on the fundamental theorem of calculus - diva...

On the fundamental theorem of calculus

Jesper Singh

VT 2015Examensarbete, 30hpMagisterexamen i matematik, 240hpInstitutionen for matematik och matematisk statistik

Abstract

The Riemann integral have many flaws, some that becomes visible in the fundamental theoremof calculus. The main point of this essay is to introduce the gauge integral, and prove a much moresuitable version of that theorem.

Sammanfattning

Riemannintegralen har manga brister. Vissa utav dessa ser man i integralkalkylens huvudsats.Huvudmalet med denna uppsats ar att introducera gauge integralen och visa en mer lamplig versionav huvudsatsen.

Contents

AbstractSammanfattning1. Introduction 12. The Riemann integral 52.1. The formal definition and some examples 52.2. Properties of the Riemann integral and integrable functions 82.3. The fundamental theorem of calculus 163. The gauge integral 213.1. Gauges and δ-fine partitions 213.2. The gauge integral and the relation to the Riemann integral 253.3. A little taste of exceptional sets, null sets and null functions 293.4. Properties of the gauge integral 323.5. The fundamental theorems of calculus for the gauge integral 36Appendix A. An open letter to authors of calculus books 49Acknowledgements 55References 57

1. Introduction

The fundamental theorem of calculus is historically a major mathematical breakthrough, andis absolutely essential for evaluating integrals. In today’s modern society it is simply difficult toimagine a life without it. The history goes way back to sir Isaac Newton long before Riemannmade the first sound foundation of the Riemann integral itself. It is argued in [1] that the firstpublished ”proof” of the fundamental theorem of calculus is due to James Gregory. The booktitled Geometriae Pars Universalis that was published in 1668 by the Scottish mathematician andastronomer contains a first version of the fundamental theorem of calculus. Later in history it isconsidered that Newton himself discovered this theorem, even though that version was publishedat a later date. For further information on the history of the fundamental theorem of calculus werefer to [1].

The main point of this essay is the fundamental theorem of calculus, and in modern notationsit is stated as follows.

The fundamental theorem of calculus for the Riemann integral:

Part I Let f, F : [a, b]→ R be two functions that satisfies the following

a) f is Riemann integrable on [a, b];

b) F is continuous on [a, b], and

c) F ′(x) = f(x) for all x ∈ [a, b].

Then we have that ˆ b

a

f = F (b)− F (a) .

Part II Let f : [a, b]→ R be a continuous function, and define

G(x) =

ˆ x

a

f .

Then G is differentiable on [a, b], and G′(x) = f(x) for all x ∈ [a, b].

In Theorem 2.11 we give a proof of Part I, and in Corollary 2.1 we have Part II. The Riemannintegral have many flaws, some that becomes visible in the above theorem. The main points arethat there are too few Riemann integrable functions, and of course it is annoying with all theassumptions. As seen in Example 2.7 there are functions f such that there exists a primitivefunction F , i.e. F ′ = f on [0, 1], but f fails to be Riemann integrable. The French mathematicianArnaud Denjoy was interested in integrating such functions. In 1912, he published a paper [5]were he defined an integral handling problems like this. Later in 1914, Oskar Perron defined anintegral seemingly different to that of Denjoy ([11]). Both these definitions are very technical anddifficult to grasp. Surprisingly, these definitions are equivalent. It was not until 1957, that JaroslavKurzweil found out an elementary definition of this integral, and later developed by Henstock ([8]).As we shall see we only need to make some minor changes to the Riemann integral, and we getsomething that is surprisingly superior to the integrals of Riemann and Lebesgue ([2]). To putthings in perspective it can be interesting to know that a function f is Lebesgue integrable if,and only if, f and |f | are integrable in the sense of Denjoy, Perron, Kurzweil, and Henstock ([6]).There are many names for the things we love. This integral goes under names like for exampleDenjoy-Perron integral, Henstock–Kurzweil integral, and the generalized Riemann integral. Weshall simply call it the gauge integral. One can only speculate why this integral is still in the

1

shadows, and we shall not do that here. For people interested in the history of the Riemann, andLebesgue integral we refer to [7], and for those interested in the movement in trying to use thegauge integral in freshmen calculus we refer to an open letter attached in Appendix A.

Before we state the fundamental theorem of calculus for the gauge integral let us recall thedefinition of the Riemann integral, and what changes that needs to be made to get the gaugeintegral. A partition of the interval [a, b] ⊂ R, is a set P = {x0, x1, ..., xn−1, xn}, xj ∈ I, thatsatisfies

a = x0 < x1 < ... < xn−1 < xn = b .

The norm of a partition P is defined by

‖P‖ = max{x1 − x0, x2 − x1, ..., xn − xn−1} .

If a point ti ∈ [xi−1, xi] has been chosen from each subinterval for each i = 1, 2, ..., n, then thepoints ti are called tags of the subinterval [xi−1, xi], and the ordered pairs

P = {([xi−1, xi], ti)}ni=1

of subintervals and corresponding tags is called a tagged partition of [a, b]. Now, for a function

f : [a, b] → R and a tagged partition P = {([xi−1, xi], ti)}ni=1, then the Riemann sum of f

corresponding to P is defined by

S(f, P ) =

n∑i=1

f(ti)(xi − xi−1)

Furthermore, we say that the function f is Riemann integrable on [a, b] if ∃ A ∈ R for every ε > 0

there exists a number δε > 0 such that if P = {[xi−1, xi], ti}ni=1 is any tagged partition of [a, b]

with ‖P‖ < δε, then we have that

|S(f, P )−A| < ε .

We then say that ˆ b

a

f = A .

