgue-Stieljes measure induced by g on [a, b]. If we let

∫[a, b]fdµg represent the Lebesgue-

Stieljes integral of f and∫ baf(x)dg be the Riemann-Stieljes integral of f with respect to g

on [a, b], then

∫[a,b]

f dµg =

∫ b

a

f(x)dg.

Proof 17.0.2For each positive integer n, let πn denote the uniform partition of [a, b] which divides the intervalinto pieces of length 1

2n . Hence, πn = a + i b−a2n : 0 ≤ i ≤ 2n. Let mi and Mi be defined asusual in Chapter 4. Define the simple functions φn and ψn as follows:

329

φn(x) =

2n∑i=1

miI[xi,xi+1)

ψn(x) =

2n∑i=1

MiI[xi,xi+1).

Then φn ≤ φn+1 ≤ f and ψn ≥ ψn+1 ≥ f for all n Further, these monotonic limits define mea-surable functions u and v so that φn ↑ u and ψn ↓ v pointwise on [a, b]. Finally, it follows thatu ≤ f ≤ v on [a, b].

Next, note that

∫φn dµg =

2n∑i=1

miµg([xi, xi+1))

=

2n∑i=1

mi(g(xi+1)− g(xi))

= L(f, g, πn)∫φn dµg =

2n∑i=1

Miµg([xi, xi+1))

=

2n∑i=1

Mi(g(xi+1)− g(xi))

= U(f, g, πn)

Since f is Riemann-Stieljes integrable with respect to g by assumption, we know L(f, g, πn) →∫ baf(x)dg and U(f, g, πn) →

∫ baf(x)dg. Hence, limn

∫φndµg ≤

∫ baf(x)dg. It follows from

Levi’s Theorem 9.5.4 that u is summable and∫φn dµg ↑

∫u dµg . We can apply Levi’s Theorem

again to −psin to conclude v is summable and∫ψndµ ↓

∫vdµg .

However, we also know that

∫(u− v) dµg = lim

n

∫(ψn − φn) dµg

= limn

(U(f, g, πn)− L(f, g, πn)

)= 0.

We conclude u = v a.e. Since u ≤ f ≤ v, this tells us u = f = v a.e. Since Lebesgue measure iscomplete and u and v are measurable, we now know f is measurable and summable with

∫f dµg = lim

n

∫φn dµg = lim

nL(f, g, πn) =

∫ b

a

f(x)dg.

330 CHAPTER 17. CONNECTIONS TO RIEMANN INTEGRATION

Comment 17.0.1 It is easy to see this theorem extends to Riemann-Stieljes integrators g that are ofbounded variation. If g is of bounded variation and continuous from the right, then we can writeg = u − v where both u and v are increasing and right continuous. The function u determines aLebesgue-Stieljes measure µu; the function v, the Lebesgue-Stieljes measure µv and finally g definesthe charge µg = µu − µv . Then we see

∫f dµg =

∫ bafdg.

Chapter 18

Differentiation

We will discuss the kinds of properties a function f needs to have so that we have a FundamentalTheorem of Calculus type result: f(b)− f(a) =

∫ baf ′dµ in our setting of measures on the real line.

18.1 Absolutely Continuous Functions

Definition 18.1.1 Absolute Continuity Of Functions

Let [a, b] be a finite interval in <. f : [a, b] → < is absolutely continuous if for eachε > 0, there is a δ > 0 such that if [an, bn] is any finite or countable collection of nonoverlapping closed intervals in [a, b] with

∑k(bk−ak) < δ then

∑k |f(bk)−f(ak)| < ε.

Theorem 18.1.1 Properties Of Absolutely Continuous Functions

Let f be absolutely continuous on the finite interval [a, b]. Then

1. f is continuous on [a, b],

2. f is of bounded variation on [a, b],

3. If E is a subset of < with Lebesgue Measure 0, then f(E) is Lebesgue measurablealso with measure 0; i.e. µ(E) = 0⇒ µ(f(E)) = 0.

Proof 18.1.1Condition (1) is clear. To prove (2), choose a δ > 0 so that if [ak, bk] is any collection of nonoverlapping closed intervals in [a, b] with

∑k(bk − ak) < δ then

∑k |f(bk)− f(ak)| < 1. If [c, d]

is any interval in [a, b] with d− c < δ, then for any non overlapping sequence of intervals ([αn, βn])inside [c, d] whose summed length is less than δ, we must have

∑|f(αn) − f(βn)| < 1. It then

follows that V (f, c, d) ≤ 1.

Now choose the integer N so that N > b−aδ . Partition [a, b] into N intervals I1 to IN . The variation

of f on each Ik is less than 1; hence, V (f, a, b) ≤ N <∞. This proves condition (2).

To prove (3), let ε > 0 be given. Choose δ > 0 so that if [ck, dk] is any collection of nonoverlapping closed intervals in [a, b] with

∑k(dk − ck) < δ then

∑k |f(dk) − f(ck)| < ε. By the

infimum tolerance lemma, there is an open set G so that 0 = µ(E) ≤ µ(G) + δ. Since G is open,

331

332 CHAPTER 18. DIFFERENTIATION

we also know we can write G as a finite or countable union of disjoint open intervals (ak, bk); i.e.G = ∪k(ak, bk). Then

f(E) ⊆ f(G)

⊆ f

(∪k[ak, bk]

)⊆

⋃k

[f(uk), f(vk)]

where uk and vk are the points in [ak, bk] where f achieves its minimum and maximum, respectively.Hence, µ∗(f(E)) ≤

∑k µ∗([f(uk), f(vk)] or µ∗(f(E)) ≤

∑k([f(vk) − f(uk)]. However, the

points uk and vk determine a closed subinterval of [ak, bk] with

∑k

|vk − uk| ≤∑k

|bk − ak| < δ.

Since the intervals (ak, bk) are disjoint, the intervals formed by uk and vk are non overlapping.Thus, we conclude

∑k(f(vk)− f(uk)) < ε. This tells us µ∗(f(E)) < ε. But since ε was arbitrary,

we see µ∗(f(E)) = 0; thus, f(E) is measurable and has measure 0.

18.2 Lebesgue-Stieljes Measures and Absolutely Continuous Func-tions

If f is continuous and non decreasing, there is a nice connection between f and the associatedLebesgue-Stieljes measure µf . First, we explore a relationship between the Lebesgue-Stieljes outermeasure and Lebesgue outer measure.

Theorem 18.2.1 µ∗f (E) = µ∗(f(E))

Let f be continuous and non decreasing on < and let µ∗f be the associated Lebesgue-Stieljes outer measure. For all sets E in <, µ∗f (E) = µ∗(f(E)).

Proof 18.2.1Let E be in <. Let G be an open set that contains f(E). We know G can be written as a countabledisjoint union of open intervals Jn. Since f is continuous and non decreasing, we must have In =f−1(Jn) is an interval as well. Letting In = (an, bn) gives Jn = (f(an), f(bn)). Noting E ⊆ ∪nInand µ(G) =

∑n µ(Jn), we find

µ∗f (E) ≤ µ∗f (∪nIn) = µf (∪nIn)

≤∑n

µf (In) =∑n

µ(Jn)

= µ(G).

This is true for all such open sets G. We therefore conclude µ∗f (f(E)) ≤ µ∗(f(E)) by Theorem12.2.4.

18.3. BOUNDED VARIATION DERIVATIVES 333

The above immediately tells us that if µ∗f (E) =∞, we also have µ∗(f(E)) =∞ and the result holds.We may thus safely assume that µ∗f (E) is finite for the rest of the argument. Let ε > 0 be given. Let((an, bn]) be a collection of half open intervals whose union covers E with

∑n (f(bn)− f(an)) <

µ∗f (E) + ε. Let Jn = f((an, bn]). Again, since f is continuous and non decreasing, each intervalJn can be written as Jn = (f(an), f(bn)]. Since f(E) ⊆ ∪nJn, we must have

µ∗(f(E)) ≤ µ∗(∪nJn)

≤∑n

(f(bn)− f(an)) < µ∗f (E) + ε.

Since ε is arbitrary, we see µ∗(f(E)) ≤ µ∗f (E). This second inequality completes the argument.

There is a very nice connection between a continuous non decreasing f and its associated Lebesgue-Stieljes measure µf .

Theorem 18.2.2 f Is Absolutely Continuous IFF µf << µ.

A continuous non decreasing function f is absolutely continuous on [a, b] if and only if itsassociated Lebesgue-Stieljes measure µf is absolutely continuous with respect to Lebesguemeasure µ.

Proof 18.2.2Let f be continuous and non decreasing on [a, b]. By Theorem 18.2.1, we have µ∗f (E) = µ∗(f(E))for all subsets E in [a, b].

If we assume f is absolutely continuous on [a, b], by Condition (3) of Theorem 18.1.1, when E hasLebesgue measure 0, we know µ(f(E)) = 0 as well. Thus, µ∗(f(E)) = 0 = µ∗f (E). But then E isµf measurable with µf measure 0. This tells us µf µ.

Conversely, if µf µ, we can apply Lemma 16.4.3. Given ε > 0, there is a δ > 0, so thatµ(E) < δ implies µf (E) < ε. Hence, for any E which is a non overlapping countable union ofintervals [an, bn] with

∑n(bn − an) < δ, we have

∑n(f(bn)− f(an)) = µf (E) < ε. This says f

is absolutely continuous on [a, b].

18.3 Derivatives of Functions of Bounded Variation

We are going to look carefully at the behavior of monotone functions and their rates of change.

Definition 18.3.1 Derived Numbers

We say the extended real number α is a derived number for a function f at the point x0 indom(f) is there is a sequence of nonzero numbers x0 + hn in dom(f) with hn → 0 sothat

limn

f(x0 + hn)− f(x0)

hn= α

We use the notation Df(x) to denote the collection of all derived numbers for a function fat the point x in its domain.


If a function f is strictly increasing on [a, b] we want to be able to look quantitatively how the outermeasure size of E compares to f(E) for any subset E. To do this, we need a technical tool called aVitali Cover.

Definition 18.3.2 Vitali Cover

Let I be a collection of non degenerate closed intervals in <. Let E be a subset of < andV be a sub collection of I. If for all x ∈ E and ε > 0, there is a V ∈ V so that x ∈ V andµ(V ) < ε, we say V is a Vitali Cover for E or a Vitali Covering for E.

We can prove this important result.

Theorem 18.3.1 Vitali Covering Theorem

Let V be a Vitali Cover of a set E in <. The there is a countable collection (Vn) in V thatis pairwise disjoint and µ(E \ ∪nVn) = 0.

Proof 18.3.1First assume E is bounded. Let J be any open interval containing E and let V0 be those intervalsin V that are contained in J . Then V0 is also a Vitali Cover for E. Let V1 be chosen from V0. Ifµ(E \ V1) = 0, we have proven our conjecture. If not, we continue this process using induction.Choose V2 this way. Let F1 = V1 and G1 = J \ F1. Then G1 is open. Define the new collection V1

by

V1 = V ∈ V0 : V ⊆ G1.

Since by assumption E \ V1 is not empty and V0 is a Vitali Cover for E, there must be sets in thefamily V1. Let

S1 = sup µ(V ) : V ∈ V1.

The members of a Vitali Cover are non degenerate which implies S1 > 0. Further, each member ofV0 is in J which tells us S1 is finite. Choose V2 from V1 so that µ(V2) > S1

2 . Then V2 ⊆ G1 implyingV1 and V2 are disjoint. If µ(E \ V1 ∪ V2) = 0 we are done. Otherwise, we continue.

We can now see how to do the induction. Suppose we have chosen the sets V1, V2, . . . , Vn so thatthey are pairwise disjoint and µ(E \ ∪ni=1Vi) 6= 0. Then, let Fn = ∪ni=1Vi and Gn = J \ Fn. ThenGn is open. Define the new collection Vn by

Vn = V ∈ V0 : V ⊆ Gn.

Since by assumption E \ Vn is not empty and Vn is also a Vitali Cover for E, there must be sets inthe family Vn. Let

Sn = sup µ(V ) : V ∈ Vn.

The members of a Vitali Cover are non degenerate which implies Sn > 0. Further, each member ofV0 is in J which tells us Sn is finite. Choose Vn+1 from Vn so that


µ(Vn+1) >Sn2. (18.1)

Then again, Vn+1 ⊆ Gn implying the new collection V1, V2, . . . , Vn, Vn+1 is pairwise disjoint.If this process terminates after a finite number of steps, we have proven the result. Otherwise, weconstruct a countable number of sets Vn which form a pairwise disjoint collection from V0. LetS = ∪nVn. We will show µ(E \ S) = 0 proving the result.

Each Vn has a midpoint. Let Wn be an interval with the same midpoint as Vn which is 5 times thelength. Then µ(Wn) = 5µ(Vn). Also, by construction S = ∪nVn ⊆ E ⊆ J and so

∑n

µ(Wn) = 5∑n

µ(Vn) ≤ 5 µ(J). (18.2)

Now let x ∈ E \ S. Then, x ∈ ∩n(E ∩ V Cn ) implying x ∈ E ∩ ECn for all n. But since E ⊆ J , thistells us x ∈ J ∩ECn always. From the definition of Gn, it follows that x ∈ Gn for all n. We concludex ∈ ∩nGn.

Now fix the positive integer i. Since x is in Gi which is open, there is a positive number r so thatB(x; r) ⊆ Gi. From the definition of of a Vitali Cover of E, it follows there is a non degenerateclosed interval V in V0 with µ(V ) < r

2 . Hence, V is contained in Gi and so V ∩ Vi = ∅.

Since x ∈ SC , we see that V can not be any of the intervals Vn we have constructed. Now all theintervals Vn are non degenerate and all live in the bounded interval J . Thus,

∑i µ(Vi) < µ(J)

which implies µ(Vi)→ 0. From Equation 18.1, it follows that Sn → 0 as well. Choose a value of Nso that SN < µ(V ).From the definition of SN , we then have that V can not be in GN and so V ∩ FN 6= ∅.

LetM = infj : V ∩Fj 6= ∅. It is clearM > i. Hence, for this indexM , V ∩FM−1 = ∅ implyingV ⊆ GM−1. Further, V ∩ FM 6= ∅ and this tells us V ∩ VM 6= ∅. From the definition of SM−1, itthen follows that µ(V ) ≤ SM−1 < 2µ(Vm). Since WM shares the same center as VM with 5 timesthe length, this means V ⊆WM .

Finally, M > i, so V ⊆ ∪∞j=i Wj . Since x ∈ V , this shows x ∈ ∪∞j=i Wj . We conclude E \ S ⊆∪∞j=1Wj . From Equation 18.2, we know the series

∑n µ(Wn) converges. Given any positive ε, there

is a P so that∑∞j=1 µ(Wj) < ε if i > P . Since µ(E \ S) ≤

∑∞j=1 µ(Wj) for all i, we conclude

µ(E \ S) = 0.The proof for the case that E is unbounded is left to you.

Lemma 18.3.2 µ∗(f(E)) ≤ p µ∗(E)

Let f be strictly increasing on the interval [a, b] and let E ⊆ [a, b]. If at each pointx ∈ E, there exists at least one derived number α in Df(x) satisfying α < p, thenµ∗(f(E)) ≤ p µ∗(E).

Proof 18.3.2Let ε > 0 be chosen and let G be a bounded open set containing E so that


µ(G) < µ∗(E) + ε. (18.3)

For any x0 in E, there is a sequence (hn), all hn 6= 0, with hn → 0 so that the interval [x0, x0 +hn](if hn > 0) or [x0 + hn, x− 0] (if hn < 0) are in G and

f(x0 + hn) − f(x0)

hn< p. (18.4)

To keep our notation simple, we will simply use the notation [x0, x0 + hn] whether hn is positiveor negative. Let In(x0) = [x0, x0 + hn] and Jn(x0) = [f(x0), f(x0 + hn)]. Since f is strictlyincreasing, f(In(x0)) ⊆ Jn(x0) and Jn(x0) is a non degenerate closed interval. We also knowµ(In(x0) = |hn| and mu(Jn(x0) = |f(x0 + hn)− f(x0)|. It follows from Equation 18.4 that

µ(Jn(x0)) < p µ(In(x0)) = p |hn|. (18.5)

Since hn → 0, we then see limn µ(Jn(x0) = 0. Let V be the collection of intervals Jn(x0) : x0 ∈E, n ∈ Z+. It is easy to see V is a Vitali Cover for f(E). Thus, by Theorem 18.3.1, there is acountable disjoint family Jni(xi) : i ∈ Z+ so that

µ

(f(E) \ ∪i Jni(xi)

)= 0. (18.6)

From Equation 18.6, we then find

µ∗(f(E)) ≤∑i

µ(Jni(xi)) < p∑i

µ(Ini(xi)). (18.7)

But f is strictly increasing and so the intervals Ini(xi) must also be pairwise disjoint. Using Equa-tion 18.3, we infer

∑i

µ(Ini(xi)) = µ

(∪iIni(xi)

)≤ µ(G) < µ∗(E) + ε. (18.8)

Combining, we have µ∗(f(E)) < pµ∗(E) + ε. Since ε is arbitrary, this proves the result.

Lemma 18.3.3 µ∗(f(E)) ≥ q µ∗(E)

Let f be strictly increasing on the interval [a, b] and let E ⊆ [a, b]. If at each pointx ∈ E, there exists at least one derived number α in Df(x) satisfying α > q, thenµ∗(f(E)) ≥ q µ∗(E).