We shall for example see that all continuous functions are Riemann integrable (Theorem 2.8), aswell as all monotone functions (Theorem 2.9). The transition to gauge integral is surprisinglytrivial. We only need to replace the constant δε > 0 above with a function δε : [a, b]→ (0,∞).

Now to the fundamental theorem of calculus for the gauge integral. We now see clearly that thegenerality of this theorem is far greater than the corresponding one for the Riemann integral.

The fundamental theorem of calculus for the gauge integral:

Part I Let f, F : [a, b]→ R be two functions that satisfies the following

a) F is continuous on [a, b], and

b) F ′(x) = f(x) for all x ∈ [a, b]\E. Here E ⊂ [a, b] is a countable set or the empty set.Then we have that f is gauge integrable, and

ˆ b

a

f = F (b)− F (a) .

2

Part II Let f : [a, b]→ R be a function that is gauge integrable, and define

G(x) =

ˆ x

a

f .

Then G is continuous on [a, b], and G′(x) = f(x) almost everywhere w.r.t. the Lebesguemeasure.

Part I in the above theorem is Theorem 3.15, and Part II is Theorem 3.21.

Section 2 is all about the Riemann integral, and this section is based on [3], and section 3 isdevoted to the gauge integral and there we have used [2]. We shall follow a pure mathematicalpath, however the reader with an undergraduate background in mathematical analysis should haveno problem to follow the material that is presented. A good prerequisite and information aboutbasic analysis can be found in [10].

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2. The Riemann integral

The structure of this section about the Riemann integral is as follows. In subsection 2.1, weintroduce the Riemann integral, prove that it is a sound definition, and of course give some el-ementary examples. Later in subsection 2.2 we prove all basic properties that is needed to givea proof of the aim of this section, namely the fundamental theorem of calculus. This is done insubsection 2.3.

2.1. The formal definition and some examples. In this subsection we shall introduce theRiemann sum (Definition 2.2), and then the concept of Riemann integral (Definition 2.3). We thenprove that the definition of Riemann integral is well-posed (Theorem 2.1). We end this subsectionwith some examples.

Definition 2.1. If I = [a, b] is a closed bounded interval that belongs to R, then a partition ofI is a finite, ordered set P = {x0, x1, ..., xn−1, xn} of points in I such that a = x0 < x1 < ... <xn−1, xn = b. The points of P is used to divide I = [a, b] into non-overlapping subintervals

I1 = [x0, x1], I2 = [x1, x2], ..., In = [xn−1, xn].

More often than not we will denote the partition P by the notation P = {[xi−1, xi]}ni=1. The normof P is defined by the number

‖P‖ = max{x1 − x0, x2 − x1, ..., xn − xn−1}.

Hence the norm is the length of the largest subinterval of I. Many partitions have the samenorm, therefore the partition is not a function of the norm. If a point ti ∈ Ii has been chosenfrom each subinterval Ii = [xi−1, xi], for i = 1, 2, ..., n, then the points ti are called tags of thesubinterval Ii. A set of ordered pairs

P = {([xi−1, xi], ti)}ni=1

of subintervals and corresponding tags is called a tagged partition of I. The dot over the Pindicates that a tag has been selected from each subinterval. The tags can be chosen in infinitelymany ways, for instance we can choose the left endpoints, or the right endpoints, or the midpointsof the subintervals. Note and endpoint of a subinterval can be used as the tag for at most twoconsecutive subintervals. The norm of a tagged partition is defined in the same matter as for anordinary partition, so the tags wont depend on the choice of tags.

Definition 2.2. let f : I → R and let P = {([xi−1, xi], ti)}ni=1 is a tagged partition of I, then thesum

S(f, P ) =

n∑i=1

f(ti)(xi − xi−1),

is called the Riemann sum of f corresponding to P .

As we know from calculus the Riemann sum under a positive function f on [a, b] is the sum ofthe areas of n rectangles whose base are the subintervals Ii = [xi−1, xi] with heights f(ti).

Definition 2.3. A function f : I → R is said to be Riemann-integrable on I if there exists anumber A ∈ R such that for every ε > 0 there exists a number δε such that if P = {Ii, ti}ni=1 is

any tagged partition of I with ‖P‖ < δε, then

|S(f, P )−A| < ε,

and we write lim‖P‖→0 S(f, P ) = A. The set of all Riemann integrable functions on [a, b] will be

denoted by R[a, b].

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Notation 2.1. The interpretation of lim‖P‖→0 S(f, P ) = A is that the integral A is ”the limit” of

the Riemann sums S(f, P ) when the norm ‖P‖ → 0. This is different than the limit of a function

since S(f, P ) is not a function of ‖P‖. Many different partitions can share the same norm.

Instead of A we often use the calculus notation

A =

ˆ b

a

f or

ˆ b

a

f(x)dx.

The letter x is just a dummy variable and can be replaced with any other letter as long it doesnot become ambiguity.

Theorem 2.1. If f ∈ R[a, b], then the value of the integral is uniquely determined.

Proof. Assume that A′ and A′′ satisfies Definition 2.3 and let ε > 0 be given. Then there existsδ′ε/2 > 0 such that if P1 is any tagged partition with ‖P1‖ < δ′ε/2, we have

|S(f, P1)−A′| < ε/2.

Similarly there exists δ′′ε/2 > 0 such that if P2 is any tagged partition with ‖P2‖ < δ′′ε/2, then

|S(f, P2)−A′′| < ε/2.

Put δε = min{δ′ε/2, δ′′ε/2} > 0 and let P be a tagged partition with ‖P‖ < δε. Since ‖P‖ < δ′ε/2

and ‖P‖ < δ′′ε/2, we get

|S(f, P )−A′| < ε/2 and |S(f, P )−A′′| < ε/2.

Hence by the Triangle Inequality it follows that

|A′ −A′′| = |L′ − S(f, P )− S(f, P )− L′′|

≤ |L′ − S(f, P )|+ |S(f, P )− L′′| < ε/2 + ε/2 = ε.