Proof 18.3.3The proof of this is left to you.

We are now ready to prove a very nice result: a function of bounded variation is differentiable µ a.e.

Theorem 18.3.4 Functions of Bounded Variation Are Differentiable a.e


Let f be a function of bounded variation on the bounded interval [a, b]. Then f has a finitederivative µ a.e.

Proof 18.3.4Since a function of bounded variation can be decomposed into the difference of two monotone func-tions, it is enough to prove this result for a non decreasing function. Further, if f is non decreasing,then g(x) = x + f(x) is strictly increasing. If we prove the result for g, we will know the result istrue for f as well. Hence, we may assume, without loss of generality, that f is strictly increasing inour proof.

Let E∞ be the set of points in [a, b] where Df(x) contains the value∞. Then f(E∞) ⊆ [f(a), f(b)]since f is strictly increasing. Now for any positive integer q, since ∞ is a derived number at xin E∞, by Lemma 18.3.3, it follows that µ∗(f(E∞)) ≥ qµ∗(E∞)). We also know µ∗(f(E∞)) ≤µ∗([f(a), f(b)] = f(b)− f(a). Thus,

q µ∗(E∞) ≤ f(b)− f(a), for all positive integers q.

Hence, µ∗(E∞) = 0.

Now choose real numbers u and v so that 0 ≤ u < v and define the set Euv by

Euv = x : ∃ α, β ∈ Df(x) 3 α < u < v < β.

Applying Lemma 18.3.2 and Lemma 18.3.3, we have

vµ∗(Euv ≤ µ∗(f(Euv) ≤ uµ∗(Euv).

Since v > u, we then have (v − u)µ∗(Euv) ≤ 0. This implies µ∗(Euv) = 0.

Now, if f is not differentiable at a point x, f either has∞ as a derived number or it has at least twodifferent finite derived numbers there. In the second case, let the two derived numbers be αx and βx.Then there are rational numbers r1 and r2 so that αx < r1 < r2 < βx. This implies x ∈ Er1 r2 .Hence, if N is the set of points where f fails to be differentiable in [a, b], we have

N ⊆ E∞ ∪ Epq : p, q rational .

But the outer measure of all these component sets is 0. Therefore, N has outer measure 0 also whichtells us N is measurable and has measure 0.

We are now getting closer to our Fundamental Theorem of Calculus extension. We can now prove avery weak form of the Recapture Theorem in Riemann integration.

Theorem 18.3.5 The Weak Monotone Recapture Theorem or Monotone Functions Have SummableDerivatives

Let f be non decreasing on [a, b]. Then f ′ is measurable and∫ baf ′dµ ≤ f(b) − f(a).

Note this tells us f ′ is summable.


Proof 18.3.5First, extend f to the interval [a, b+ 1] by setting f(x) = f(b) on [b, b+ 1]. For convenience, we willcall this extended f by the same name. Then, let the functions fn be defined by

fn(x) =f(x + 1

n ) − f(x)1n

= n

(f(x +

1

n) − f(x)

).

Then, at each point where f ′ exists, fn → f ′. Since each fn is measurable as f is monotone, wehave the pointwise limit f ′ is measurable a.e. Now apply Fatou’s Lemma to see

∫ b

a

lim infnfn dµ =

∫ b

a

f ′ dµ ≤ lim infn

∫ b

a

fn dµ.

Looking at the definition of the limit inferior, we see lim infn ≤ supn always. Hence, for all n, wehave

∫ b

a

f ′ dµ ≤∫ b

a

fn dµ.

Since f is monotone, each fn is Riemann integrable and so the Lebesgue integrals here are Riemannintegrals. We will use substitution to finish our argument. Note

∫ b

a

fn(x) dx = n

∫ b

a

(f(x+1

n)− f(x)) dx

= n intb+ 1

n

a+ 1n

f(x) dx − n intba f(x) dx

= n

∫ b+ 1n

b

f(x) dx − n

∫ a+ 1n

a

f(x) dx

= f(b) − n

∫ a+ 1n

a

f(x) dx,

because on [b, b+ 1n ], f(x) = f(b). Finally, on [a, a+ 1

n ], f(x) ≥ f(a). Hence, the last integral isbounded below by n f(a) 1

n = f(a). Hence, supn∫ bafn dµ ≤ f(b)− f(a) always. We conclude∫ b

af ′ dµ ≤ f(b)− f(a) too.

If f is absolutely continuous, we can say more. First, we show an absolutely continuous function canbe rewritten as the difference of two non decreasing absolutely continuous functions.

Lemma 18.3.6 AC Functions Are Difference of AC Non Decreasing Functions

Let f be absolutely continuous on [a, b]. Then there are absolutely continuous non decreas-ing functions u and v so that f = u − v. In fact, u = vf and v = vf − f are the usualchoices.

Proof 18.3.6For any a ≤ α < β ≤ b, from Theorem 3.4.1 we know the total variation function is additive onintervals. Recall the total variation function of f is defined by


Vf (x) =

0, x = aV (f ; a, x), a < x ≤ b

Thus, it follows that

Vf (β)− Vf (α) = V (f, α, β).

Let ε > 0 be chosen. Then there is a δ > 0 so that if ((an, bn)) is a collection of non overlappingintervals with

∑n (bn−an) < δ, then

∑n |f(bn)− f(an)| < ε

2 . Let πn be any partition of [an, bn].Then its component intervals form a non overlapping collection of intervals whose summed lengthis smaller than δ. Call this collection In and let its intervals be labeled [xin, y

in] for convenience.

In fact, the component intervals in each In can be glued together via a union to form a collectionwhose summed length is also less than δ. Call this larger collection I . Then, the absolute continuitycondition for f says

∑In

|f(yin)− f(xin)| <ε

2.

However, the choice of partitions πn is arbitrary. Hence, it follows that

∑n

V (f, an, bn) < ε.

But, we know the total variation is additive. Hence, we can rewrite the inequality using V (f, an, bn) =Vf (bn)− Vf (an) to get

∑n

Vf (bn)− Vf (an) < ε.

Thus, vf is absolutely continuous on [a, b]. Moreover, v = vf − f is also absolutely continuous assums and differences of absolutely continuous functions are also absolutely continuous. It is thenclear that u = vf and v = vf − f is a suitable decomposition.

We can now prove a reasonable recapture theorem.

Theorem 18.3.7 The Absolutely Continuous Recapture Theorem

Let f be absolutely continuous on [a, b]. Then

∫ b

a

f ′ dµ = f(b) − f(a).

Proof 18.3.7We have proven most of the requisite pieces for this result. From Lemma 18.3.6, we know f canbe written as the difference of two absolutely continuous functions. From Theorem 18.3.4, we thenknow f is differentiable a.e. From Theorem 18.3.5, we know f ′ is summable. It is now clear, we canassume without loss of generality that our absolutely continuous function f is non decreasing. To


finish, look at the proof of Theorem 18.3.5. Extend f to [b, b + 1] as before and consider the samesequence of functions (fn).

fn(x) =f(x + 1

n ) − f(x)1n

= n

(f(x +

1

n) − f(x)

).

Since f is absolutely continuous, it is also continuous and so this time, we know each fn is continu-ous. From the proof of Theorem 18.3.5, we know

∫ b

a

fn(x) dx = f(b) − n

∫ a+ 1n

a

f(x) dx.

The integral term above converges to f(a) using the standard Fundamental Theorem of Calculus forcontinuous integrands. Thus, we know

∫ b

a

fn(x) dx → f(b) − f(a).

Let ε > 0 be chosen. Our extension of f to [b, b + 1] is still absolutely continuous and so there isa δ1 > 0 so that given any collection of non overlapping intervals ([an, bn]) from [a, b + 1] whosesummed length is less than δ1, we have

∑n |f(bn)− f(an)| < ε

3 .

Further, since f ′ is summable, by the absolutely continuity of the integral, there is a δ2 > 0 so that∫F|f ′| dµ < ε

3 whenever µ(F ) < δ2. We can choose δ2 < δ1.

Let E be the set of points in [a, b] where f is differentiable. Then EC is measure zero and we knowfn → f ′ pointwise a.e. From Egoroff’s Theorem, Theorem 15.2.1, we then know fn → f ′ almostuniformly. Apply this theorem for tolerance δ2. Then, there is a measurable set G with measure lessthan δ2 so that fn → f ′ uniformly on GC . Thus, we also know

∫Gf ′ dµ < ε

3 . From the definition ofuniform convergence of GC , we know there is a positive integer N so that

|f ′(x)− fn(x)| <ε

3(b − a, ∀ n > N.

It is clear that GC does not contain points of EC!) It then follows that for all n > N ,

∫GC|f ′ − fn| dµ < µ)GC)

ε

3(b − a)<ε

3.

Now let’s combine our pieces. We have for n > N ,

∣∣∣∣∫ b

a

(fn − f ′) : dµ

∣∣∣∣ =

∣∣∣∣∫G

(fn − f ′) : dµ +

∫GC

(fn − f ′) : dµ

∣∣∣∣≤

∫GC|fn − f ′| : dµ +

∫G

|fn| : dµ +

∫G

|f ′| : dµ

< 2ε

3+ +

∫G

fn dµ,


where we can drop the absolute values of fn because f is non decreasing by assumption and so eachfn is non negative.

It remains to show∫Gfn dµ < ε

3 also. Since µ(G) < δ2 < δ1, we can find an open subset V of(a, b) so that G ⊆ V and µ(V ) < δ1. Express V as a countable union of disjoint open intervals((ci, di)). Pick any x from [0, 1]. Then the collection ([ci+x, di+x]) is a non overlapping collectionof intervals whose summed length is less than δ1. Hence,

∑i

|f(di + x)− f(ci + x)| < ε

3.

From the proof of Theorem 18.3.5, we find

∫ di

ci

fn(x) dx = n

∫ di+1n

di

f(x) dx − n

∫ ci+1n

ci

f(x) dx

= n

∫ 1n

0

(f(di + x) − f(ci + x)) dx.

Summing over i, we have

∑i

∫ di

ci

fn(x) dx = n

∫ 1n

0

(∑i

(f(di + x) − f(ci + x))

)dx.

But the inner summation adds up to less than ε3 . We conclude

∑i

∫ di

ci

fn(x) dx < n

∫ 1n

0

ε

3dx =

ε

3.

Thus,

∫G

fn dµ ≤∫V

fn dµ =∑i

∫ di

ci

fn(x) dx <ε

3.

This is the last piece to complete the proof of this result.

We can now establish the linkage between the Lebesgue-Stieljes charges induced by an absolutelycontinuous function f on [a, b] and the derivative f ′.

Theorem 18.3.8 Characterizing Lebesgue-Stieljes Measures Constructed From Absolutely Con-tinuous Functions

Let f be absolutely continuous on [a, b]. Let νf be the finite charge induced by f in theLebesgue-Stieljes construction process. Then, for all measurable E, we have

νf (E) =

∫E

f ′ dµ


Proof 18.3.8Since f is absolutely continuous, νf µ. From the Radon-Nikodym theorem, there is therefore asummable g so that

νf (E) =

∫E

g dµ

for all measurable E. In particular, for any x ∈ (a, b],

νf ((a, x]) = f(x) − f(a) =

∫(a,x]

g dµ.

However, we also know

f(x) − f(a) =

∫(a,x]

f ′ dµ.

Hence,∫ xa

(g − f ′) dµ = 0. In fact,∫I(g − f ′) dµ = 0 for any interval in [a, b]. Let h = g − f ′

which is summable. Now let’s assume there is a measurable set E in [a, b] with∫Eh dµ > 0 and let

ε =∫Eh dµ. From the continuity of the integral, for this ε, there is a δ > 0 so that

∫Gh dµ < ε for

any measurable set G with µ(G) < δ.

Choose an open set U in < which contains E with µ(U) < µ(E) + δ. Write the open set U as acountable union of open intervals In = (an, bn). Let Vn = In ∩ [a, b]. Then,

∫Vn

h dµ = 0 as Vnis an interval. From this we conclude, if V = ∪nVn, that

∫Vh dµ = 0 as well. By construction,

E ⊆ V ⊆ U which implies V \ E ⊆ U . Thus

µ(E) + µ(EC ∩ V ) + µ(V C ∩ U) = µ(U)

< µ(E) + δ.

It follows that µ(V \ E) < δ and hence,∫V \E h dµ < ε. However,

0 =

∫V

h dµ =

∫E

h dµ +

∫V \E

h dµ

It follows immediately that

ε =

∣∣∣∣∫E

h dµ

∣∣∣∣ =

∣∣∣∣∫V \E

h dµ

∣∣∣∣ < ε.

This is a contradiction. We conclude∫Eh dµ = 0 for all measurable E. In particular, this is true

for x | h(x) > 0 implying h+ = 0 a.e. and for x | h(x) ≤ 0 implying h− = 0 a.e. Thus, h = 0a.e. We therefore have shown that f ′ = g a.e. and we can conclude

νf (E) =

∫E

f ′ dµ.

Part VII

Summing It All Up

345

Chapter 19

Summing It All Up

We have now come to the end of this set of notes. There is, of course, much more we could havediscussed. We have not covered all of the things we wanted to but we view that as a plus: there ismore to look forward to! In particular, in (Peterson (4) 2019) we discuss representation theoremsfor bounded linear functionals on certain spaces. For example, the dual space, (C([a, b]))′ has a nicerelationship to certain types of measures.

347

348 CHAPTER 19. SUMMING IT ALL UP

Part VIII

References

349

References

[1] A. Bruckner, J. Bruckner, and B. Thomson. Real Analysis. Prentice - Hall, 1997.

[2] S. Douglas. Introduction To Mathematical Analysis. Addison-Wesley Publishing Company,1996.

[3] W. Fulks. Advanced Calculus: An Introduction to Analysis. John Wiley & Sons, third edition,1978.

[4] J. Peterson. Basic Analysis V: Linear Functional Analysis and Topology. CRC Press, a Divisionof the Taylor and Francis Group, 6000 Broken Sound Parkway NW, Suite 300, Boca Raton,Florida 33487, 2019. Contract December 15, 2016, 450 pages, 100 illustrations.

[5] J. Peterson. Basic Analysis IV: Abstract Measure Theory. CRC Press, a Division of the Taylorand Francis Group, 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, Florida 33487,2019. Contract December 15, 2016, 450 pages, 100 illustrations.

[6] J. Peterson. Basic Analysis I: Functions of a Real Variable. CRC Press, a Division of the Taylorand Francis Group, 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, Florida 33487,2019. Contract December 15, 2016, 450 pages, 100 illustrations.

[7] J. Peterson. Basic Analysis III: Mappings on Infinite Dimensional Spaces. CRC Press, a Di-vision of the Taylor and Francis Group, 6000 Broken Sound Parkway NW, Suite 300, BocaRaton, Florida 33487, 2019. Contract December 15, 2016, 450 pages, 100 illustrations.

[8] J. Peterson. Basic Analysis II: Functions of Multiple Variables. CRC Press, a Division of theTaylor and Francis Group, 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, Florida33487, 2019. Contract December 15, 2016, 450 pages, 100 illustrations.

[9] H. Sagan. Advanced Calculus of real valued functions of a Real Variable and Vector - ValuedFunctions of a Vector Variable. Houghton Mifflin Company, 1974.

[10] G. Simmons. Introduction to Topology and Modern Analysis. McGraw-Hill Book Company,1963.

[11] K. Stromberg. Introduction To Classical Real Analysis. Wadsworth International Group andPrindle, Weber and Schmidt, 1981.

[12] A. Taylor. General Theory of Functions and Integration. Dover Publications, Inc., 1985.

351

352 REFERENCES

Part IX

Detailed Indices

353

Index

DefinitionRS[g, a, b], 102Absolute Continuity Of A Measure, 172Absolute Continuity Of Charges, 312Abstract Darboux Integrability, 185Additive Set Function, 227Algebra Of Sets, 216Almost Uniform Convergence, 284Caratheodory Condition, 216Cauchy Sequence In Norm, 199Cauchy Sequences In Measure, 284Charges, 154Common Refinement Of Two Partitions, 34Complete Measure, 166Complete NLS, 199Conjugate Index Pairs, 195Content Of Open Interval, 235Continuity Of A Function At A Point: ε − δ

Version, 13Continuity Of A Function At A Point: Limit

Version, 14Continuous Almost Everywhere, 94Convergence In Measure, 284Convergence Pointwise and Pointwise a.e., 283Convergence Uniformly, 284Darboux Integrability, 61Darboux Lower And Upper Integrals, 61Darboux Upper and Lower Sums, 58Derived Numbers, 333Differentiability of A Function At A Point, 14Equivalent Conditions For The Measurability

of a Function, 139Equivalent Conditions For The Measurability

of an Extended Real Valued Function, 143Essentially Bounded Functions, 205Extended Real Number System, 137Functions

Absolute Continuity, 331Functions Of Bounded Variation, 45Inner Product Space, 211Integral Of A Non-negative Measurable Func-

tion, 161Integral Of A Simple Function, 160Lebesgue Outer Measure, 236Limit Inferior And Superior Of Sequences Of

Sets, 157

Measurability of a Function, 139Measurability Of Extended Real Valued Func-

tions, 143Measures, 154

Borel Measure, 251Metric On A Set, 191Metric Outer Measure, 221Monotone Function, 36

Associated Saltus Function, 40Norm Convergence, 192Norm On A Vector Space, 191Outer Measure, 215Partition, 34Positive and Negative Sets For a Charge, 303Premeasures and Covering Families, 226Propositions Holding Almost Everywhere, 159Pseudo-Measure, 229Refinement Of A Partition, 34Regular Outer Measures, 228Rewriting Lebesgue Outer Measure Using Edge

Length Restricted Covers, 245Riemann - Stieljes Criterion For Integrability,

110Riemann - Stieljes Darboux Integral, 110Riemann - Stieljes Sum, 101Riemann Integrability Of a Bounded f , 55Riemann Integrability Of A Bounded Func-

tion, 18Riemann Sum, 16, 55Riemann’s Criterion for Integrability, 61Set of Extended Real Valued Measurable Func-

tions, 143Sets Of Content Zero, 94Sigma - Algebra Generated By Collection A,

135Sigma Algebra, 133Simple Functions, 160Singular Measures, 321Space Of p Summable Functions, 195Spaces of Essentially Bounded Functions, 206Step Function, 104Summable Functions, 174The Continuous Part Of A Monotone Func-

tion, 41The Discontinuity Set Of A Monotone Func-

tion, 37

355

356 INDEX

The Generalized Cantor Set, 263The Positive and Negative Parts Of a Charge,

303Upper and Lower Riemann - Stieljes Darboux

Sums, 108Upper and Lower Riemann - Stieljes Integrals,

109Variation of a Charge, 309Vitali Cover, 334

Differentiability Implies Continuity, 14

FunctionsAntiderivative, 15Bounded, 15Continuity, 14Differentiable, 14Primitive, 15

IntegrationAntiderivatives Of Simple Powers, 24Antiderivatives of Simple Trigonometric Func-

tions, 24Cauchy Fundamental Theorem of Calculus, 22Definite Integrals Of Simple Powers, 24Definite Integrals Of Simple Trigonometric Func-

tions, 25Evaluation Set, 15Functions With Jump Discontinuities, 30Functions With Removable Discontinuities, 28Norm of a partition, 18Partition, 15Primitive, 15Riemann Integrable, 21Riemann Sum, 15, 18Sequences of partitions, 17Symbol For The Antiderivative of f is

∫f , 24

Symbol For The Definite Integral of f on [a, b]

is∫ baf(t) dt, 24

The indefinite integral of f is also the antideriva-tive, 24

LemmaMδ = µ∗, 246µ∗(f(E)) ≥ q µ∗(E), 336µ∗(f(E)) ≤ p µ∗(E), 335f = g on (a, b) Implies Riemann Integrals

Match, 84f Zero On (a, b) Implies Zero Riemann Inte-

gral, 83Outer Measure Of The Closure Of Interval Equals

Content Of Interval, 244Absolute Continuity Of The Integral, 313AC Functions Are Difference of AC Non De-

creasing Functions, 338Approximate Finite Lebesgue Covers Of I .,

245

Characterizing Limit Inferior And SuperiorsOf Sequences Of Sets, 157

Condition For Outer Measure To Be Regular,228

Continuity Of The Integral, 296Continuous Functions Of Finite Measurable Func-

tions Are Measurable, 148Continuous Functions Of Measurable Functions

Are Measurable, 150De Morgan’s Laws, 133Disjoint Decompositions Of Unions, 158Epsilon - Delta Version Of Absolute Continu-

ity Of a Charge, 313Essentially Bounded Functions Bounded Above

By Their Essential Bound a.e, 207Essentially Bounded Functions That Are Equiv-

alent Have The Same Essential Bound,207

Extended Valued Measurability In Terms OfThe Finite Part Of The Function, 144

Extending τg To Additive Is Well - Defined,270

Finite Jump Step Functions As Integrators, 108Function f Zero a.e. If and Only If Its Integral

Is Zero, 172Function Measurable If and Only If Positive

and Negative Parts Measurable, 142Fundamental infimum and supremum equali-

ties, 64Hahn Decomposition Characterization of a Charge,

309Infimum Tolerance Lemma, 35Lebesgue - Stieljes Outer Premeasure Is a Pseudo-

Measure, 272Limit Inferiors And Superiors Of Monotone

Sequences Of Sets, 157Measure Of Monotonic Sequence Of Sets, 155Monotonicity, 155Monotonicity Of The Abstract Integral For Non

Negative Functions, 164Non Measurable Set Lemma 1, 255Non Measurable Set Lemma 2, 256One Jump Step Functions As Integrators One,

104One Jump Step Functions As Integrators Three,

107One Jump Step Functions As Integrators Two,

106Outer Measure Of Interval Equals Content Of

Interval, 244p-Summable Cauchy Sequence Condition I, 295p-Summable Cauchy Sequence Condition II,

297p-Summable Functions Have p-Norm Arbitrar-

ily Small Off a Set, 294p-Summable Inequality, 295

INDEX 357

Pointwise Infimums, Supremums, Limit Infe-riors and Limit Superiors are Measurable,146

Products of Measurable Functions Are Mea-surable, 147

Properties of Extended Valued Measurable Func-tions, 145

Properties of Measurable Functions, 141Properties Of Simple Function Integrations, 161Radon - Nikodym Technical Lemma, 314Real Number Conjugate Indices Inequality, 195Sums Over Finite Lebesgue Covers Of I Dom-

inate Content Of I , 236Supremum Tolerance Lemma, 35The Upper And Lower Darboux Integral Is Ad-

ditive On Intervals, 71The Upper And Lower Riemann - Stieljes Dar-

boux Integral Is Additive On Intervals,112

MeasureBorel, 274Lebesgue - Stieljes, 274Measurable Cover, 228

Monotone FunctionContinuous at x From Left If and Only If u(x) =

0, 39Continuous at x From right If and Only If v(x) =

0, 39Continuous at x If and Only If u(x) = v(x) =

0, 39Left Hand Jump at x, u(x), 39Right Hand Jump at x, v(x), 39Total Jump at x, u(x) + v(x), 39

Monotone FunctionsSaltus Function

Properties, 40

PartitionsGauge or Norm, 35

PropositionRefinements and Common Refinements, 34

Theorem, 187DAI(f(x) ≡ c) is cµ(X), 187DAI(f,m0,M0) is independent of the choice

of m0 and M0, 186L(f,M0) ≤ U(f,M0), 186L(f,π1) ≤ U(f,π2), 61, 185L(f, g,π1) ≤ U(f, g,π2), 109L1 Semi-norm, 192Lp Is A Vector Space, 198Lp Semi-Norm, 198, 208RI[a, b] Is A Vector Space and RI(f ; a, b) Is

A Linear Mapping, 56U(f,M0) ≤ L(f,M0), 186

Vf and Vf − f Are Monotone For a Functionf of Founded Variation, 51

π′ refines π Implies L(f,π) ≤ L(f,π′) andU(f,π) ≥ U(f,π′), 185

π π′ Implies L(f,π) ≤ L(f,π′) andU(f,π) ≥ U(f,π′), 58

π π′ Implies L(f, g,π) ≤ L(f, g,π′) andU(f, g,π) ≥ U(f, g,π′), 109

L1 Is Separable, 254µ∗f (E) = µ∗(f(E)), 332f ∈ BV [a, b] ∩ C[a, b] If and Only If Vf and

Vf − f Are Continuous and Increasing,54

f ∈ RS[g, a, b] Implies f ∈ RS[Vg, a, b] andf ∈ RS[Vg − g, a, b], 113

f Bounded Variation and g Continuous Im-plies Riemann - Stieljes Integral Exists,121

f Bounded and Continuous At All But FinitelyMany Points Implies f is Riemann Inte-grable, 86

f Bounded and Continuous At All But OnePoint Implies f is Riemann Integrable,85

f Continuous and g Bounded Variation Im-plies Riemann - Stieljes Integral Exists,120

f Continuous and g Riemann Integrable Im-plies f g is Riemann Integrable, 92

f Is Absolutely Continuous IFF µf << µ.,333

f2, f1f2 and 1/f Riemann Stieljes IntegrableWith Respect To g Of Bounded Varia-tion, 115

µ∗ Measurable Sets Form Algebra, 216µ∗ Measurable Sets Properties, 218A Function Of Bounded Variation Is The Dif-

ference of Two Increasing Functions, 51A Monotone Function Has A Countable Num-

ber of Discontinuities, 37Abstract Darboux Integral Absolute Inequal-

ity, 189Abstract Darboux Integral Is Additive, 188Abstract Darboux Integral Is Monotone, 188Abstract Darboux Integral Is Scalable, 189Abstract Darboux Integral Measures, 187Abstract Darboux Integral Measures Are Ab-

solutely Continuous, 190Abstract Darboux Integral Zero Implies f = 0

a.e., 190Abstract Integral Darboux Lower and Upper

Bounds, 187Abstract Integration Is Additive, 169Almost Uniform Convergence Implies Conver-

gence In Measure, 291

358 INDEX

Alternate Characterization Of Essentially BoundedFunctions, 206

Approximation Of A Summable Function WithA Continuous Function, 253

Approximation Of Non negative MeasurableFunctions By Monotone Sequences, 149

Approximation Of The Riemann Integral, 79Average Value For Riemann Integrals, 77Bounded Differentiable Implies Bounded Vari-

ation, 46Bounded Variation Implies Riemann Integrable,

71Cauchy - Schwartz Inequality, 196Cauchy Fundamental Theorem Of Calculus,

24Cauchy In Measure Implies A Convergent Sub-

sequence, 285Cauchy In Measure Implies Completeness, 288Cauchy Schwartz Inequality: Sequence Spaces,

205Cauchy’s Fundamental Theorem, 78Characterizing Lebesgue-Stieljes Measures Con-

structed From Absolutely Continuous Func-tions, 341

Conditions For OMI-F Measures, 228Conditions For OMI-FE Measures, 229Constant Functions Are Riemann Integrable,

70Constructing Measures From Non Negative Mea-

surable Functions, 171Constructing Outer Measures Via Premeasures,

226Continuous Approximation Of A Characteris-

tic Function, 252Continuous Approximation Of A Simple Func-

tion, 253Continuous Implies Riemann Integrable, 70Convergence Relationships On Finite Measure

Space, 299Convergence Relationships On General Mea-

surable Space, 299Convergence Relationships With p-Domination,

300, 301Convergent Subsequences Exist, 301Egoroff’s Theorem, 292Equality a.e. Can Imply Measurability Even If

The Measure Is Not Complete, 166Equality a.e. Implies Measurability If The Mea-

sure Is Complete, 166Equivalent Absolute Continuity Conditions For

Charges, 312Every Riemann Integrable Function Is Lebesgue

Integrable and The Two Integrals Coin-cide, 327

Existence Of The Riemann Integral, 19Extended Monotone Convergence Theorem, 169

Extended Monotone Convergence Theorem Two,173

Fatou’s Lemma, 170Function Of Bounded Variation Continuous If

and Only If Vf Is Continuous, 52Functions

Properties Of Absolutely Continuous Func-tions, 331

Functions Of Bounded Variation Always Pos-sess Right and Left Hand Limits, 52

Functions Of Bounded Variation Are Bounded,45

Functions Of Bounded Variation Are ClosedUnder Addition, 46

Functions of Bounded Variation Are Differen-tiable a.e, 336

Functions Of Bounded Variation Have Count-able Discontinuity Sets, 52

Fundamental Abstract Integration Inequalities,178

Fundamental Riemann Integral Estimates, 57Fundamental Riemann Stieljes Integral Esti-

mates, 111Fundamental Theorem Of Calculus, 21Fundamental Theorem Of Calculus Reversed,

21Holder’s Inequality, 196Holder’s Inequality: p = 1, 211Holder’s Inequality: Sequence Spaces, 205Hahn Decomposition Associated With A Charge,

307Inner Product On The Space of Square Summable

Equivalence Classes, 211Integrals Of Summable Functions Create Charges,

178Integrand Continuous and Integrator Continu-

ously Differentiable Implies Riemann -Stieljes Integrable, 121

Integrand Riemann Integrable and IntegratorContinuously Differentiable Implies Rie-mann - Stieljes Integrable, 122

Integration By Parts, 81Inverses Of Functions Of Bounded Variation,

47Lebesgue Decomposition Theorem, 322Lebesgue Measure Is Regular, 248Lebesgue Measure Is Translation Invariant, 255Lebesgue Outer Measure Is Metric Outer Mea-

sure, 247Lebesgue’s Criterion For The Riemann Inte-

grability of Bounded Functions, 94Lebesgue’s Dominated Convergence Theorem,

181Lebesgue-Stieljes Measures

Properties, 276Leibnitz’s Rule, 82

INDEX 359

Levi’s Theorem, 176Limit Interchange Theorem For Riemann - Stiel-

jes Integral, 129Linearity of the Riemann - Stieljes Integral,

102Measurability and Approximation Conditions,

249Measure Induced By Outer Measure, 220Measure Induced By Outer Measure Is Com-

plete, 221Measures and Metric Spaces

Approximating Sets With A Lebesgue-StieljesOuter Measure, 275

Finite Measure Borel Sets Can Be Approx-imated By Closed Sets, 251

Finite Measure Borel Sets Can Be Approx-imated By Open Sets, 252

Minkowski’s Inequality, 196Minkowski’s Inequality: Sequence Spaces, 205Monotone Convergence Theorem, 167Monotone Functions

A Partition Sum Estimate, 36Monotone Functions Are Of Bounded Varia-

tion, 45Monotone Implies Riemann Integrable, 71Non Lebesgue Measurable Set, 256Open Set Characterization Lemma, 137Open Sets in a Metric Space Are OMI Mea-

surable, 222Open Sets In Metric Space µ∗ Measurable If

and Only If µ∗ Metric Outer Measure,225

p-Norm Convergence Implies Convergence inMeasure, 290

Pointwise a.e. Convergence Plus DominationImplies p-Norm Convergence, 293

Pointwise Limits of Measurable Functions AreMeasurable, 147

Products Of Functions Of Bounded VariationAre Of Bounded Variation, 47

Properties of fc, 41Properties Of The Riemann Integral, 66Properties Of The Riemann Stieljes Integral,

111Radon - Nikodym Theorem: Finite Charge Case,

315Radon - Nikodym Theorem: Sigma-Finite Charge

Case, 319Representing The Cantor Set, 264Riemann - Lebesgue Lemma, 94Riemann - Stieljes Integral, 101Riemann Stieljes Fundamental Theorem Of Cal-

culus, 117Riemann Stieljes Integral Is Additive On Subin-

tervals, 116Riemann Stieljes Integration By Parts, 103

Riemann-Stieljes Integrable Functions Are Lebesgue-Stieljes Integrable and The Two IntegralsCoincide: One, 328

Sequences Of Equivalence Classes in Lp ThatConverge Are Cauchy, 199

Sequences That Converge In p - Norm PossessSubsequences Converging Pointwise a.e.,204

Space of Square Summable Equivalence ClassesIs A Hilbert Space, 212

Substitution In Riemann Integration, 81Summable Function Equal a.e. To Another

Function With Measure Complete ImpliesThe Other Function Is Also Summable,176

Summable Function Equal a.e. To AnotherMeasurable Function Implies The OtherFunction Is Also Summable, 175

Summable Function Form A Linear Space, 179Summable Implies Finite a.e., 175The Absolutely Continuous Recapture Theo-

rem, 339The Antiderivative of f , 77The Fundamental Theorem Of Calculus, 74The Jordan Decomposition Of A Charge, 307The Jordan Decomposition Of A Finite Charge,

303The Mean Value Theorem For Riemann Inte-

grals, 76The Metric Space of All Finite Measure Sets

Is Complete, 258The Metric Space of Finite Measurable Sets,

257The Recapture Theorem, 78The Riemann Integral Equivalence Theorem,

62The Riemann Integral Exists On Subintervals,

73The Riemann Integral Is Additive On Subin-

tervals, 73, 113The Riemann Integral Is Order Preserving, 57The Riemann Integral Limit Interchange The-

orem, 87The Riemann Stieljes Integral Equivalence The-

orem, 110The Riemann Stieljes Integral Exists On Subin-

tervals, 112The Riemann Stieljes Integral Is Order Pre-

serving, 112The Set Of Equivalence Classes of L1 Is A

Normed Linear Space, 193The Set Of Equivalence Classes of L∞ Is A

Normed Linear Space, 209The Set Of Equivalence Classes of Lp Is A

Normed Linear Space, 198

360 INDEX

The Space of Equivalence Class of L∞ Is ABanach Space, 209

The Space of Equivalence Class of Lp Is ABanach Space, 200

The Total Variation Is Additive On Intervals,49

The Total Variation Of A Function Of BoundedVariation, 46

The Upper And Lower Abstract Darboux In-tegrals Are Finite, 185

The Upper And Lower Darboux Integral AreFinite, 61

The Upper And Lower Riemann - Stieljes Dar-boux Integral Are Finite, 109

The Variation Function Of a Function f OfBounded Variation, 50

The Weak Monotone Recapture Theorem orMonotone Functions Have Summable Deriva-tives, 337

Two Riemann Integrable Functions Match AtAll But Finitely Many Points Implies In-tegrals Match, 85

Variation of a Charge In Terms of The Plus andMinus Parts, 311

Variation of a Charge is a Measure, 310Vitali Convergence Theorem, 297Vitali Covering Theorem, 334Weierstrass Approximation Theorem, 89

Worked Out SolutionsIntegration Substitution∫

(t2 + 1) 2dt, 25∫(t2 + 1)3 4dt, 26∫sin(t2 + 1) 5t dt, 27∫ √t2 + 1 3t dt, 27∫ 5

1(t2 + 2t + 1)2 (t + 1) dt, 28

Part X

Glossary Of Terms

361

Part XI

Appendix: Undergraduate AnalysisExaminations

363

Appendix A

Senior Advanced Calculus I StudyGuides

Presented below are study guides and coursedescription material from some of the times wehave taught the first semester senior level un-dergraduate analysis course here. At ClemsonUniversity, it is called MTHSC 453. Anothersource of information is our web-based discus-sion board material. The MTHSC 453 discus-sion board can be accessed either through myweb page by following the links or by the di-rect routehttp://www.ces.math.clemson.edu/discus. Onthat page you will see exercises, worked outproblems and so forth in an easy to follow for-mat.