Since ε > 0 is arbitrary, it follows that L′ = L′′. �

If Definition 2.3 is used to show that a function is Riemann integrable we must know (or guess)the value A of the integral, also we must construct a δε that will do for an arbitrary ε > 0. Thedetermination of the value A can be done by calculating Riemann sums and guessing the valueA, the construction of a sufficient δε will often give us a hard time. Later on we will prove sometheorems that will be very handy when it comes to determine if f ∈ R[a, b]. For now lets have alook at some instructive examples.

Example 2.1. If f : [a, b]→ R and f = c, where c is constant, then f ∈ R[a, b] and´ baf = c(b−a).

Let f(c) = c for all x ∈ [a, b]. If P = {([xi−1, xi], ti)}ni=1 is any tagged partition of [a, b], then

S(f, P ) =

n∑i=1

f(ti)(xi − xi−1) =

n∑i=1

c(xi − xi−1) = c(b− a).

Thus, for any ε > 0 we can choose δε with ease, lets put δε = 1 so that if ‖P‖ < 1, then

|S(f, P )−A| = |S(f, P )− c(b− a)| = 0 < ε.

Since ε > 0 is arbitrary, the conclusion is that f ∈ R[a, b] and´ baf = c(b− a). �

6

Example 2.2. Let g : [0, 1] → R be defined by g(x) = x for all x ∈ [0, 1]. Then g ∈ R[0, 1] with´ 10g = 1

2 . Now the ”trick” of using particular points as tags will be a good guess of the integralvalue. If Q = {Ii}ni=1 is any partition of [0, 1], choose the tag ti ∈ Ii = [xi−1, xi] to be the midpoint

qi = 12 (xi−1 + xi) we now have a tagged partition Q = {(Ii, qi)}ni=1. The Riemann sum becomes

S(g, Q) =

n∑i=1

g(qi)(xi − xi−1 =

n∑i=1

1

2(xi + xi−1)(xi − xi−1) =

n∑i=1

1

2(x2i − x2i−1) =

1

2(12 − 02) =

1

2.

Now let P = {(Ii, ti)}ni=1 be an arbitrary tagged partition of [0, 1] with ‖P‖ < δ so that xi−xi−1 < δ

for i = 1, ..., n. Also let Q have the same partition points, but choose the tag qi to be the midpointof the interval Ii. The tags of P can be anywhere between xi−1 and xi. Since ti, qi ∈ [xi−1, xi], wehave |ti − qi| < δ. Using the Triangle Inequality, we deduce

|S(g, P )− S(g, Q)| =

∣∣∣∣∣n∑i=1

ti(xi − xi−1)−n∑i=1

qi(xi − xi−1)

∣∣∣∣∣=

∣∣∣∣∣n∑i=1

(ti − qi)(xi − xi−1)

∣∣∣∣∣ ≤n∑i=1

|ti − qi| (xi − xi−1)

< δ

n∑i=1

(xi − xi−1) = δ(xn − x0) = δ(1− 0) = δ.

Since S(g, Q) = 12 , we infer that if P is any tagged partition with ‖P‖ < δ, then∣∣∣∣S(g, P )− 1

2

∣∣∣∣ < δ.

Therefore any δε ≤ ε will do. If we choose δε = ε and works backwards we conclude that g ∈ R[0, 1]

and´ 10g =´ 10xdx = 1

2 . �

Changing a function at a finite number of points does not affect its integrability nor the valueof the integral.

Theorem 2.2. Let g : [a, b] → R be integrable on [a, b] and if g(x) = f(x) except for a finite

number of points in [a, b], then f is Riemann integrable and´ baf =´ bag.

Proof. Lets prove the case of one exceptional point, then the extension to a finite number of pointscan be done by mathematical induction. Assume c ∈ [a, b] and f(x) = g(x) for all x 6= c and let

A =´ bag. For any tagged partition P the Riemann sums S(f, P ) and S(g, P ) are identical with

exception of at most two terms, that is the case when c = xi or c = xi−1. Therefore

|S(f, P )− S(g, P )|

=

∣∣∣∣∣n∑i=1

(f(xi)− g(xi)) (xi − xi−1)

∣∣∣∣∣ = |[f(c)− g(c)] (xi − xi−1)|

≤ (|f(c)|+ |g(c)|) (xi − xi−1) ≤ (|f(c)|+ |g(c)|) ‖P‖.

Now, given ε > 0, let δ1 > 0 satisfy δ1 < ε/2(|f(c)|+ |g(c)|), then |S(f, P )− S(g, P )| < ε/2. Since

g ∈ R[a, b] there exists a δ2 > 0 such that when ‖P‖ < δ2 we have |S(g, P ) − A| < ε/2. Now let

δ = min{δ1, δ2}. Then if ‖P‖ < δ we obtain

|S(f, P )−A| = |S(f, P )−S(g, P )+S(g, P )−A| ≤ |S(f, P )−S(g, P )|+|S(g, P )−A| < ε/2+ε/2 = ε.

Hence f ∈ R[a, b] with´ baf = A. �

7

The interpretation of Theorem 2.2 is that a singleton set has no length so the ”area” under thecurve is zero.

Example 2.3. Let g : [a, b]→ R be defined by

g(x) =

{c if x ∈ (a, b)

0 if x ∈ {a, b} .

Since the function f in Example 2.1 is integrable with integral c(b − a), and g = f except at the

finite points a or b, then g ∈ R[a, b] with´ bag = c(b− a) by Theorem 2.2. �

2.2. Properties of the Riemann integral and integrable functions. In this subsection westart by proving some basic properties of the Riemann integral, like the linearity in Theorem 2.3,that every Riemann integrable function is bounded on [a, b] (Theorem 2.4), but the most importantfor the rest of this section on the Riemann integral is that continuous functions defined on a compactinterval [a, b] is Riemann integrable (Theorem 2.8).