A-1 Course Structure

This course is about learning how to think care-fully about mathematics. This course (and ourcompanion one Abstract Algebra: here MTHSC412 are the first courses where we really in-sist that you begin to master one of the bigtools of our trade: abstraction. Now, of coursethis is not easy as most of you have not reallyneeded to learn how to do this in your previ-ous courses. However, the time has come, andin this class, we will work very hard to helpyou learn how to think well and deep aboutmathematical issues. We will learn how to doproofs of many propositions and learn a lot offacts about how functions of a real variable xbehave. A lot of what we learn can be gen-eralized (and for good useful reasons as wewill see) to more abstract settings and we’ll setpointers to the future about that as much as we

can. But most of all, we will consistently chal-lenge you to think about the how and why ofall that we do. It should be fun!

To help you with all of this new and inter-esting stuff, we will be using web-based dis-cussion pages. You can check out details aboutthis later in this syllabus.

In broad outline, we will discuss:

• Sequences

• Subsequences

• Bolzano-Weierstrass theorem

• Limit inferiors and Limit superiors

• Basics about sets of real numbers: callthis point set topology

• Limits of functions

• Continuity of functions

• Consequences of Continuity

• Differentiation of functions

• Consequences of Differentiation

A-2 Study GuideThe final exam is a closed book and closednotes test. It is also cumulative for all of thematerial in this class. Here is a list of what istypically done in this class.

Give precise mathematical definitions of thefollowing mathematical concepts using the ap-propriate mathematical formalism:

365

366 APPENDIX A. ADVANCED CALCULUS I

Basics: 1. Mathematical Induction

2. Bounded sets

3. Infimum and supremum of a boundedset

4. Theorem: If S is bounded aboveand M = sup S, then for any y <M , there is an x ∈ S such that y <x ≤M . (Similar type theorem forinfimums).

5. Theorem: Every nonempty set boundedabove has a supremumEvery nonempty set bounded be-low has an infimum

6. Theorem: Triangle Inequality andReverse Triangle Inequality

Functions, Sequences and Limits: 1. Functionsf : S → <, where S is a subset of<.

2. Let f : S → <; limx→x0 f(x)

3. Let f : S → <;

(a) S is unbounded above: limx→∞ f(x),convergence or divergence

(b) S is unbounded below: limx→−∞ f(x),convergence or divergence

4. Sequences

(a) f : N0 → <; N0 is the set ofintegers starting at n0.

(b) Let f(n) = an. Then the se-quence can be written as an∞n0

≡an, where it is usually un-derstood where the sequencebegins.

(a) limn→∞ an = A, convergenceor divergence.

(b) Squeezing Lemma(c) Theorem: Uniqueness of Lim-

its

5. Operations with Sequences; limn→∞ ≡lim an

(a) Theorem: If sequence convergesto A, so does any subsequence

(b) Theorem: If sequence converges,sequence forms bounded set

(c) Theorem: lim(an + bn)

(d) Theorem: lim(an − bn)

(e) Theorem: lim(anbn)

(f) Theorem: lim(anbn )

(g) Theorem: lim |an| = |lim(an)|(h) Theorem: If an ≥ bn, then

lim an ≥ lim bn

6. Limits of Functions

(a) Limits from the right and theleft

(b) Operations with Limits of Func-tions: (All of the items underoperations with sequences trans-lated to the function setting)

7. Monotone Sequences

(a) Increasing and Decreasing Se-quences

(b) Non-increasing and non-decreasingSequences

(c) Completeness Axiom Non-decreasingsequences bounded above con-verge and lim an = sup an.Theorem: (Similar statementfor non-increasing sequences)

8. Monotone Functions

(a) Increasing and decreasing func-tions

(b) Non-decreasing and non-increasingfunctions

Continuity and Limits Revisited: 1. Continuityof f : S → < at a ∈ S.

2. Discontinuity of f : S → < at a ∈S.

3. Uniform continuity of f : S → <on S.

4. Operations with continuous func-tions

(a) Same items as in operations withfunctions

(b) Theorem: Intermediate ValueTheorem

(c) Theorem: Inverse Functions

5. Cluster Points and Accumulation Points

(a) Cluster Point of a sequence(b) Cluster point of a function

A-3. EXAMS VERSION A 367

(c) Theorem: Bolzano-WeierstrassTheorem Every bounded infi-nite set has at least one accu-mulation point and every boundedsequence has at least one clus-ter point

(d) Theorem: Cauchy Criterion Se-quences

(e) Theorem: Cauchy Criterion Func-tions

6. Limit Inferior

7. Limit Superior

8. Theorem: Let [a, b] be finite. If fcontinuous on on [a, b], then f isbounded on [a, b].

9. Theorem: Let [a, b] be finite. Thenthere is point x0 ∈ [a, b] with f(x0) ≥f(x)∀x ∈ [a, b].(Continuous function achieves max-imum on finite interval).

10. Theorem: Let [a, b] be finite. Thenf continuous on [a, b] implies f isuniformly continuous on [a, b].

Derivatives: 1. Derivative of a function

2. Theorem: f has derivative at a im-plies f is continuous at a.

3. Theorem: Chain Rule

4. Secant Lines, Tangent Lines

5. Theorem: If f attains an extremumat a, then if f

′(a) exists, f

′(a) =

0.

6. Theorem: Rolle’s Theorem

7. Theorem: Mean Value Theorem

8. Theorem: If f′(x) = g

′(x), then

f(x) g(x) + c, where c is a con-stant.

9. Theorem: f non-decreasing impliesf′(a) ≥ 0 at every point a where

f is differentiable.

10. Theorem: f′(x) ≥ 0 on interval

implies f is non-decreasing on theinterval. If f

′(x) > 0, then f is

increasing on the interval.

11. Theorem: Cauchy Mean value The-orem

12. Theorem: L’Hopital’s Rule

13. Theorem: Taylor’s Theorem withRemainder

14. Maximum and Minimum Values off

(a) Critical Points of f(b) Theorem: First Derivative Test(c) Theorem: Second Derivative

Test

A-3 Sample Exams Version A

A-3.1 Exam 1AInstructions:This is a closed book and closed notes test.You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.

Part 1: Definitions (24 Points)

1. (3 Points) Give a precise mathemat-ical definition of the meaning of asequence of real numbers.

2. (3 Points) Give a precise mathemat-ical definition of the phrase ”the se-quence an converges to a”.

3. (9 Points) Let S be a nonempty setof real numbers. Give a precise math-ematical definition of the follow-ing phrases:

(a) U is an upper bound of S(b) U is the least upper bound of

S(c) u is a maximal element of S

4. (3 Points) State the Principle of Math-ematical Induction.

5. (3 Points) State the Triangle Inequal-ity for Real Numbers.

6. (3 Points) Give a precise mathemat-ical definition of the phrase ”the limitof the function f(x) as x tends to in-finity is A”

Part 2: Short Answer (36 Points)You must determine whether or not thesestatements are true. If the statement istrue, you MUST give us the reason why


it is true; if the statement is false, youmust give us a counterexample.

1. (4 Points) Is it true that all sequenceswhich converge are bounded?

2. (4 Points) Is it true that if a sequenceis bounded it must converge?

3. (4 Points) If | an |→| a |, is it truethat an → a?

4. (4 Points) If an → a, is it true that| an |→| a |?

Show all your work on the short calcu-lational exercises below. You may use acalculator if you wish.

1. (5 Points) Find the inf and sup ofthe set

S = x ∈ (0,π

2) | tan(x) < exp(−x)

You can indicate these points graph-ically.

2. (5 Points) Give an example of a func-tion whose domain is the real num-bers which has a finite limit as xtends to infinity.

3. (5 Points) Give an example of a se-quence of real numbers which doesnot converge.

4. (5 Points) Give an example of a nonemptyset of real numbers for which theinf and sup are the same.

Part 3: Proofs (40 Points)Provide careful proofs of the followingpropositions. You will be graded on themathematical correctness of your argu-ments as well as your use of language,syntax and organization in the proof.

1. (20 Points)Proposition: For all n ≥ 1,

1

1 · 2+ . . .+

1

n · (n+ 1)

=n

n+ 1

2. (20 Points)Proposition: If the sequence anis defined by

an =−4n2 − 2n+ 3

n2 − 3n, n ≥ 2,

then the sequence converges.


Part 1: Definitions (36 Points)Give precise mathematical definitions ofthe following mathematical concepts us-ing the ε− δ formalism:

1. (3 Points) ”The sequence an con-verges to a”.

2. (4 Points) ”The function f : D ⊆R → R has a limit A at the pointa ∈ D”.

3. (4 Points) ”The function f : D ⊆R → R is continuous at the pointa ∈ D”.

4. (4 Points) ”The function f : D ⊆R → R is uniformly continuousin D”.

Give precise mathematical definitions ofthe following mathematical concepts:

1. (3 Points) ”The sequence an is amonotone sequence”.

2. (3 Points) ”The function f : [0, 1]→R is monotone increasing”.

State precisely the following mathemat-ical theorems and/or axioms:

1. (5 Points) The Completeness Ax-iom

2. (5 Points) The Intermediate ValueTheorem

3. (5 Points) The Bolzano WeierstrassTheorem

Part 2: Short Answer (30 Points)You must answer the following questions.

A-3. EXAMS VERSION A 369

If the answer is YES, you MUST give usthe reason why; if the answer is NO, youmust also give a reason.

1. (3 Points) If the sequence an isincreasing and bounded above byM , does the sequence converge?

2. (3 Points) If the sequence an isincreasing and bounded above byM , does the sequence converge toM?

3. (3 Points) If the sequence an boundedabove by M and below by L, doesthe sequence converge?

4. (3 Points) If the sequence an boundedabove by M and below by L, doesthe sequence have at least one con-vergent subsequence?

5. (4 Points) Consider f : [0, 1]→ R.Is it possible for f2 to be continu-ous on [0, 1] and f to be discontin-uous at each point in [0, 1]?

6. (4 Points) Consider f : (0, 1) →R. Is it possible for f to be contin-uous only at x = .5 and discontin-uous everywhere else in (0, 1)?

7. (4 Points) Consider f : (0, 1) →R. Is it possible for f to be contin-uous on the irrational numbers in(0, 1) and discontinuous on the ra-tional numbers in (0, 1)?

Show all your work on the short calcu-lational exercise below. You may use acalculator if you wish.

1. (6 Points) Prove that there is a so-lution to the equation

x + 10sin(x) = 0

on the interval [1.0, 4.5].


1. (17 Points)Proposition: If the function f :

D ⊆ R → R is continuous ata ∈ D, then there exists δ > 0and M > 0 such that |f(x)| < Mif |x− a| < δ.


a1 = 1,

an+1 =2an + 3

4, n > 1



Part 1: Definitions (28 Points)Give precise mathematical definitions ofthe following mathematical concepts us-ing the appropriate mathematical formal-ism:

1. (4 Points) ”The Cauchy Criterionfor a sequence an”.

2. (4 Points) ”A is a cluster point ofthe sequence an”.

3. (4 Points) ”The limit superior ofthe sequence an”.

4. (4 Points) ”The function f : D ⊆R→ R, where D is the domain ofthe function f , has a derivative atx0 ∈ D”.

State precisely the following mathemat-ical theorems:

1. (4 Points) The The Mean ValueTheorem

2. (4 Points) The Bolzano-WeierstrassTheorem

3. (4 Points) The Rolle’s Theorem

Part 2: Short Answer (38 Points)You must answer the following questions.


If the answer is YES or NO, you MUSTgive us the reason why; ( e.g. the com-plete statement of a relevant theorem, acounterexample etc.)

1. (4 Points) Is it possible for a func-tion to be continuous at a point butnot differentiable there?

2. (4 Points) Is it possible for a func-tion to be differentiable at a pointbut not continuous there?

3. (4 Points) Is f(x) =√

(x) uni-formly continuous on [0, 1]?

4. (4 Points) Is is possible for a func-tion that is continuous on [2, 5] tobe unbounded?

5. (4 points) Does every continuousfunction on [0, 1] achieve a maxi-mum value?


1. (6 Points) If

f(x) = x2, x ≤ 1

= 2x− 1, x > 1

find f′(x) at all points where the

derivative exists.

2. (6 points) Find all cluster points ofthe sequence an, where

an = (−1)nsin(nπ

4) +

1

3n, n ≥ 1,

and verify that each is indeed a clus-ter point.

3. (6 points) Use the Mean Value The-orem to show that if x < y, then

ex(y − x) < ey − ex < ey(y − x).


1. (17 Points)Proposition: Let f : (a, b) → Rand assume that f

′(x) < 0 on (a, b).

Then, f is strictly decreasing on(a, b).

2. (17 Points)Proposition: If the sequence ansatisfies

|an+1 − an| <1

3n, n > 0,


A-3.4 Final AInstructions:This is a closed book and closed notes test.You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.


1. (6 points) f : [a, b]→ R is contin-uous at a ∈ (a, b).

2. (6 points) f : [a, b] → R is uni-formly continuous in [a, b].

3. (6 points) The Cauchy Criterionfor convergence of a sequence an.

4. (6 points) A is a cluster point ofthe sequence an.

5. (6 points) µ is the least upper boundof the nonempty and bounded setS.

6. (6 points) The lim inf an, whenan is a bounded sequence.

State precisely the following mathemat-ical theorems or axioms:

1. (6 Points) The Bolzano WeierstrassTheorem

2. (6 Points) The Completeness Prop-erty of the real numbers

3. (6 Points) Rolle’s Theorem

A-4. EXAMS VERSION B 371

4. (6 points) The The First Deriva-tive Test

5. (6 points) The Intermediate ValueTheorem

Part 2: Short Answer (66 Points)Answer the following questions. If theanswer is YES or NO, you MUST giveus the reason why; ( e.g. the completestatement of a relevant theorem, a coun-terexample etc.)

1. (4 points) Is it possible for a con-tinuous function on [1,∞) to be un-bounded?

2. (4 points) Is it necessarily true thatthe limit inferior of a sequence is acluster point of the sequence?

3. (4 points) Is it possible for a func-tion to be continuous and differen-tiable at only one point?

4. (4 points) If a sequence is bounded,does it have to converge?

5. (4 points) If the first derivative of afunction is zero at a point, does thefunction have to have a maximumor minimum at that point?

6. (4 points) If the second derivativeof a function exists at a point, isthe first derivative of the functioncontinuous at that point?

7. (10 points)Use the Intermediate Value Theo-rem to prove that x5 + 2x3 + x −3 has at least one real root. UseRolle’s Theorem to show that theroot is unique.

8. (12 points)Let the sequence an be recursivelydefined by the formula

a1 = 1,

an+1 =

(1− 1

(n+ 1)2

)an.

Show that an converges.

9. (10 points)Find limx→0+

tan(x)−xx3

10. (10 points)Show that∣∣∣∣ sin(x)−

(x− x3

6+

x5

120

)∣∣∣∣<

1

5040, for |x| ≤ 1.


1. (17 points)If an → a, bn → b and cn →c use an ε − δ argument to provethat an + bn + cn → a+ b+ c.

2. (17 points)If f : R→ R is defined by f(x) =3x2−2x+1, use an ε−δ argumentto prove that f is continuous at x =2.

3. (17 points)Prove that if f : [a, b] → R has abounded derivative f

′(x) on (a, b),

then f is uniformly continuous on(a,b).

4. (17 points)Prove for all positive integers

d

dx(xn) = n xn−1

A-4 Sample Exams Version B

A-4.1 Exam 1B

Instructions:This is a closed book and closed notes test.You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.


1. (3 Points) Give a precise mathemat-ical definition of the meaning of asequence of real numbers.


2. (3 Points) Give a precise mathemat-ical definition of the phrase ”the se-quence an converges to a”.

3. (3 Points) Let S be a nonempty setof real numbers. Give a precise math-ematical definition of the supremumof S.

4. (6 Points) State precisely the fol-lowing theorems:

(a) The limit of the sum of sequencestheorem.

(b) The limit of product of sequencestheorem.

5. (3 Points) State the Triangle andReverse Triangle Inequality for RealNumbers.

Part 2: Short Answer (44 Points)You must determine whether or not thesestatements are true. If the statement istrue, you MUST give us the reason whyit is true; if the statement is false, youmust give us a counterexample.

1. (6 Points) Let the sequence an →a, a 6= 0, as n→∞. Then there isan integer N such that if n > N ,then |a|2 ≤ |an| ≤ |a|+ 1.