The obstacle involved when determining the value of the integral and of δε suggest that it wouldbe very nice and useful to have some general theorem. The next result enables us to performcertain algebraic manipulations of functions that belong to R[a, b].

Theorem 2.3. Suppose f, g ∈ R[a, b] and k ∈ R. Then

a) kf ∈ R[a, b] and´ bakf = k

´ baf ;

b) (f + g) ∈ R[a, b] and´ ba

(f + g) =´ baf +´ bag;

c) if f(x) ≤ g(x) for all x ∈ [a, b], then´ baf ≤´ bag.

Proof. a) For k = 0 there is nothing to prove so we assume k 6= 0. Let ε > 0 be given. If

P = {([xi−1, xi], ti)}ni=1 is a tagged partition of [a, b] and since f ∈ R[a, b] there exists a δε > 0such that ∣∣∣∣∣S(f, P )−

ˆ b

a

f

∣∣∣∣∣ < ε

|k|whenever ‖P‖ < δε.

Thus, if ‖P‖ < δε then∣∣∣∣∣S(kf, P )− kˆ b

a

f

∣∣∣∣∣ =

∣∣∣∣∣kS(f, P )− kˆ b

a

f

∣∣∣∣∣ = |k|

∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣ < |k| ε|k| = ε.

Hence lim‖P‖→0 S(kf, P ) = k´ baf , therefore kf ∈ R[a, b] and

´ bakf = k

´ baf by Definition 2.3.

b) Let ε > 0 be given. Since f, g ∈ R[a, b] there exists δ′ε/2 > 0 such that if P1 is any tagged

partition with ‖P1‖ < δ′ε/2, then ∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣ < ε

2,

and there exists δ′′ε/2 > 0 such that if P2 is any tagged partition with ‖P2‖ < δ′′ε/2, then∣∣∣∣∣S(g, P )−ˆ b

a

g

∣∣∣∣∣ < ε

2.

8

Let δε = min{δ′ε/2, δ′′ε/2} > 0 and let P be a tagged partition with ‖P‖ < δε. Since both ‖P‖ < δ′ε/2

and ‖P‖ < δ′′ε/2, we get∣∣∣∣∣S(f + g, P )−

(ˆ b

a

f +

ˆ b

a

g

)∣∣∣∣∣ =

∣∣∣∣∣S(f, P ) + S(g, P )−ˆ b

a

f −ˆ b

a

g

∣∣∣∣∣≤

∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣+

∣∣∣∣∣S(g, P )−ˆ b

a

g

∣∣∣∣∣ < ε

2+ε

2= ε.

Hence lim‖P‖→0 S(f + g, P ) =´ baf +´ bag, so (f + g) ∈ R[a, b] and

´ baf + g =

´ baf +´ bag by

Definition 2.3.

c) Let ε > 0 be given. Since f, g ∈ R[a, b] then for any tagged partition P with P < δε we have∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣ < ε

2and

∣∣∣∣∣S(g, P )−ˆ b

a

g

∣∣∣∣∣ < ε

2.

Removing the absolute values and using the fact that S(f, P ) ≤ S(g, P ) gives usˆ b

a

f − ε

2< S(f, P ) ≤ S(g, P ) <

ˆ b

a

g +ε

2,

adding ε/2 yields ˆ b

a

f < S(f, P ) +ε

2≤ S(g, P ) +

ε

2<

ˆ b

a

g + ε.

Since´ bafz´ bag + ε for every ε > 0 the conclusion is that

´ baf ≤´ bag. �

Now when some algebraic combinations been established lets look at certain demands on thefunction f to be required for the Riemann integral to exist. An unbounded function for sure is notRiemann integrable. Let B[a, b] denote the set of all bounded functions on [a, b].

Theorem 2.4. If f ∈ R[a, b], then f ∈ B[a, b].

Proof. The proof is by contradiction. Assume f /∈ B[a, b] and´ baf = A. Then there exists δ > 0

and ε > 0 such that if P is any tagged partition of [a, b] with ‖P‖ < δ, we have |S(f, P )−A| < ε,which gives

A− ε < S(f, P ) < A+ ε,

hence|S(f, P )| < |A+ ε| ≤ |A|+ ε. (2.1)

Now let Q = {[xi−1, xi]}ni=1 be a partition of [a, b] with ‖Q‖ < δ. Since |f | is unbounded on[a, b], there exists at least one subinterval in Q, say [xk−1, xk], where |f | is not bounded. If |f |where bounded on each [xi−1, xi] ∈ [a, b] by say Mi, then |f | would be bounded on [a, b] bymax{M1, ...,Mn}. The strategy now is to tag Q in such way that equation 2.1 is contradicted. Todo so tag Q by ti = xi for i 6= k and tk ∈ [xk−1, xk] such that

|f(tk)(xk − xk−1)| > |A|+ ε+

∣∣∣∣∣∣∑i 6=k


∣∣∣∣∣∣ ,this is always possible since f is not bounded, thus

|f(tk)(xk − xk−1)| −

∣∣∣∣∣∣∑i 6=k


∣∣∣∣∣∣ > |A|+ ε.

9

By the inequality form |A+B| ≥ |A| − |B| we find

|S(f, Q)| =

∣∣∣∣∣∣f(tk)(xk − xk−1) +∑i 6=k


∣∣∣∣∣∣≥ |f(tk)(xk − xk−1)| −

∣∣∣∣∣∣∑i 6=k


∣∣∣∣∣∣ > |A|+ ε.

Hence we have found a tagged partition of [a, b] which contradicts equation 2.1, therefore anunbounded function cannot be integrable. Since f ∈ R[a, b] the conclusion is that f ∈ B[a, b]. �

Note that there is no guarantee that a bounded function is Riemann integrable(it may or maynot be). Lets have a look at another important steppingstone that will be needed to establish theintegrability of several classes of functions such as step, monotone and continuous functions. Thisresult is known as the Cauchy criterion. The major gain by the Cauchy criterion is that it makesit possible to prove that a function is Riemann integrable without the need to know the integralsvalue. There is a price to be paid thou, we need to consider two Riemann sums instead of just one.

Theorem 2.5 (Cauchy criterion). A function f : [a, b] → R belongs to R[a, b] if and only if for

every ε > 0 there exists ηε > 0 such that if P and Q are any tagged partitions of [a, b] with ‖P‖ < ηεand ‖Q‖ < ηε, then

|S(f, P )− S(f, Q)| < ε.

Proof. Let f ∈ R[a, b] with´ baf = A, let ηε = δε/2 > 0 be such that if P , Q are tagged partitions

of [a, b] such that ‖P‖ < ηε and ‖Q‖ < ηε, then

|S(f, P )−A| < ε/2 and |S(f, Q)−A| < ε/2.

Hence

|S(f, P )− S(f, Q)| = |S(f, P )−A+A− S(f, Q)|

≤ |S(f, P )−A|+ |A− S(f, P )| < ε/2 + ε/2 = ε.

Conversely for each n ∈ N let δn > 0 be such that if P and Q be tagged partitions of [a, b] with

‖P‖ < δn and ‖Q‖ < δn, then

|S(f, P )− S(f, Q)| < 1/n.

We can assume that δn ≥ δn+1 for each n ∈ N, if not just substitute δn with δ′n = min{δ1, ..., δn}.For each n ∈ N, let Pn be a tagged partition of [a, b] with ‖Pn‖ < δn. Hence, if m > n then both

‖Pn‖ < δn and ‖ ˙Pm‖ < δn, therefore

|S(f, Pn)− S(f, ˙Pm)| < 1/n if m > n,

consequently {S(f, ˙Pm)}∞m=1 is a Cauchy sequence in R and therefore this sequence converges in

R to say A, thus sending m to the limit gives us limm→∞ S(f, ˙Pm) = A, and

|S(f, Pn)− L| ≤ 1/n for all n ∈ N.

To be convinced that´ baf = A, given ε > 0, let K ∈ N satisfy K > 2/ε. If Q is any tagged

partition of [a, b] with ‖Q‖ < δK , then

|S(f, Q)−A| = |S(f, Q)− S(f, ˙PK) + S(f, ˙PK)−A|

≤ |S(f, Q)− S(f, ˙PK)|+ |S(f, PK)−A| ≤ 1/K + 1/K = 2/K < ε.

Since ε > 0 is arbitrary, f ∈ R[a, b] and´ baf = A. �

10

Theorem 2.5 is a practical tool to determine if either f ∈ R[a, b] or f /∈ R[a, b].

Example 2.4. Let f : [0, 1]→ R be defined by

f(x) =

{1 if x ∈ Q ∩ [0, 1]

0 if x ∈ (R\Q) ∩ [0, 1] .

Then f /∈ R[a, b]. Hence we have to show that there exists ε0 > 0 such that for any δε0 > 0 there

exists partitions P and Q with ‖P‖ < ε0 and ‖Q‖ < ε0 such that

|S(f, P )− S(f, Q)| ≥ ε0.

In order to do so we choose ε0 = 1/2. If P is any partition with tags ti ∈ Q ∩ [0, 1] then S(f, P ) = 1,

on the other hand if Q is any tagged partition with tags tk ∈ (R\Q) ∩ [0, 1] then S(f, P ) = 0. Hence

|S(f, P )− S(f, Q)| = 1 > 1/2,

since P , Q are tagged partitions with arbitrary small norms, the conclusion is that f /∈ R[a, b]. �

The definition of the Riemann integral gives us two types of obstacles. The first one is thatthere is infinitely ways to choose tags ti. Secondly there is infinitely many partitions with normless than δε. Below we will establish a useful theorem that will ease us from some of the difficultieswhen proving that a function f is integrable, the theorem is called the squeeze theorem. The basicidea is that a function can be ”squeezed” between two functions known to be integrable. It wasthis idea that led the French mathematician Gaston Darboux to develop a different approach tointegration by introducing upper and lower integrals. For a fully treatment of the Darboux integralwe refer to [10].

Theorem 2.6. Let f : [a, b] → R. Then f ∈ R[a, b] if and only if for every ε > 0 there existsfunctions αε, ωε ∈ R[a, b] with

αε(x) ≤ f(x) ≤ ωε(x) for all x ∈ [a, b],

such that ˆ b

a

(ωε − αε) < ε.

Proof. Choose αε = ωε = f for all ε > 0, then´ ba

(ωε − αε) = 0 < ε.

Conversely, let ε > 0 be given. Since αε, ωε ∈ R[a, b], there exists δε > 0 such that if P is any

tagged partition with ‖P‖ < δε, then∣∣∣∣∣S(αε, P )−ˆ b

a

αε

∣∣∣∣∣ < ε and

∣∣∣∣∣S(ωε, P )−ˆ b

a

ωε

∣∣∣∣∣ .These inequalities givesˆ b

a

αε − ε < S(αε, P ) and S(ωε, P ) <

ˆ b

a

ωε + ε.

Since αε ≤ f ≤ ωε we have S(αε, P ) ≤ S(ωε, P ), henceˆ b

a

αε − ε < S(f, P ) <

ˆ b

a

ωε + ε.