2. (4 Points) Is it true that if two se-quences diverge, their product mustdiverge?

3. (4 Points) Is it true that the infi-mum of a set is always achieved bysome element of the set?

4. (4 Points) If |x| < ε for all ε > 0,can x be nonzero?

Show all your work on the short calcula-tional and/or discussion exercises below.You may use a calculator if you wish.

1. (8 Points) The sequence nn+1. clearly

has limit 1. Discuss what happensin an attempted convergence proofif you try to prove the limit is 2.Where does the proof fail?

2. (5 Points) What method of proofwould be required to prove the propo-sition d

dx (xn) = nxn−1?

3. (4 Points) Find limit, if it exists, forthe sequence whose nth term is:

an =n(n+ 4)

4n(n+ 1).

4. (3 Points) Find limit, if it exists, forthe sequence whose nth term is:

an =2n

nn.

5. (6 Points) Find the inf and sup ofthe set

S = i2j|i, j are pos. int.


1. (19 Points)Proposition: For all n ≥ 1,

1 + 4 + 7 + . . .+ (3n− 2) =1

2n(3n− 1)


an =3n2 + 12n− 13

2n2 − 5n, n ≥ 1,


A-4.2 Exam 2BInstructions:This is a closed book and closed notes test.You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.


1. Give precise mathematical defini-tions of the following mathemati-


cal concepts using either the ε, δ orε,N formalism:

(a) (4 Points) ”The function f :D ⊆ R → R has a limit A atthe point a ∈ D”.

(b) (4 Points) ”The function f :D ⊆ R→ R is continuous atthe point a ∈ D”.

(c) (4 Points) ”The function f :D ⊆ R → R is uniformlycontinuous in D”.

2. Give precise mathematical defini-tions of the following mathemati-cal concepts:

(a) (3 Points) A is a cluster pointof the sequence an.

(b) (3 Points) A is a cluster pointof the function f(x) at x = a.

(c) (3 Points) A is an accumula-tion point of the nonempty setB.

3. State precisely the following math-ematical theorems and/or axioms:the

(a) (5 Points) Completeness Ax-iom

(b) (5 Points) Intermediate ValueTheorem

(c) (5 Points) Bolzano WeierstrassTheorem

Part 2: Short Answer (31 Points)

1. You must answer the following ques-tions. If the answer is YES, youMUST give us the reason why; ifthe answer is NO, you must alsogive a reason or a counterexample.

(a) (3 Points) If the sequence anis increasing and bounded be-low by M , does the sequencenecessarily converge?

(b) (3 Points) If the sequence anis decreasing and bounded be-low by M , does the sequenceconverge to M?

(c) (3 Points) Is it possible for thefunction f + g to be continu-ous for all real numbers even

though both f and g are con-tinuous at only one point each?

2. Show all your work on the shortcalculational exercises below. Youmay use a calculator if you wish.

(a) (6 Points) Prove that there is asolution to the equation

2 +x

2 + sin2(x)+

x3

x2 + 3

= 0

on the interval [−5, 0].(b) (8 Points) Find the set of clus-

ter points for the the functionf(x) = 1

4sin( 5x ) at x = 0.

(c) (8 Points) Find the set of clus-ter points for the sequence an,where an = cos(nπ3 ) and ex-hibit a subsequence which con-verges to each point.


1. (15 Points) Use an ε, δ argument toprove that if the functions f, g :D ⊆ R → R are continuous ata ∈ D, then the function f + 2g :D ⊆ R → R is also continuous ata ∈ D.

2. (18 Points) Let the sequence anbe defined by

a1 = 1,

an+1 =5(1 + an)

5 + an, n > 1.

It can be shown that√

(5) < an <5 for all n ≥ 2. Prove that this se-quence converges and find the valueof the limit.

BONUS: (5 Points)Show that

√(5) < an < 5 for all

n ≥ 2.


A-4.3 Exam 3BInstructions:This is a closed book and closed notes test.You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.


1. Give precise mathematical defini-tions of the following mathemati-cal concepts using the appropriatemathematical formalism:(a) (4 Points) ”The Cauchy Cri-

terion for a sequence an”.(b) (4 Points) ”The function f [a, b] ⊆

R→ R, where [a, b] is a closedand finite interval, has a deriva-tive at x0 ∈ (a, b)”.

2. State precisely the following math-ematical theorems: the(a) (4 Points) Completeness Ax-

iom(b) (4 Points) Bolzano-Weierstrass

Theorem(c) (4 Points) Rolle’s Theorem(d) (4 Points) Intermediate Value

Theorem(e) (4 Points) The Mean Value The-

orem

Part 2: Short Answer (20 Points)You must answer the following questions.If the answer is YES or NO, you MUSTgive us the reason why; ( e.g. the com-plete statement of a relevant theorem, acounterexample etc.)

1. (4 Points) Is it possible for the max-imum and minimum of a functionto be the same?

2. (4 Points) Let f and f′

denote afunction and its derivative. Is it pos-sible for f

′(c) to exist even though

f′

is not continuous at c?3. (4 Points) Is f(x) = 1

x+2 uni-formly continuous on [0, 1]?

4. (4 Points) Can a function f be dif-ferentiable on an interval I and beunbounded on I?

5. (4 points) If f and g are contin-uous functions on [0, 1], does thefunction f−g2 achieve a minimumvalue on [0, 1]?

Part 3: Calculational Exercises (22 Points)Show all your work on the short calcu-lational exercises below. You may use acalculator if you wish.

1. (8 Points) If

f(x) = x2 + a, x ≤ −1

= bx, −1 < x < 1

= −cx2 + d, x ≥ 1

(a) Find values of a, b, c, d that willmake f

′exist for all x.

(b) Find the set of x where f′

iscontinuous.

2. (8 points) Use the Mean Value The-orem to approximate (28)

13 .

3. (6 Points) Find the following limit:

limx→0+

− ln(x)

csc(x)

Part 3: Proofs (30 Points)

Provide careful proofs of the followingpropositions. You will be graded on themathematical correctness of your argu-ments as well as your use of language,syntax and organization in the proof.

1. (15 Points)Prove that the function f(x) = 3x3+4x+ 9 has one and only one root.

2. (15 Points)Proposition: If the sequence ansatisfies

|an+1 − an| <(

2

5

)n, n > 0,


A-4.4 Final BInstructions:This is a closed book and closed notes test.


You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.


1. Give precise mathematical defini-tions of the following mathemati-cal concepts using the appropriatemathematical formalism: (24 Points)

(a) (4 Points) The sequence an con-verges to a.

(b) (4 points) The Cauchy Crite-rion for convergence of a se-quence an.

(c) (4 points) A is a cluster pointof the sequence an.

(d) (4 points) f : [a, b] → R iscontinuous at a ∈ (a, b).

(e) (4 points) f : [a, b] → R isuniformly continuous in [a, b].

(f) (4 Points) A is a cluster pointof the function f at the pointa.

2. (32 Points) State precisely the fol-lowing mathematical theorems oraxioms: the

(a) (4 Points) Bolzano WeierstrassTheorem

(b) (4 Points) Completeness Prop-erty of the real numbers

(c) (4 Points) Rolle’s Theorem(d) (4 points) Intermediate Value

Theorem(e) (4 Points) Mean Value Theo-

rem(f) (8 Points) Taylor Polynomial

and Remainder Theorem(g) (4 points) First Derivative Test


Answer the following questions. If theanswer is YES or NO, you MUST giveus the reason why; ( e.g. the completestatement of a relevant theorem, a coun-terexample etc.)

1. (4 points) Is it possible for a con-tinuous function on (0, 1) to be un-bounded?

2. (4 points) Is it possible for a con-tinuous function on [0, 1] to be un-bounded?

3. (4 points) Is it possible for a dif-ferentiable function on (0, 1) to beunbounded at the point .5?

4. (4 points) Is it possible for a se-quence an to have two cluster pointswhen its limit exists?

5. (4 points) Is it possible for a func-tion to be continuous at only onepoint?

6. (4 points) Is it possible for a func-tion to be continuous and differen-tiable at only one point?

7. (4 points) If a function is bounded,does it have to be continuous?

8. (4 Points) If a function is bounded,does it necessarily have a maximumand a minimum?

9. (4 Points) If the first derivative of afunction f is zero at the point x0,is it necessarily true that f attainsa local maximum value at x0?

Part 3: Calculational Exercises (39 Points)

1. (9 points)

Find the cluster points of the se-quence defined by an = (−1)n sin(nπ3 )for all n > 0 and exhibit explicitsubsequences for each cluster point.

2. (10 points)

Find the cluster points of the func-tion defined by f(x) = sin( 1

x2 ) atx = 0 and exhibit an explicit sub-sequence for each cluster point.

3. (10 points)Estimate the largest interval [−a, a]for which the

| sin(x)− x |< 10−6,


4. (10 points)Find the smallest order of Taylorpolynomial, pn(x), centered at x =0 for which

| cos(x)− pn(x) |< 10−3,


1. (9 points)Show that if the sequence an is aCauchy sequence, then it is bounded.

2. (15 points)Let the sequence an be definedby an = n2+2

−3(n2)+5 , for all n > 1.Use an ε, δ argument to prove thelimit as n goes to infinity exists.

3. (15 points)If f : R→ R is defined by f(x) =4(x)2 − 2x+ 10, use an ε, δ argu-ment to prove that f is continuousat x = 1.

4. (15 points)If f : R→ R is defined by f(x) =3(x)2 + 3x + 8, use an ε, δ argu-ment to prove that f is uniformlycontinuous on the interval [−2, 3].

5. (15 points)If the sequence an is defined by

an = 1, n = 1

=√

2 an−1, n > 1

,prove that the limit of this sequenceexists and find its value.

A-5 Sample Exams Version C

A-5.1 Exam 1CInstructions:This is a closed book and closed notes test.

You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.


1. (3 Points) Let S be a nonempty setof real numbers. Give a precise math-ematical definition of the supremumof S.

2. (3 Points) Let p be an accumula-tion point of the nonempty set S.Give a precise mathematical def-inition of the term accumulationpoint.

3. (3 Points) State the Triangle andReverse Triangle Inequality for RealNumbers.

4. (4 Points) State the Bolzano-WeierstrassTheorem.

5. (4 Points) State the Heine-Borel The-orem.

6. (3 Points) Give a precise mathemat-ical definition of the meaning of thephrase limx−→a f(x) = L.


1. (4 Points) Is it true that the supre-mum of a set is always achieved bysome element of the set?

2. (4 Points) Is it true that every opencover of the interval [2, 5] must havea finite subcover?

3. (4 Points) Let f(x) = 4λx(1 −x), λ ∈ (0, 1) and consider the se-quence of points generated by theprocedure x0 = .1, x1 = f(x0),x2 = f(x1) etc. What does theBolzano-Weierstrass theorem say aboutthis sequence?

4. (4 Points) Is it possible for a set tohave no accumulation points?

A-5. EXAMS VERSION C 377

Show all your work on the short calcula-tional and/or discussion exercises below.You may use a calculator if you wish.

1. (8 Points) Let S = (−1, 10]. Findan open cover of this set that hasno finite subcover.

2. (12 Points) Let S = 14n : n =

1, 2, · · · , ∪ (2, 5).(a) (6 Points) Find all accumula-

tion points of S.(b) (6 Points) Find the inf and sup

of S.3. (8 Points) Finish the following proof:

Proposition A-5.1 If |x| < ε, ∀ε > 0, then x = 0.

Proof:(a) We will prove by contradiction.(b) Assume x 6= 0.(c) Since x < ε for any positive ε,

we are free to choose the par-ticular value ε = |x|

2 .(d) SHOW THIS LEADS TO A

CONTRADICTION.


1. (12 Points) Provide an ε − δ proofof the following proposition:

Proposition A-5.2

limx−→ 4

(x2 + 6) = 22.

2. (12 Points)

Proposition A-5.3 For all n > 1,

n

n+ 1=

1

1 · 2+

1

2 · 3

+1

3 · 4+ · · ·+ 1

n · (n+ 1).

3. (12 Points)

Proposition A-5.4√

5 is irrational.

A-5.2 Exam 2C



1. (3 Points) Let S be a set of realnumbers, let p ∈ S and let f :S → < be a function. Give a full,complete and precise definition ofthe continuity of the function fat the point p.

2. (3 Points) Let S be a set of realnumbers, let p ∈ S and let f :S → < be a function. Give a full,complete and precise definition ofthe uniform continuity of f on theset S.

3. (3 Points) Let S be a set of realnumbers, let p ∈ S and let f :S → < be a function. Give a full,complete and precise definition ofthe derivative of f at the point p.

4. (3 Points) Let S be a set of realnumbers, let p ∈ S and let f :S → < be a function. Give a full,complete and precise definition ofthe limx→∞ f(x) = −∞.

Part 2: Theorems: (12 Points)Give full, complete and precise statementsof the following theorems:

1. (4 Points) The Intermediate ValueTheorem

2. (4 Points) Rolle’s Theorem

3. (4 Points) The Mean Value Theo-rem



1. (4 Points) Let f be a function de-fined on the finite interval [a, b] with0 ∈ (a, b). Assume f

′(0) = 0 and

f′(0) = 0. Is it possible for f to

have a minimum or maximum at0?

2. (4 Points) Let f be a function de-fined on the finite set S. If f isbounded and continuous on S, is itnecessarily true that f is uniformlycontinuous on S?

3. (4 Points) Is is possible for a func-tion to be continuous at only onepoint?

4. (4 Points) Let f be a function de-fined on the finite set S and assumef′

and f′

exist on S. Let p ∈ Swith f

′(p) > 0. Is it true that there

is a δ > 0 so that f′> 0 on the

interval (p− δ, p+ δ)?

Part 4: Calculations (34 Points)Show all your work on the short calcula-tional and/or discussion exercises below.You may use a calculator if you wish.

1. (6 Points)

limx→∞

x ln2x+ 1

2x− 10

2. (6 Points) Prove

√1 + x < 1 +

x

2, ∀x > 0

3. (6 Points) Is f(x) = 1x uniformly

continuous on (0, 1)?

4. (6 Points) Is the function f definedby

f(x) =2x− 1, x ∈ [0, 1]x3 − 5x2 + 5, x ∈ (1, 2]

uniformly continuous on [0, 2]?

5. (10 Points) For any positive integern, let gn be the function

gn(x) =

n∑i=0

(1

2)i cos((13)iπx)

(a) (2 Points) Let n = 7. Illus-trate what the graph of g7 lookslike on the intervals [−.1, .1],and [−10−8, 10−8].

(b) (2 Points) Let n = 12. Il-lustrate what the graph of g12

looks like on the intervals [−.1, .1],and [−10−28, 10−28].

(c) (2 Points) As n → ∞, whathappens to the graphs of gn asthe interval you graph gn onshrinks?

(d) (4 Points) If we let f(x) =limn→∞ gn(x), what kind offunction is f?

Part 4: Proofs (26 Points)Provide careful proofs of the followingpropositions. You will be graded on themathematical correctness of your argu-ments as well as your use of language,syntax and organization in the proof. Hint:These proofs are short!

1. (13 Points)

Proposition A-5.5 Let [a, b] be abounded closed interval of real num-bers and f, g : [a, b] → < be con-tinuous functions on [a, b]. Assumef(a) > g(a) and f(b) < g(b).Then there exists a point c ∈ (a, b)with f(c) = g(c).

2. (13 Points)

Proposition A-5.6 Let [a, b] be abounded closed interval of real num-bers and f : [a, b] → < be a con-tinuous function on [a, b]. Let Adenote the range of the function fon [a, b]; i. e., A = y : y =f(x), x ∈ [a, b]. ThenA is a closedand bounded interval, i. e., A =[c, d], for some finite numbers c andd.

A-5.3 Exam 3CInstructions:This is a closed book and closed notes test.You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.


Part 1: Definitions (18 Points):Define the following terms in full detail.

1. (3 Points) A Partition P of the fi-nite interval [a, b].

2. (3 Points) The Lower Sum L(P, f)for the bounded functionf : [a, b] → < and the partition Pof [a, b].

3. (3 Points) The Upper Sum L(P, f)for the bounded functionf : [a, b] → < and the partition Pof [a, b].

4. (3 Points) The Lower integral∫ baf(t)dt.

5. (3 Points) The Upper integral∫ baf(t)dt.

6. (3 Points) The integral∫ baf(t)dt

Part 2: Theorems (14 Points):State carefully the following theorems.the

1. (4 Points) First Fund. Theoremof Calculus

2. (5 Points) Second Fund. Theo-rem of Calculus

3. (5 Points) Taylor’s Theorem withLagrange Form of Remainder

Part 3: Short Answer: (18 Points):

1. (6 Points)If the function f is differ-entiable on the finite interval [a, b],is it necessarily true that the func-tion F defined on [a, b] by F (x) =∫ xaf(t)dt is twice differentiable on

[a, b]?

2. (6 Points) Give an example of a func-tion which is integrable on [0, 1] butnot continuous on [0, 1]. Explainyour example carefully.

3. (6 Points) If f is bounded on [0, 1],is it necessarily true that f is inte-grable?

Part 4: Calculational: (20 Points):

1. (8 Points) Compute

d

dx

∫ x3−1

x2

cos(ln(x)) dx

2. (12 Points) Find the Taylor poly-nomial pn0(x) which will approxi-mate the function f(x) = cos(2x)on the interval [−π, π] to accuracy10−5.