If Q is another tagged partition with ‖Q‖ < δε, then we also haveˆ b

a

αε − ε < S(f, Q) <

ˆ b

a

ωε + ε,

11

by subtracting these two inequalities and using the assumtion´ ba

(ωε − αε) < ε , we conclude that

|S(f, P )− S(f, Q)| <

∣∣∣∣∣ˆ b

a

ωε + ε−

(ˆ b

a

αε − ε

)∣∣∣∣∣=

ˆ b

a

ωε −ˆ b

a

αε + 2ε =

ˆ b

a

(ωε − αε) + 2ε < ε+ 2ε = 3ε.

Since ε > 0 is arbitrary f ∈ R[a, b] by Theorem 2.5.�

Step functions are important in the nature of Riemann integrability, and will be used to provethat every continuous function over and interval [a, b] is also Riemann integrable over the sameinterval. First we prove a special case in the following lemma, then we prove the general statementin Theorem 2.7.

Lemma 2.1. If J ⊆ [a, b] having endpoints c < d and if ϕJ(x) = 1 for x ∈ J and ϕJ(x) = 0 for

x ∈ [a, b] \ J , then ϕJ ∈ R[a, b] and´ baϕJ = d− c.

Proof. With c = a and d = b it is just Example 2.1 all over again so assume a < c and d < b. IfP = {([xi−1, xi], ti)}ni=1 is any tagged partition of [a, b], then it is clear that

S(ϕJ , P ) =

n∑i=1

ϕJ(ti)(xi − xi−1)

= ϕJ(t1)(a− c) + ϕJ(t2)(d− c) + ϕJ(t3)(b− d) = ϕJ(t2)(d− c) = d− c.

Hence, for any ε > 0, we can choose δε = 1 so if ‖P‖ < δε, then

|S(ϕJ , P )− (d− c)| = 0 < ε .

Since ε > 0 is arbitrary, the conclusion is that ϕJ ∈ R[a, b] and´ baϕJ = d− c. �

Theorem 2.7. If ϕ : [a, b]→ R is a step function, then ϕ ∈ R[a, b].

Proof. Step functions like in Lemma 2.1 are called ”elementary step functions”, for further studiessee [2]. Any step function ϕ can be written as a linear combination of elementary step functions:

ϕ =

m∑j=1

kjϕJj ,

where Jj has endpoints cj < dj . Lemma 2.1 and Theorem 2.3 implies that ϕ ∈ R[a, b] and that

ˆ b

a

ϕ =

ˆ b

a

(k1ϕJ1 + ...+ kmϕJm) = k1

ˆ b

a

ϕJ1 + ...+ km

ˆ b

a

ϕJm

= k1(d1 − c1) + ...+ km(dm − cm) =m∑j=1

kj(dj − cj).

�

Some examples regarding step functions and the squeeze theorem would be appropriate.

Example 2.5. Let g : [0, 1]→ R be defined by

g(x) =

{2 if x ∈ [0, 1]

3 if x ∈ (1, 3] .

12

To show that g ∈ R[0, 3] using Definition 2.3 would be half complicated and require some ”tricks”,see [2] Example 7.1.4, page 202. However since

g =

m∑j=1

kjϕJj =

2∑j=1

kjϕJj ,

is a step function and therefore by Theorem 2.7 we haveˆ 1

0

g =

2∑j=1

kjϕJj = 2(1− 0) + 3(3− 1) = 2 + 6 = 8.

�

The next example combines step functions with the useful squeeze theorem.

Example 2.6. Let h(x) = x on [0, 1] and let Pn = {0, 1/n, 2/n, ..., (n − 1)/n, n/n = 1}. Defineαn : [0, 1]→ R by

αn(x) =

{h(k−1n

)= k−1

n , if x ∈ [k−1n , kn ), k = 1, 2, ..., n− 1

h(n−1n

)= n−1

n , if x ∈ [n−1n , 1], k = n .

Hence αn ≤ h on [0, 1/n), [1/n, 2/n), ..., [(n − 2)/n, (n − 1)/n), [(n − 1)/n, 1]. Similarly we defineωn : [0, 1]→ R by

ωn(x) =

{h(kn

)= k

n , if x ∈ [k−1n , kn ), k = 1, 2, ..., n− 1

h(nn ) = 1

)= n−1

n , if x ∈ [n−1n , 1], k = n .

Hence ωn ≥ h on [0, 1/n), [1/n, 2/n), ..., [(n − 2)/n, (n − 1)/n), [(n − 1)/n, 1]. By Theorem 2.7 wehaveˆ 1

0

αn =

n∑k=1

(k − 1)

n· 1

n=

1

n2

n∑k=1

k − 1 =1

n2· (0 + 1 + 2 + ...+ n− 1)

=1

n2· n(n− 1)

2=

1

2

(1− 1

n

).

Alsoˆ 1

0

ωn =

n∑k=1

k

n· 1

n=

1

n2

n∑k=1

k =1

n2· (1 + 2 + ...+ n− 1 + n)

=1

n2· n(n+ 1)

2=

1

2

(1 +

1

n

).

Thus we have αn(x) ≤ h(x) ≤ ωn(x) for x ∈ [0, 1] andˆ 1

0

ωn −ˆ 1

0

αn =

ˆ 1

0

(ωn − αn) =1

n.

Therefore for a given ε > 0 we can choose n ∈ N so large that 1n < ε, Theorem 2.6 tells us that

h ∈ R[0, 1] and since limn→∞∑nk=1 αn · 1/n = limn→∞

∑nk=1 ωn · 1/n = 1/2 we also know that´ 1

0h = 1/2 since

´ 10αn ≤

´ 10h ≤´ 10ωn. �

We are now equipped to prove that all continuous functions over [a, b] belongs to R[a, b], to doso we will use Theorem 2.6 and some properties about continuous functions. Let C[a, b] denote allcontinuous functions on [a, b].