Part 5: Proofs (30 Points):Provide careful proofs of the followingpropositions. You will be graded on themathematical correctness of your argu-ments as well as your use of language,syntax and organization in the proof.

1. (15 Points)

Proposition A-5.7 Let the functionf be defined on the finite interval[a, b]. Assume f is continuous on[a, b] and f(x) ≤ 0 for all x in[a, b] and there exists a point x0

such that f(x0) < 0. Then∫ baf(t)dt <

0.

2. (15 Points)

Proposition A-5.8 Use the defini-tion of the integral to prove that

∫ 1

03t2dt =

1.0.

A-5.4 Final C



1. Give precise mathematical defini-tions for the following concepts us-ing the appropriate mathematical for-malism: (18 Points)

(a) (3 points) A is an accumula-tion point of the set S.


(b) (3 Points) m is the inf of theset S.

(c) (3 Points) The sequence an con-verges to a.

(d) (3 Points) limx→p f(x) = Lfor p ∈ (a, b) where the func-tion f : [a, b]→ <.

(e) (3 points) f : [a, b] → < iscontinuous at p ∈ (a, b).

(f) (3 points) f : [a, b] → < isuniformly continuous in [a, b].

(g) Integration Theory: (12 Points)f : [a, b] → < is a boundedfunction on the finite interval[a, b]. Define the following con-cepts:

i. (3 Points) Q is a refine-ment of the partition P of[a, b]

ii. (3 Points) The Upper sumU(P, f) for partition P .

iii. (3 Points) The Lower In-tegral

∫ baf(t)dt.

iv. (3 Points) The integral off on [a, b].

(h) Theorems (24 Points)State precisely the followingmathematical theorems or ax-ioms:

i. (3 Points) The Heine-BorelTheorem

ii. (3 Points) The Least Up-per Bound Axiom

iii. (3 Points) Rolle’s Theo-rem

iv. (3 Points) The Interme-diate Value Theorem

v. (3 Points) The Mean ValueTheorem

vi. (3 Points) The Weak Prim-itive Theorem

vii. (3 Points) The First Fun-damental Theorem of Cal-culus

viii. (3 Points) The Second Fun-damental Theorem of Cal-culus


Answer the following questions. If theanswer is YES or NO, you MUST giveus the reason why; ( e.g. the completestatement of a relevant theorem, a coun-terexample etc.)

1. (4 points) Is it possible for a func-tion to be continuous nowhere?

2. (4 points) Is it possible for a func-tion to be continuous at one point?

3. (4 points) Is it possible for a func-tion to be differentiable at one point?

4. (4 points) If f(x) = sin(x)+∫ x−1g(t)dt,

where g is integrable on [−1, 2], isit necessarily true that f is uniformlycontinuous on [−1, 2]?

5. (4 points) Is it necessarily true thata function that is twice differentiableon the interval (0, 1) has a boundedderivative?

6. (4 points) Is it possible for a se-quence to have 16 subsequential lim-its?

7. (4 points) Is it necessarily true thatan integrable function is continu-ous?

8. (4 points) If a function has a min-imum value on the finite interval[a, b], does the function have to bedifferentiable at the point where theminimum occurs?

Part 3: Calculational Exercises (50 Points)

1. (10 Points)Find an open cover of the interval(1, 2) that has no finite subcover.

2. (10 Points)Use the definition of the derivativeto show that the function f definedby

f(x) = x2 sin(1

x), x 6= 0

= 0, x = 0

satisfies f′(0) = 0.

3. (10 Points)Show that cos(x) = x3 + x2 + 4xhas exactly one root in [0, π2 ].


4. (10 Points)

(a) (5 Points) Using base point 2,find the second order TaylorPolynomial, p2,2(x) for the func-tion f(x) =

√x.

(b) (5 Points) Using the Lagrangeform of the remainder, estimatethe error r2,2(x) over the in-terval [1, 3].

5. (10 Points)Let the function f be defined by

f(x)

= x2 +

∫ ex

sin(x)

sin13 (t+ t2) dt

Compute f′(x).


1. (16 points)Let the sequence an be definedby an = 2n2+2

5n2+5n+2 , for all n >1. Use an ε, N argument to provelimn→∞ exists.

2. (16 points)If f : (a, b) → R is differentiableon (a, b) and there exist a positiveB such that | f ′(x) | ≤ B for all xin (a, b), prove that f is uniformlycontinuous on (a, b).

3. (16 points)Let the function f be defined by

f(x) = 1, x > 1

= 0, x ≤ 1

Use an ε, δ argument to prove thatf is not continuous at x = 1.

4. (16 points)Let f be continuous for all non-negative x. Suppose further that

f(x) 6= 0 for all positive x. Provethat if f2(x) = 2

∫ x0f(t)dt, for

all x ≥ 0, then f(x) = x for allx ≥ 0.

Appendix B

Senior Advanced Calculus II StudyGuides

Presented below are study guides and coursedescription material from some of the times wehave taught the second semester senior levelundergraduate analysis course here. At Clem-son University, it is called MTHSC 453. An-other source of information is our web-baseddiscussion board material. The MTHSC 454discussion board can be accessed either throughmy web page by following the links or by thedirect routehttp://www.ces.math.clemson.edu/discus.On that page you will see exercises, workedout problems and so forth in an easy to followformat.

B-1 MTHSC 454

B-2 Course StructureThe intent of this course is to continue to teachyou the basic concepts of analysis. We assumethat you have all had a careful and completegrounding in the in the necessary skills to read,write and understand the process of provinga mathematical proposition. We also assumethat you have covered the equivalent of the firstfour chapters of the Fulks’ text. In this course,we will introduce you to integration theory, in-finite series, sequences and series of functionsand uniform convergence Taylor series and ad-ditional other concepts as time permits. Hereare some details:

M454/M454H/M654 Outline:

Integration:f : [a, b]→ < is bounded functionon the finite interval [a, b].

1. Partition P of [a, b]

2. Refinements of Partition P′

ofP .

3. Norm of Partition P , ‖P‖4. Upper sum SP (f) ≡ U(P, f)

5. Lower sum SP (f) ≡ L(P, f)

6. Fundamental InequalitiesL(P, f) ≤ U(P, f)L(P, f) ≤ L(P

′, f)

U(P′, f) ≤ U(P, f)

L(Q, f) ≤ U(P, f) for anytwo partitionsP andQ of [a, b]

7. Upper Integral∫ baf(x)dx = inf U(P, f) :

P is a partition of[a, b]8. Lower integral∫ b

af(x)dx = sup L(P, f) :

P is a partition of[a, b]

9.∫ baf(x)dx ≤

∫ baf(x)dx

10. infx∈[a,b](f(x))(b−a) ≤∫ baf(x)dx ≤∫ b

af(x)dx ≤ supx∈[a,b](f(x))(b−

a)

11. Algebra of upper and lower in-tegrals

(a) Theorem:∫ ba(f(x)+c)dx =∫ b

af(x)dx+ c(b− a)

(Similar for lower integrals)

383

384 APPENDIX B. ADVANCED CALCULUS II

(b) Theorem:∫ baf(x)dx =∫ c

af(x)dx +

∫ bcf(x)dx,

for any c ∈ [a, b](Similar for lower integrals)

12. Theorem: (Weak Fundamen-tal Theorem)ddx (∫ xaf(t)dt)|c = d

dx (∫ xaf(t)dt)|c =

f(c),at any point c ∈ [a, b] where fis continuous.

13. Theorem: f is Riemann-integrable

if∫ baf(x)dx =

∫ baf(x)dx

This common value is denoted∫ baf(x)dx

14. What functions are Riemann-integrable?(a) Theorem: f monotone⇒

f Riemann-integrable(b) Theorem: f continuous⇒ f Riemann-integrable

15. Theorem: Necessary and Suf-ficient Condition for integralto exist(a) Theorem: f Riemann-integrable⇔ ∀ε > 0 ∃ partition Psuch thatU(P, f)−L(P, f) <ε.

16. Upper and Lower integral es-timates

Theorem: f bounded on [a, b]⇒ ∀ε > 0 ∃δ(ε) so that

U(P, f) <∫ baf(x)dx + ε

and L(P, f) >∫ baf(x)dx−ε

for any partitionP with ‖P‖ <δ(ε).

17. Riemann SumS(P, f, ξ)

18. Theorem: J = lim‖P‖→0 S(P, f, ξ)

19. Theorem: f Riemann-integrable⇔ J =

∫ baf(x)dx

20. Properties of Riemann-Integrable(Section 5.4)(a) Theorem: f Riemann-integrable⇒ F (x) =

∫ xaf(t)dt is

continuous on [a, b]

(b) Theorem: Subset Property

(c) Theorem: Intermediate PointProperty

(d) Theorem: f(x) ≥ g(x)

⇒∫ baf(x)dx ≥

∫ bag(x)dx

(e) Theorem: Integral of lin-ear combinations of func-tions is linear combinationof integral of functions

(f) Theorem: f integrable⇒f+ and f− integrable

(g) Theorem: f integrable⇒|f | integrable

(h) Theorem: Integral Esti-mate|∫ baf(x)dx |≤

∫ ba| f(x) |

dx

(i) Theorem: f, g integrable⇒ fg integrable

21. Fundamental Theorem of Cal-culus

(a) Primitive (antiderivative, in-definite integral) of f

(b) Theorem: Fundamental The-orem of Calculusf continuous and F prim-itive of f⇒

∫ baf(x)dx =

F (b)− F (a)

(c) When does integrating f′

recapture f?Theorem: f differentiableand f

′integrable⇒

∫ baf′(x)dx =

f(b)− f(a)

(d) Theorem: Integration bySubstitution

(e) Theorem: ddx (∫ v(x)

u(x)f(t)dt) =

v′(x)f(v(x))−u′(x)f(u(x))

22. Theorem: Integration by PartsInfinite Series: 1. Definition of a se-

ries:an∞n0

is a sequence.

(a) Partial Sums: Sp =∑n0+p−1n0

anSum of the first p terms ofthe sequence

(b) Theorem: If limp→∞ Spexists, series converges.let∑∞n0an ≡ limp→∞ Sp,

where∑∞n0an denotes the

sum of the series.

B-2. COURSE STRUCTURE 385

(c) Theorem: If limp→∞ Spdoes not exist, series di-verges.

(d) Absolute convergence(e) Conditional convergence

2. Summing a series(a) telescoping terms(b) partial fractions

3. Tests for Convergence and/ordivergence(a) Theorem: nth term test(b) Theorem: Cauchy Crite-

rion specialized to series(c) Theorem: Absolute con-

vergence implies originalseries converges

4. Theorem: Tests for Conver-gence and/or divergence:series of non-negative terms(a) Theorem: p series(b) Theorem: geometric se-

ries(c) Theorem: comparison test(d) Theorem: limit compari-

son test(e) Theorem: integral test(f) Theorem: root test(g) Theorem: ratio test

5. Tests for Series with Variablesigns(a) Theorem: Alternating Se-

ries Test (AST)(b) Apply any of tests for se-

ries of non-negative termsto series of absolute values–checks for absolute con-vergence. If this fails, onlyoption is AST

Sequence and Series of Functions:fn ≡ fn∞n0

is a sequence offunctions defined on [a, b]

1. Pointwise convergence of fnto a function f on [a, b]

2. Uniform convergence of fnon [a, b]

3. Theorem: fn converges uni-formly to f on [a, b]⇔Mn =supx∈[a,b] | fn(x)− f(x) |→0 as n→∞.

(a) Find f via pointwise limitfirst

(b) Once pointwise limit fknown, computeMn limit.If limit nonzero, conver-gence not uniform.

4. Theorem: Cauchy Criterion foruniform convergence

5. Theorem: Weierstrass K-TestThis is for series of functions∑∞n0un(x).

LetKn = supx∈[a,b] | un(x) |.Then

∑∞n0Kn converges⇒ se-

ries of functions converges uni-formly on [a, b]

6. Consequences of Uniform Con-vergence(a) Theorem: If a sequence

of continuous functions con-verges uniformly on [a, b],then limit function is con-tinuous

(b) Theorem: If series of con-tinuous functions convergesuniformly on [a, b], then itssum is continuous function

(c) Theorem: Interchange ofintegration and limitfn → uniformlyf on [a, b]where each fn is integrable.Then f is also integrableand lim

∫ bafn =

∫ ba

lim fn(also Theorem: (each fnis continuous), Theorem:(each fn is monotone), The-orem: (applied to seriesof functions) )

(d) When does f′

n → f′?

Theorem: Very technicalstatement: –nice theorembecause it pulls together alot of results)

Taylor Series: 1. Power Series(a) Radius of convergence(b) Interval of convergence(c) What happens at endpoints

of interval of convergence?2. Power series converges abso-

lutely inside interval of con-vergence and diverges outside


interval of convergence. Mustcheck behavior at endpoints ex-plicitly.

3. Finding Radius of Convergence(a) Theorem: Ratio Test:

q = lim | an+1

an|

R = 0 if q =∞R =∞ if q = 0R = 1

q if q finite(b) Root test

4. Properties of Power Series(a) Theorem: Power series con-

verges uniformly on anyclosed subinterval containedin the interval of conver-gence

(b) Theorem: Let interval ofconvergence be (−R,R).Then power series convergesat R implies power seriesconverges uniformly on [0, R].(Similar result for conver-gence at other endpoint)

(c) Theorem: If power serieshas interval of convergence(−R,R), so does derivedseries.

(d) Theorem: IfR > 0, deriva-tive of a power series isgiven by the derived seriesand the interval of conver-gence is same for both se-ries. (same type result forintegral of power series)

(e) Theorem: IfR > 0, powerseries has derivatives of allorders and interval of con-vergence is always same

(f) Theorem: f(x) =∑anx

n

and radius of convergenceR > 0. Then an = f(n)(0)

k!

(g) Theorem: Power series ex-pansions are unique

5. Operating on Power Series(a) Theorem: Integrating se-

ries(b) Theorem: Differentiating

series(c) Theorem: Summing Se-

ries

6. Taylor Series(a) Theorem: Taylor’s theo-

rem with remainder revis-ited

(b) Theorem: Taylor series ex-pansion of arbitrary func-tion

7. Finding Taylor Series(a) Finding pattern in deriva-

tives(b) Using geometric series tricks(c) Using recursion trick for

ratios of functions8. Using Taylor Series

(a) Integrating a function(b) Evaluating a limit

B-3 Sample Exams Version A

B-3.1 Exam 1A



1. (15 Points) Carefully define and dis-cuss the integral of a bounded func-tion f : [a, b] → R, where [a, b]is a closed and finite interval. Youwill need to explain in detail all rel-evant symbols and terms.


1. (7 Points) Let [a, b] be a closed andfinite interval and assume f, g : [a, b]→R are not integrable on [a, b]. Is itnecessarily true that f + g is notintegrable on [a, b]?

B-3. EXAMS VERSION A 387


1. (17 Points) Consider f(x) = x2 onthe interval [0, 1]. Let Pn be theuniformly spaced partition of [0, 1]consisting of n equal subintervals.Find SPn − SPn and show that

SPn − SPn → 0 as n→∞.

Does this prove that f(x) = x2 isintegrable on [0, 1]? Explain youranswer carefully.

2. Determine whether or not the fol-lowing functions are integrable on[0, 1] giving full reasons for youranswers.

(a) (7 points)

f(x) = x2 − 3x + 7,

(b) (7 points)

f(x) =π

2, 0 ≤ x < .25

= 1.0, x = .25

= .6, .25 < x < .75

= .17, x = .75

= .038, .75 < x ≤ 1.0

(c) (7 points)

f(x) = 2, x ∈ Q= −9, x ∈ I

whereQ and I indicate the ra-tional and irrational numbersrespectively.

(d) Differentiate the function f :[1, 10]→ R defined by,

i. (3 points)

f(x) =∫ x

1

(1 + t2)−2 dt

ii. (3 points)

f(x) =

∫ x2

1

(2 + t3)−4 dt


1. (17 Points)Proposition: Let [a, b] be a finiteinterval and let f : [a, b] → Rbe continuous and non-negative on[a, b], i.e. f(x) ≥ 0, a ≤ x ≤ b.Then if ∃ x0, a < x0 < b 3f(x0) > 0, this implies that

∫ baf(x)dx >

0.

2. (17 Points)Carefully state and prove a theo-rem for calculating

d

dx

(∫ u(x)

a

f(t) dt

)

B-3.2 Exam 2AInstructions:This is a closed book and closed notes test.You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.

Part 1: Definitions (24 Points)Carefully and precisely define the fol-lowing mathematical phrases:

1. (6 Points)

∞∑n=0

an = S.

In the phrases below, I is a finite inter-val, fn is a sequence of real valuedfunctions defined on I and f is a realvalued function defined on I .

1. (6 Points)

fn(x) → f(x) pointwise on I.


2. (6 Points)

fn(x) → f(x) uniformly on I.

3. (6 Points)

∞∑n=0

fn(x)

converges uniformly on I to the func-tion f(x).


1. (5 Points) Let I be a finite intervaland fn a sequence of real valuedfunctions defined on I . If fn →f pointwise on I , is it necessarilytrue that f is continuous on I?

2. (5 Points) If an → 0 is it necessar-ily true that

∑∞n=0 an converges?

For each of the following series, tellwhether or not it is absolutely conver-gent, conditionally convergent or di-vergent,giving full reasons for your answers.