Theorem 2.8. If f : [a, b]→ R and f ∈ C[a, b], then f ∈ R[a, b].

13

Proof. f is continuous on [a, b] and therefore f is uniformly continuous on [a, b]. For a furtherexplanation about uniformly continuity see [10]. Hence, for given ε > 0 there exists a δε such thatif u, v ∈ [a, b] and |u−v| < δε, then we have |f(u)−f(v)| < ε/(b−a). Let P = {Ii}ni=1 be a partitionsuch that ‖P‖ < δε. Since f ∈ C[a, b], there exists points u, v ∈ [a, b] such that f(u) ≤ f(x) ≤ f(v)see [10] Theorem 4.5, page 104. Let ui ∈ Ii be the point such that f(ui) ≤ f(x) for all x ∈ Ii, andlet vi be the point such that f(x) ≤ f(vi) for x ∈ Ii. Define a step function αε : [a, b]→ R by

αε(x) =

{f(ui) for x ∈ [xi−1, xi), i = 1, ..., n− 1

f(un) for x ∈ [xn−1, xn] .

Similarly define ωε : [a, b]→ R by

ωε(x) =

{f(vi) for x ∈ [xi−1, xi), i = 1, ..., n− 1

f(vn) for x ∈ [xn−1, xn] .

This gives usαε(x) ≤ f(x) ≤ ωε(x) for all x ∈ [a, b].

Hence

0 ≤ˆ b

a

(ωε − αε) =

n∑i=1

(f(vi)− f(ui)) (xi − xi−1)

<

n∑i=1

(ε

b− a

)(xi − xi−1) =

(ε

b− a

)(xn − xo) = ε.

Therefore it follows from Theorem 2.6 that f ∈ R[a, b]. �

Monotone functions are not always continuous over an interval [a, b], however they are alwaysintegrable on [a, b].

Theorem 2.9. If f : [a, b]→ R is monotone on [a, b], then f ∈ R[a, b].

Proof. Assume f(xi−1) ≤ f(xi) on [a, b] (the case when f is decreasing is handled similarly).Partition [a, b] into n subintervals Ik = [xk−1, xk] with equal lenght, thus xk − xk−1 = (b − a)/n,k = 1, ..., n. Since f is increasing on Ik, the minimum value is attained at the left endpoint xk−1and the maximum value is attained at the right endpoint xk. Define αε : [a, b]→ R by

αε(x) =

{f(xk−1) for x ∈ [xk−1, xk), k = 1, ..., n− 1

f(xn−1) for x ∈ [xn−1, xn] .

Similarly define ωε : [a, b]→ R by

ωε(x) =

{f(xk) for x ∈ [xk−1, xk), i = 1, ..., n− 1

f(xn) for x ∈ [xn−1, xn] .

Then we haveαε(x) ≤ f(x) ≤ ωε for all x ∈ [a, b].

Theorem 2.7 givesˆ b

a

αε =

n∑k=1

f(xk−1)(xk − xk−1) =b− an

(f(x0) + f(x1) + ...+ f(xn−1)) ,

and ˆ b

a

ωε =

n∑k=1

f(xk)(xk − xk−1) =b− an

(f(x1) + f(x2) + ...+ f(xn)) .

14

Subtracting and cancellation of terms yields

ˆ b

a

(ωε − αε) =

n∑k=1

(f(xk)− f(xk−1)) (xk − xk−1)

=b− an

(f(xn)− f(x0)) =b− an

(f(b)− f(a)) .

Hence for a given ε > 0, we choose n ∈ N so large that n > (b− a)(f(b)− f(a))/ε. Consequently´ ba

(ωε − αε) < ε and therefore f ∈ R[a, b] by Theorem 2.6. �

Lets return to arbitrary functions that belong to R[a, b]. The next result is quite obvious,nevertheless it requires a proof.

Theorem 2.10. Let f : [a, b]→ R and let c ∈ (a, b). Then f ∈ R[a, b] if and only if its restrictionf1 of f to [a, c] is in R[a, c] and f2 of f to [c, b] is in R[a, b]. In this caseˆ b

a

f =

ˆ c

a

f +

ˆ b

c

f.

Proof. Suppose that f1 ∈ R[a, c] with´ caf1 = A1 and f2 ∈ R[c, b] with

´ bcf1 = A2. Then for

given ε > 0 there exists δ′ > 0 such that if P1 is a tagged partitions of [a, c] with ‖P1‖ < δ′, then

|S(f1, P1)−A1| < ε/3. Also there exist δ′′ > 0 such that if P2 is any tagged partition of [c, b] with

‖P2‖ < δ′′, then |S(f2, P2)−A2| < ε/3. If M is a bound for |f |, define δε = min{δ′, δ′′, ε/6M} and

let Q be a tagged partition of [a, b] with ‖Q‖ < δ. We have to prove that

|S(f, Q)− (A1 +A2)| < ε.

There are two cases to consideri) If c ∈ Q and we split Q into a partition Q1 of [a, c] and a partition Q2 of [c, b]. Since S(f, Q) =

S(f, Q1) + S(f, Q2) and since ‖Q1‖ < δ′ and ‖Q2‖ < δ′′, we have

|S(f, Q)− (A1 +A2)| = |S(f, Q1) + S(f, Q2)− (A1 +A2)|

≤ |S(f, Q1)−A1|+ |S(f, Q2)−A2| <ε

3+ε

3=

2ε

3< ε.

ii) If c /∈ Q = {(Ik, tk)}mk=1, then there exists k ≤ m such that c ∈ (xk−1, xk). Just add c to Q and

let Q1 be the tagged partition of [a, c] defined by

Q1 = {(I1, t1), ..., (Ik−1, tk−1), ([xk−1, c], c)}.

similarly let Q2 be the tagged partition of [c, b] defined by

Q2 = {([c, xk], c), (Ik+1, tk+1), ..., (Im, tm)}.Subtracting many terms shows that

S(f, Q)− S(f, Q1)− S(f, Q2)

=

m∑k=1

f(tk)(xk − xk−1)−k−1∑k=1

f(tk)(xk − xk−1)−m∑

k=k−1

f(tk)(xk − xk−1)

= f(tk)(xk − xk−1)− f(c)(c− xk−1)− f(c)(xk − c)= f(tk)(xk − xk−1)− f(c)(xk − xk−1) = (f(tk)− f(c)) (xk − xk−1).