1. (6 Points)

∞∑n=0

2n

3 (5n+1)

2. (6 Points)

∞∑n=3

n3 + 6n− 1

n4 − 17

3. (6 Points)

∞∑n=2

n!

(2n− 2)!

4. (6 Points)

∞∑n=1

1

(2n)n

5. (6 Points)

∞∑n=0

(−1

2)n

6. (6 Points)

1− 1

2− 1

4+

1

8

− 1

16− 1

32+

1

64− . . .

(plus, then 2 minuses)


1. (15 Points) Consider for all n ≥ 1,

fn(x) =n(x2)

1 + n(x2)

Discuss the pointwise and uniformconvergence of fn on the intervals:

(a) [−1, 1]

(b) [1, 4]


1. (15 Points)If∑an and

∑bn both converge

absolutely, then∑

(an + bn) con-verges absolutely.

B-3.3 Exam 3A


Part 1: Convergence of Series (36 Points)Find all the values of x at which the fol-lowing series converge giving full rea-sons for your answers.

B-3. EXAMS VERSION A 389

1. (12 Points)

∞∑n=1

x2n

n 3n

2. (12 Points)

∞∑n=1

n! xn

nn

3. (12 Points)

∞∑n=1

an xn, an = 1, if n is even

=1

n, if n is odd

Part 2: Taylor Series (48 Points)Show all your work on the short calcula-tional exercises below.

1. (16 Points) Let f(x) = ex−1x , for

x 6= 0.

(a) Find the power series that rep-resents f for x 6= 0.

(b) Find the power series that rep-resents f

′for x 6= 0.

2. (16 points) Sum the series

∞∑n=1

xn+1

n+ 1

3. (16 Points) Let f(x) =∫ x

0tsin(t)dt.

Find the Taylor series centered atx = 0 for f .

Part 3: Proofs (16 Points)Provide careful proofs of the followingproposition. You will be graded on themathematical correctness of your argu-ments as well as your use of language,syntax and organization in the proof.

(16 Points)Let fn and gn be two sequences ofreal valued functions defined on the fi-nite interval I. If fn → f uniformly on Iand gn → g uniformly on I, then (fn +gn)→ (f + g) uniformly on I.

B-3.4 Final AInstructions:This is a closed book and closed notes test.You will need to give us all the details of yourarguments as that is the only way we can de-cide if partial credit is warranted.


1. Convergence IssuesLet [a, b] be a finite interval of thereal line with fn denoting a se-quence of real-valued functions, fn :[a, b] → < and f indicating an-other real-valued function, f : [a, b]→<. Define the following concepts:

(a) (3 points) fn → f pointwisefor x ∈ [a, b]

(b) (3 points) fn → f uniformlyon [a, b]

2. Elementary Topology in <n: De-fine the following terms:

(a) (3 points) S is an open set in<n

(b) (3 points) S is an closed set in<n

(c) (3 points) | x | for x ∈ <n

(d) (3 points) The boundary of aset S in <n

3. Integration Theory:f : [a, b] → < is a bounded func-tion on the finite interval [a, b]. De-fine the following concepts:

(a) (3 points) P is a partition of[a, b]

(b) (3 Points) P′

is a refinementof the partition P of [a, b]

(c) (3 Points) The Upper sum S(P, f)of the bounded function f on[a, b] for partition P .

(d) (3 Points) The Lower sum S(P, f)of the bounded function f on[a, b] for partition P .

(e) (3 Points) The Upper Integralof f on [a, b]


(f) (3 Points) The Lower Integralof f on [a, b]

(g) (3 Points) The integral of f on[a, b]

4. Series:Let

∑∞n0an denote an infinite se-

ries: Define the following concepts:

(a) (3 Points) The nth partial sumof the series

(b) (3 Points) The convergence ofthe series

5. TheoremsState precisely the following math-ematical theorems or axioms:

(a) (5 Points) The Cauchy SchwartzInequality

(b) (5 Points) The Fundamental The-orem of Calculus

(c) (5 Points) Taylor’s Theorem withRemainder

Part 2: Short Answer (50 Points)Answer the following questions. If theanswer is YES or NO, you MUST giveus the reason why; ( e.g. the completestatement of a relevant theorem, a coun-terexample etc.)

1. Let f : [a, b] → < for the finiteinterval [a, b].

(a) (5 Points) If f is not continu-ous on [a, b], does that neces-sarily imply f is not integrableon [a, b]?

(b) (5 Points) If f is continuouson [a, b], does that necessarilyimply that f is integrable on[a, b]?

(c) (5 Points) If f is bounded on[a, b], does that necessarily im-ply that f is integrable on [a, b]?

(d) (5 Points) If f2 is integrableon [a, b], does that necessarilyimply that f is integrable on[a, b]?

2. Let∑an denote an infinite series

(a) (5 Points) If limn→∞ an = 0,does that necessarily imply thatthe series

∑an converges?

(b) (5 Points) If limn→∞an+1

an=

1, does that necessarily implythat the series

∑an diverges?

3. Let f(x) =∑anx

n denote a powerseries and R denote its radius ofconvergence.

(a) (5 Points) Let R > 0. If theinterval [a, b] is strictly containedin the interval (−R,R), is itnecessarily true that the powerseries

∑anx

n converges uni-formly in [a, b]?

(b) (5 Points) Is it possible for theradius of convergence of thef′

series to be different fromR?

4. Let [a, b] be a finite interval and letf : [a, b] → < denote a functionwith derivatives of all orders. LetR be the radius of convergence ofthe Taylor series of f .

(a) (5 Points) Is it possible for fto have a Taylor series expan-sion with R = 0?

(b) (5 Points) Suppose f is definedon the interval [−3, 3] and ffails to be continuous at the points−1.7 and 2.2. Is it possible forR to be 2.5?

Part 3: Calculational Exercises (56 Points)Provide complete solutions with all ap-propriate detail to the following compu-tational exercises.

1. (14 Points) Let

fn(x) =sin(nx)

nx, x 6= 0

= 1, x = 0

Determine whether the sequence fnconverges uniformly on the follow-ing intervals:

(a) [1, 6]

(b) [−1, 1]

2. (14 Points) Determine whether ornot the function

B-4. EXAMS VERSION B 391

f(x, y) =x3 tan(x (y4))

x4 + y4

is continuous at the point (0, 0).

3. (14 Points) Let

f(x) =

∫ x

0

1

1 + t8dt

(a) Find f′(x)

(b) Express f(.5) without using in-tegrals.

4. (14 Points) Find the Taylor seriesexpansion of f(x) = tan−1(x) aboutbase point 0 and determine its ra-dius of convergence.

Part 4: Proofs (34 Points)Provide a careful proofs of the followingpropositions. You will be graded on themathematical correctness of your argu-ments as well as your use of language,syntax and organization in the proof.

1. (17 points)Let [a, b] be a finite interval of thereal line with fn denoting a se-quence of real-valued functions, fn :[a, b] → <. Assume fn → 0 uni-formly on [a, b] and the functiong : [a, b]→ < is bounded on [a, b].Then gfn → 0 uniformly on [a, b]

2. (17 points)Let f : R → R be continuous onthe finite interval [a, b]. If

∫ xaf(x)dx =

0 for all x ∈ [a, b], then f(x) = 0for all x ∈ [a, b].

B-4 Sample Exams Version B

B-4.1 Exam 1B



1. (15 Points) Carefully define and dis-cuss the integral of a bounded func-tion f : [a, b] → R, where [a, b]is a closed and finite interval. Youwill need to explain in detail all rel-evant symbols and terms.


1. (7 Points) Let [a, b] be a closed andfinite interval and assume f, g : [a, b]→R are not integrable on [a, b]. Is itnecessarily true that f + g is notintegrable on [a, b]?


1. (17 Points) Consider f(x) = x2 onthe interval [0, 1]. Let Pn be theuniformly spaced partition of [0, 1]consisting of n equal subintervals.Find SPn − SPn and show that

SPn − SPn → 0 as n→∞.

Does this prove that f(x) = x2 isintegrable on [0, 1]? Explain youranswer carefully.

2. Determine whether or not the fol-lowing functions are integrable on[0, 1] giving full reasons for youranswers.

(a) (7 points)

f(x) = x2 − 3x + 7,

(b) (7 points)

f(x) =π

2, 0 ≤ x < .25

= 1.0, x = .25

= .6, .25 < x < .75

= .17, x = .75


= .038, .75 < x ≤ 1.0

(c) (7 points)

f(x) = 2, x ∈ Q= −9, x ∈ I

whereQ and I indicate the ra-tional and irrational numbersrespectively.

(d) Differentiate the function f :[1, 10]→ R defined by,

i. (3 points)

f(x) =∫ x

1

(1 + t2)−2 dt

ii. (3 points)

f(x) =∫ x2

1

(2 + t3)−4 dt


1. (17 Points)Proposition: Let [a, b] be a finiteinterval and let f : [a, b] → R benon-negative on [a, b], i.e. f(x) ≥0, a ≤ x ≤ b. Then if ∃ x0, a <x0 < b 3 f(x0) > 0, this impliesthat

∫ baf(x) dx > 0.

2. (17 Points)Carefully state and prove a theo-rem for calculating

d

dx

(∫ u(x)

a

f(t) dt

)

B-4.2 Exam 2B

Instructions:This is a closed book and closed notes test.You will need to give us all the details of your

arguments as that is the only way we can de-cide if partial credit is warranted.

Part 1: Definitions (18 Points)Carefully and precisely define the fol-lowing mathematical phrases:

1. (6 Points)

∞∑n=0

an = S.

In the phrases below, I is a finite inter-val, fn is a sequence of real valuedfunctions defined on I and f is a realvalued function defined on I .

1. (3 Points) f is uniformly continu-ous on I .

2. (3 Points)

fn(x) → f(x) pointwise on I.

3. (3 Points)

fn(x) → f(x) uniformly on I.

4. (3 Points)

∞∑n=0

fn(x)

converges uniformly on I to the func-tion f(x).

Part 2: Theorems (12 Points)Carefully and precisely state the fol-lowing theorems:

1. (3 Points) The Comparison Test2. (3 Points) The Limit Comparison

Test3. (3 Points) The Ratio Test4. (3 Points) The Weierstrass K Test


B-4. EXAMS VERSION B 393

1. (5 Points) Let I be a finite intervaland fn a sequence of real valuedfunctions defined on I . If fn →f uniformly on I , is it necessarilytrue that fn → f pointwise on I?

2. (5 Points) If an → 1 is it necessar-ily true that

∑∞n=0 an converges?

For each of the following series, tellwhether or not it is absolutely conver-gent, conditionally convergent or di-vergent,giving full reasons for your answers.

1. (6 Points)

∞∑n=0

4n+1 + 5n

5n+1 + n2 + 5

2. (6 Points)

∞∑n=3

(−1)n(n2)

n3 + n− 2

3. (6 Points)

∞∑n=2

ln(1

n2)

Show all your work on the exercises be-low. You may use a calculator if youwish.

1. (15 Points) Consider for all n ≥ 1,

fn(x) =n2(x6)

n+ n2(x6)

Discuss the pointwise and uniformconvergence of fn on the intervals:

(a) [−1, 1]

(b) [ 12 , 4]


1. (13 Points)If fn → f and gn → g uniformly

on the interval I , then fn + 3gn →f + 3g uniformly on I .

2. (14 Points)If∑an converges absolutely, then∑ an

1+(an)2 converges absolutely.

B-4.3 Exam 3BInstructions:This is a closed book and closed notes test.You will need to give me all the details of yourarguments as that is the only way I can decideif partial credit is warranted.

Part 1: Definitions (12 Points):Let fn∞n=1 be a sequence of functionsand f be a function defined on the finiteinterval [a, b]. Define the following con-cepts:

1. fn → f pointwise on [a, b].

2. fn → f uniformly on [a, b].

3.∑∞

1 fn(x) converges pointwise tof on [a, b].

4.∑∞

1 fn(x) converges uniformlyto f on [a, b].

Part 2: Theorems (16 Points):Let fn∞n=1 be a sequence of functionsand f be a function defined on the fi-nite interval [a, b]. State carefully thefollowing theorems.

1. (6 Points) The interchange of limitand integration for

(a) For fn → f

(b) For∑∞

1 fn(x) = f(x)

2. (10 points) The interchange of limitand differentiation for

(a) For fn → f

(b) For∑∞

1 fn(x) = f(x)

Part 3: Short Answer: (20 Points):Let fn∞n=1 be a sequence of functionsand f be a function defined on the finiteinterval [a, b].

1. (7 Points) If fn → f uniformly on[a, b], each fn continuous on [a, b],is it necessarily true that f is con-tinuous on [a, b]?


2. (7 Points) If fn → f on [a, b] andlimn

∫ bafn(x)dx =

∫ baf(x)dx,

does it necessarily follow that fn →f uniformly on [a, b]?

3. (6 Points) If∑∞

1 an(x− x0)n hasradius of convergence R > 0, is itnecessarily true that the series con-verges at x0 +R?

Part 4: Calculational: (24 Points):

1. (8 Points) Compute

lim(x,y)→(0,0)

tan(√x2y2 + x4y4)

sin(|xy|)

if it exists.

2. (8 Points) Sum the series∑∞

1xn−1

n+1

3. (8 Points) Find the Taylor series ex-pansion about 0 for f(x) =

∫ x0e−t

2

dt

Part 5: Proofs (28 Points):Provide careful proofs of the followingpropositions. You will be graded on themathematical correctness of your argu-ments as well as your use of language,syntax and organization in the proof.

1. (14 Points) Provelim(x,y)→(0,0)

√x+ y + 1 = 1.

2. (14 Points) Let f(x) =∑∞

1 nxn.Provide all the arguments and de-tails that prove that for x ∈ (−1, 1)f′(x) =

∑∞1 n2xn−1.

Part XII

Appendix: Linear AnalysisExaminations

395

Appendix C

Linear Analysis Study Guide

Presented below are study guides and coursedescription material from some of the times wehave taught the graduate course on linear anal-ysis here, MTHSC 821 Another source of in-formation is our web-based discussion boardmaterial. The MTHSC 821 discussion boardcan be accessed either through my web pageby following the links or by the direct routehttp://www.ces.math.clemson.edu/discus.On that page you will see exercises, workedout problems and so forth in an easy to followformat.

C-1 Course StructureThe intent of this course is to develop yourability to think abstractly using a standard ele-ment of the practicing analyst’s toolkit–abstractspaces. In addition, we will continue to de-velop your abilities to read and write good math-ematics at the professional level.

C-2 Sample Exams Version A

C-2.1 Exam 1AInstructions:This is a closed book and closed notes test.You will need to give all the details of yourarguments as that is the only way partial creditmay be warranted.

Part 1: Definitions (30 Points)Carefully and precisely define the fol-lowing mathematical concepts:

1. (6 Points) Let X be a set.

(a) (3 Points) d : X ×X → < isa metric on X .

(b) (3 Points) d : X → < is anorm on X .


(a) (3 Points) X is a vector spaceover the field of real numbers.

(b) (3 Points) X is a metric spaceover the field of real numbers.

(c) (3 Points)X is a normed spaceover the field of real numbers.


(a) (3 Points)X is a separable met-ric space over the field of realnumbers.

(b) (3 Points) xn is a CauchySequence of elements in the normedspace X over <.

(c) (3 Points)X is a complete met-ric space over the field of realnumbers.

4. (6 Points) LetX be a normed spaceset and M be a subset of X .

(a) (3 Points)M is a compact sub-set of X .

(b) (3 Points) M is a subspace ofX properly contained inX andY is the normed space X/Mwith the usual induced norm.

Part 2: Short Answer (44 Points)There are several types of questions here:

True or False: You must determine whetheror not these statements are true. Ifthe statement is true, you MUST

397

398 APPENDIX C. LINEAR ANALYSIS I

give us the reason why it is true;if the statement is false, you mustgive us a counterexample.

Discussion: A careful discussion and rea-soned answer to the given questionis required.

1. (3 Points) LetX be a normed spaceand assume that the setx ∈ X | || x ||≤ 1 is compact.Is it possible for X to be finite di-mensional?

2. (3 Points) LetX be a normed spaceand let Y be a finite dimensionalsubspace of X that is not closed.Is this possible?

3. (8 Points) Let X be the set of allcontinuous functions defined on thefinite interval [a, b]. Describe a normonX which makeX complete anda norm on X which makes X notcomplete.

4. (3 Points) Is there a procedure whichenables us to complete an arbitrarymetric space?

5. (3 Points) Is there a procedure whichenables us to complete an arbitrarynormed space?

6. (6 Points) Give an example of aninfinite dimensional metric space whichis not separable.

7. (6 Points) Is necessarily true thatthe metric in an arbitrary metric spaceX comes from a norm on X?

8. (12 Points) Let X = <7. Let L =< x1, x2, x3, x4, x5, x6, x7 >

′∈<7 | x1 + x5 + 23.0x7 = 0.(a) (2 Points) Show L is a vector

space.(b) (2 Points) What is the dimen-

sion of L?(c) (2 Points) What is the dimen-

sion of X/L?(d) (2 Points) What are the elements

of X/L?(e) (2 Points) Is L closed?(f) (2 Points) Is L compact?


1. ( 13 Points) Let X be the set ofall continuous functions defined onthe finite closed interval [a, b] withthe maximum metric. Let φ be agiven element ofX and define TX →X by

(T (x))(t) =

∫ t

a

φ(s)x(s) ds

If xn is a Cauchy sequence inX ,prove that there exists an element yin X such that T (xn) → y.

2. ( 13 Points) Let X be the set ofall continuous functions defined onthe finite closed interval [0, 1] withthe maximum metric. Let S be theset of functions

S =et, e2t, e3t

Prove that S is a linearly indepen-dent set in X .

C-2.2 Exam 2A

Instructions:This is a closed book and closed notes test.You will need to give all the details of yourarguments as that is the only way partial creditmay be warranted.

Part 1: Definitions and Short Answers(34 Points)

Carefully and precisely define the fol-lowing mathematical concepts and an-swer the short questions.

1. (6 Points)

(a) Precisely define the meaningof a metric.

C-2. EXAMS VERSION A 399

(b) Precisely define the metric spaceX over the field of real num-bers.

(c) Is it required that X be a vec-tor space?

2. (10 Points)

(a) Precisely define the meaningof a norm.

(b) Precisely define the NormedSpace X over the real num-bers.

(c) Is it required that X be a vec-tor space?

(d) Define the metric onX inducedby the norm.

(e) Is it true that all metrics can bederived from a norm?

3. (10 Points)

(a) Precisely define the meaningof an inner product.

(b) Precisely define the inner prod-uct spaceX over the real num-bers.

(c) Define the norm onX inducedby the inner product.

(d) Define the metric onX inducedby the inner product.

(e) State the Schwartz Inequal-ity for this inner product space.

4. (8 Points)

(a) Let X be a vector space. Pre-cisely define the meaning ofthe algebraic dual space of X ,X∗.

(b) LetX be a normed space. Pre-cisely define the meaning ofthe dual space of X , X

′.

(c) Precisely define the norm ofan element of X

′.

(d) LetX be a normed space. Pre-cisely define the meaning ofthe dual space of X

′, X

′′.

Part 2: Short Answer (30 Points)A careful discussion and reasoned an-swer to the given question is required.

1. (15 Points) Characterize the boundedlinear functionals on `1.

2. (15 Points) Characterize the boundedlinear functionals on c0.


1. ( 18 Points) Let X be the set ofall continuous functions defined onthe finite closed interval [a, b] withthe maximum metric. Let φ be agiven element of X that satisfiesfor all positive integers n

|| φn || ≤ || φ ||n

Define for all positive integers nTn X → X by

(Tn(x))(t) =

∫ t

a

φn(s)x(s) ds

Prove that the sequence of linearoperators Tn converges to the zeroelement in the space of all boundedlinear operators from X to X .

2. ( 18 Points) Assume for all positiveintegers n,

xn = ξn1 , ξn2 , . . . , ∈ `1

and

x = ξ1, ξ2, . . . , ∈ `1

satisfy

f(xn) → f(x), ∀f ∈ (`1)′

Prove that for all positive integersk

ξnk → ξk


C-2.3 Exam 3AInstructions:This is a closed book and closed notes test.You will need to give all the details of yourarguments as that is the only way partial creditmay be warranted.



1. (4 Points) Define what it means fora inner product spaceX to be com-plete.

2. (15 Points) Completion (No DetailsHere–Just Sketch!!)

(a) (5 Points) State, without proof,the basic ideas behind complet-ing a metric space.

(b) (5 Points) State, without proof,the extra things that need tobe done to complete a normedspace.

(c) (5 Points) State, without proof,the extra things that need tobe done to complete an innerproduct space.

3. (16 Points)

(a) (4 Points) Define what it meansfor the set M to be orthonor-mal in the inner product spaceX .

(b) (4 Points) Define what it meansfor M to be a total orthonor-mal set in the inner product spaceX .

(c) (4 Points) State Bessel’s Inequal-ity.

(d) (4 Points) Discuss the condi-tions under which Parseval’s Re-lation holds.


1. (5 Points) Let x be an element ofthe Hilbert space H . Let M be the

orthonormal sequence (ei) in H .Consider the sum

∑∞i=0 < x, ei >

ei. Is it necessarily true that thissum converges in H?

2. (5 Points) Let x be an element ofthe Hilbert space H . Let M be theorthonormal sequence in H . Con-sider the sum

∑∞i=0 < x, ei >

ei. Is it necessarily that if this sumconverges, the sum is x?

3. (5 Points) Let Y be a subspace ofthe Hilbert space H . Is it necessar-ily true that

H = Y ⊕ Y ⊥ ?

4. (5 Points) Is it necessarily true thatan orthonormal set M in an innerproduct space X is linearly inde-pendent?

5. (5 Points) If M is an uncountableorthonormal set in an inner prod-uct space X , is it possible that <x,m > 6= 0 for all m in M?

6. (5 Points) If M is a total subsetin the inner product space X , is itnecessarily true that M⊥ = 0?

7. (5 Points) Let M be a nonempty,convex and complete subset of theinner product space X . Let x bean element of X not in M . Is itpossible for the minimum distancefrom x toM to be achieved by twodifferent points of M?


1. ( 12 Points) Let M be a total or-thonormal set in the Hilbert spaceH . Assume x inH satisfies< x,m >= 0 for all m in M . Prove thatx = 0.

2. ( 18 Points) LetX be the inner prod-uct space C[0, 1] with the usual L2

inner product. Recall that this means

C-3. EXAMS VERSION B 401

|| x ||2 =

√∫ 1

0

| x(s) |2 ds

Let f and φ be chosen arbitrarilyfromX and define the operator T :X → X by and

(T (x))(t) =

< x, f > +

∫ t

0

x(s)φ(s)ds

Prove that T is a linear operatorand that

(a) for all t in [0, 1],

| (T (x))(t) |≤|| x ||2 (|| f ||2 + || φ ||2).

(b)

|| T ||≤|| f ||2 + || φ ||2

C-3 Sample Exams Version B

C-3.1 Exam 1BInstructions:This is a closed book and closed notes test.You will need to give all the details of yourarguments as that is the only way partial creditmay be warranted.

Part 1: Definitions (33 Points)Carefully and precisely define the fol-lowing mathematical concepts:


(a) (2 Points) X′, the closure of

X .(b) (2 Points) X is a closed set.(c) (2 Points) X is an open set.

2. (12 Points) LetX be a normed space.

(a) (3 Points) M contained in Xis dense in X .

(b) (3 Points)M is a compact sub-set of X .

(c) (3 Points)M is a Hamel Basisfor X .

(d) (3 Points)M is a Schauder Ba-sis for X .

3. (15 Points)

(a) (3 Points) X is a vector spaceover the field of real numbers.

(b) (3 Points) X is a metric spaceover the field of real numbers.

(c) (3 Points)X is a normed spaceover the field of real numbers.

(d) (3 Points)X is a complete met-ric space over the field of realnumbers.

(e) (3 Points) X is a completenormed space over the field ofreal numbers.

Part 2: Short Answer (43 Points)There are several types of questions here:

True or False: You must determine whetheror not these statements are true. Ifthe statement is true, you MUSTgive us the reason why it is true;if the statement is false, you mustgive us a counterexample.

Discussion: A careful discussion and rea-soned answer to the given questionis required.

1. (9 Points) Let X be a an incom-plete metric space. Discuss the pro-cess by which we can complete X .

2. (5 Points) Let X be the set of allcontinuous functions defined on theinterval [0, 1] with the maximum norm.Is the setx ∈ X | || x || ≤ 1 compact inX?

3. (5 Points) LetX be a normed spaceand let Y be a subspace of X thatis closed. Does Y have to be finitedimensional?

4. (9 Points) Let Y be the set of allcontinuous functions defined on thefinite interval [a, b] such that x(a) =x(b) and letX be the set of all con-tinuous functions defined on [a, b].

(a) Show Y is a subspace of X .


(b) Is Y complete with respect tothe maximum norm?

5. (9 Points) Let Y be the set of allcontinuous functions defined on thefinite interval [a, b] such that x(a) =2 and let X be the set of all contin-uous functions defined on [a, b].

(a) Is Y a subspace of X?

(b) Is Y a closed set relative to themaximum norm?

6. (6 Points) Give an example of aninfinite dimensional metric space whichis separable.


1. ( 12 Points) Let X be the set ofall continuous functions defined onthe finite closed interval [a, b] withthe maximum metric. Let φ be agiven element of X and define T :X → X by

(T (x))(t) = x(t) +

∫ t

a

φ(s)x(s) ds

If xn is a Cauchy sequence inX ,prove that T (xn) is also a Cauchysequence.

2. ( 12 Points) Let A denote an arbi-trary n × n matrix over the reals.Let X be the set of all such matri-ces endowed with the norm:

|| A ||2 =

n∑i=1

n∑j=1

A2ij

Prove

|| A2 || ≤ || A ||2

C-3.2 Exam 2BInstructions:This is a closed book and closed notes test.You will need to give all the details of yourarguments as that is the only way partial creditmay be warranted.



1. (16 Points)

(a) (8 Points) Precisely define whatit means for f : [0, 1] → <to be of bounded variation.

(b) (8 Points) Let f, g : [0, 1] →< be continuous functions on[0, 1]. Precisely define the mean-ing of the Riemann-Stieljes in-tegral

∫ 1

0f dg.

2. (6 Points)

(a) (3 Points) Precisely define whatit means for the set X to be aninner product space over thereal numbers.

(b) (3 Points) State the SchwartzInequality for this inner prod-uct space.

3. (6 Points)

(a) (3 Points) Let X be a vectorspace. Precisely define what itmeans forX to be algebraicallyreflexive.

(b) (3 Points) Let X be a normedspace. Precisely define what itmeans for X to be reflexive inthe context of bounded linearfunctionals.

4. (6 Points) State the Riesz’s Lemma.


1. (4 Points) Is it necessarily true thatevery norm arises from an inner prod-uct?


2. (8 Points) For the following prob-lem, you must give examples of var-ious types of spaces; no space shouldbe used as an answer twice.

(a) (2 Points) Give an example ofa finite dimensional normed space.

(b) (2 Points) Give an example ofan infinite dimensional normedspace.

(c) (2 Points) Give an example ofa finite dimensional inner prod-uct space.

(d) (2 Points) Give an example ofan infinite dimensional innerproduct space.

3. (8 Points) For the following prob-lem, you must give examples of var-ious types of operators. Let X andY be the infinite dimensional normedspaces and let T : X → Y be anoperator. You will have to tell mewhat X and Y are for your exam-ple.

(a) (2 Points) Give an example ofT with ker(T) equal to 0.

(b) (2 Points) Give an example ofT with ker(T) not equal to 0.

(c) (2 Points) Give an example ofT that is not bounded.

(d) (2 Points) Give an example ofT that is not linear.

4. (4 Points) Is it necessarily true thatproper subspaces of any normed spaceare finite dimensional?

5. (10 Points)

(a) (4 Points) Characterize(`p)

′), p ≥ 1.

(b) (6 Points) Characterize(C[0, 1])

′).


1. ( 16 Points) Let X be the set ofall continuous functions defined on

the finite closed interval [a, b] withthe maximum metric. Let φ be acontinuous real-valued function whosedomain is all of < that satisfies forsome positive number K:

| φ(u)− φ(v) |≤ K | u− v | .

Define the (possibly nonlinear op-erator T ) by T : X → X by

(T (x))(t) =

∫ t

a

φ(x(s)) ds

Prove using the standard ε− δ for-malism that the operator T is con-tinuous as a mapping fromX toX .

2. ( 16 Points) Let X be the normedspace `1 with the usual norm. Thenany x in X is a sequence

x = ξ1, ξ2, . . . ,

with∑∞i=1 | ξi |< ∞.

Define the operator T : `1 → `2

by

T (x) =

∞∑i=1

| ξi |2 .

Prove that T is a well-defined lin-ear operator and that || T ||= 1.

C-3.3 Final BInstructions:This is a closed book and closed notes test.You will need to give all the details of yourarguments as that is the only way partial creditmay be warranted.




1. (5 Points) A metric space X .

2. (5 Points) A normed space X .

3. (5 Points) An inner product spaceX .

4. (5 Points) A complete inner prod-uct space X .

5. (5 Points) A function of boundedvariation on the interval [0, 1].

6. (5 Points) The Riemann–Stieljes In-tegral of functions in C[0, 1].

7. (5 Points) The algebraic dual of avector space X , X∗.

8. (5 Points) The continuous dual ofnormed space X , X

′.

9. (5 Points) An orthonormal sequencein an inner product space X .

10. (5 Points) A total orthonormal se-quence in a Hilbert space X .

11. (5 Points) A sesquilinear form.

12. (5 Points) The adjoint of an opera-tor in a Hilbert Space H .

Short Answer: (80 Points)

1. (4 Points) Is it necessarily true thata projection operator P defined forthe closed subspaceX of the HilbertspaceH satisfies the equationP (I−P )x = 0 for all x in H?

2. (6 Points) For each of the follow-ing spacesX , we know what spaceis isomorphic to its dual space X

′.

What is that space?

(a) lp, 1 < p < ∞(b) l1

(c) c

3. (15 Points) For each of the follow-ing spacesX , we know how to char-acterize a given linear functional inX′. Describe that characterization

carefully:

(a) f ∈ (lp)′, 1 < p < ∞

(b) f ∈ (l1)′

(c) f ∈ (c0)′

(d) f ∈ (C[0, 1])′

(e) f ∈ H ′ , whereH is any Hilbertspace.

4. (5 Points) LetH be a Hilbert spaceand T : H → H an operator. Is itnecessarily true that the adjoint ofT always exists?

5. (5 Points)Let H be a Hilbert spaceand T : H → H an operator. Is itpossible for the distinct operatorsS1 and S2 on H to both satisfy forall x, y in H

< Tx, y > = < x, S1y >

< Tx, y > = < x, S2y > ?

6. (15 Points:) We seek solutions xfromC[0, 1] to the differential equa-tion below where f is an arbitraryelement of C[0, 1].

x′(t) + 3(x′(t))2 + 2

∫ t

0

x(s) ds

= f(t)

Convert this mapping into the form

T : dom(T ) ⊂ X → X

by specifying the information be-low:

(a) A definition of T .(b) A definition of the domain of

T , dom(T ).

Finally, answer the following ques-tions:

(a) Is dom(T ) a vector space?(b) Is T a linear operator?(c) Is T a linear functional?

7. (15 Points:) Let Π be the partitionof the interval [0, 1] consisting ofP + 1 points given by

Π = t0 = 0 < t1 < . . . < tP = 1


Let A denote the 2x2 matrix

A =

[5 22 2

]For a given pair f and g fromC[01, ],let V (f, f, ti) ≡ V (f, g, i) be therow vector f(ti), g(ti). considerthe problem of finding f and g tominimize:

P∑i=0

V (f, g, i)A V (f, g, i)′

Convert this mapping into the form

T : dom(T ) ⊂ X → X



T , dom(T ).



8. (15 Points:) Let P be a photographwhich has been discretized into anarray P of size 640×480. Each el-ement Pij in the array can take onone of 256 discrete values whichrepresents a gray scale image of theoriginal photograph. We need tocompress this information into a ma-trix of size 320 × 240. Startingin the upper left hand corner of P ,we can compress the original im-age by looking at 2×2 sub-matricesof P and applying the compressionoperation we choose to convert eachof these sub-matrices into some scalar.The 1, 1 entry in the compresseddigitized image thus comes from ap-plying our compression operator to

the 2 × 2 sub-matrix coming formthe upper left corner

P11, P12

P21, P22

the 1, 2 entry in the compressed ar-ray comes from compressing the 2×2 sub-matrix

P13, P14

P23, P24

and so on. For example, given thearray

a, b

c, d

a primitive compression operationconsists of applying the averagingoperation given by computing thereal number

a+ b+ c+ d

4.0

and then mapping that real numberto the closest integer in the range0, 1, . . . , 255.Convert this averaging mapping ofa given discretized photograph Pin

X = <640×480256

(the subscript 256 reminds us thatthese are matrices that can only haveentries out of the integers 0, 1, . . . , 255)into the form:

T : dom(T ) ⊂ X → X



T , dom(T ).





1. ( 15 Points)Let X be the set of all 640 × 480matrices whose entries lie in the setof integers 0, 1, . . . , 255. Definethe function d : X×X → < by as-signing to each matrix p and q fromX the number d(p, q) defined by

d(p, q) =

max| pij − qij |

Prove that d is a metric on X .

2. ( 15 Points)Let H be a Hilbert space and T :H → H be an operator with non-trivial kernel K. Let P be the pro-jection operator onto K and Q bethe projection operator onto K⊥.Prove that

(a) H = K ⊕K⊥

(b) The operator S = TQ is in-vertible on H

3. ( 15 Points)Let (en) be a total orthonormal se-quence in the Hilbert space H . Leta given nonzero element f have theexpansion

f = a0 + a1e1 + a2e2 + . . .

=

∞∑n=0

anen

Prove that

f

|| f ||=

∞∑n=0

bnen

bn =an√∑∞n=0 a

2n

4. ( 15 Points)Let (fn) be the sequence of func-tions in C[0, 1] defined by fn(t) =tn for all t in the interval [0, 1].

(a) Prove that fn converges to thefunction f defined by f(t) = 0for all t in [0, 1] when C[0, 1]is endowed with the L2 innerproduct.

(b) Prove that fn does not convergeto f when we simply use point-wise convergence on the inter-val [0, 1].

basic analysis iv: measure theory and...

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