Also since f(tk)− f(c) ≤ 2M and xk − xk−1 < ε/6M it follows that

|S(f, Q)− S(f, Q1)− S(f, Q2)| ≤ 2M(xk − xk−1) < 2Mε

6M=ε

3.

15

Now since ‖Q1‖ < δε ≤ δ′ and ‖Q2‖ < δε ≤ δ′′, it follows that

|S(f, Q)− (A1 +A2)|

= |S(f, Q)− S(f, Q1)− S(f, Q2 + S(f, Q1) + S(f, Q2 − (A1 +A2)|

≤ |S(f, Q)− S(f, Q1)− S(f, Q2|+ |S(f, Q1 −A1|+ |S(f, Q2 −A2|

<ε

3+ε

3+ε

3= ε.

Since ε > 0 is arbitrary, we conclude that f ∈ R[a, b] and´ baf =´ caf +´ bcf .

Conversely, suppose that f ∈ R[a, b]. For given ε > 0, let ηε > 0, by Theorem 2.5 we have

|S(f, P )− S(f, Q| < ε,

for any tagged partitions P and Q with norm less than ηε. Let f1 be the restriction of f to [a, c].

Also let P1, Q1 be tagged partitions of [a, c] with ‖P1‖, ‖Q1‖ < ηε. By adding additional partition

points and tags from [c, b], we can extend P1 and Q1 to tagged partitions P and Q of [a, b] that

satisfy ‖P‖, ‖Q‖ < ηε. If we add the same additional points and tags in [c, b] for both P and Q,then

S(f1, P1)− S(f1, Q1) = S(f, P )− S(f, Q).

Since ‖P‖, ‖Q‖ < ηε, we have

|S(f1, P1)− S(f1, Q1)| < ε.

Therefore the restriction f1 of f to [a, c] belongs to R[a, c] by Theorem 2.5. A repetition of theargument above shows that the restriction f2 of f belongs to R[c, d], only difference is that thepartition we are ”borrowing” points from is [a, c], instead of [c, b]. Now since since f ∈ R[a, c] and

f ∈ R[c, b] has been established the equality´ caf +´ bcf =

´ baf follows from the first part of the

theorem. �

2.3. The fundamental theorem of calculus. This is the final part of the Riemann integral.We shall state and prove the fundamental theorem of calculus. Furthermore, we give one exampleto show some of its flaws.

Theorem 2.11. Let f, F : [a, b]→ R such that

a) F ∈ C[a, b];

b) F ′(x) = f(x) for all x ∈ [a, b];

c) f ∈ R[a, b].

Then ˆ b

a

f = F (b)− F (a).

Proof. Since f ∈ R[a, b], given ε > 0, there exists δε > 0 such that if P is any tagged partition

with ‖P‖ < δε, then ∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣ < ε.

Since F is continuous on [a, b] and if [xi−1, xi] ∈ P then the mean-value theorem applied to F on[xi−1, xi] implies that there exists ui ∈ (xi−1, xi), such that

F (xi)− F (xi−1) = F ′(ui) · (xi − xi−1) for i = 1, ..., n.16

For more information on the Mean-value theorem of differential caclulus see [10] Theorem 5.6, page134. Adding, using the fact that F ′(ui) = f(ui) and cancellation of terms gives

F (b)− F (a) =

n∑i=1

(F (xi)− F (xi−1)) =

n∑i=1

f(ui)(xi − xi−1).

Let Pu = {([xi−1, xi], ui)}ni=1, thus S(f, Pu) =∑ni=1 f(ui)(xi − xi−1), therefore∣∣∣∣∣S(f, Pu)−

ˆ b

a

f

∣∣∣∣∣ =

∣∣∣∣∣F (b)− F (a)−ˆ b

a

f

∣∣∣∣∣ < ε.

Since ε > 0 is arbitrary, we conclude that´ baf = F (b)− F (a). �

Even if F ′ exist for all x ∈ [a, b] it is not always true that f ∈ R[a, b].

Example 2.7. Let F : [0, 1]→ R be defined by

F (x) =

{x2 cos(1/x2) if x ∈ (0, 1]

0 if x = 0 .

Differentiating yields

F ′(x) =

{2x cos(1/x2) + (2/x) sin(1/x2) for x ∈ (0, 1],

0 for x = 0 .

For x ∈ (0, 1] the chain and product rules applies and for x = 0 we have∣∣∣∣F (x)− F (0)

x− 0− F ′(0)

∣∣∣∣ =

∣∣∣∣x2 cos(1/x2)

x

∣∣∣∣ =∣∣x cos(1/x2)

∣∣ ≤ x < ε.

Hence for |x − 0| < δ choose ε = δ, this proves that F ′(0) = 0. Therefore F is continuous anddifferentiable at every point x ∈ [0, 1], but since F ′ fails to be bounded on [0, 1] it follows thatF ′ /∈ R[0, 1]. So condition c) cannot be omitted in Theorem 2.11. In section 3 we will see thatf ∈ G[0, 1] and Theorem 3.15 will even provide us with a antiderivative of f in some sense. �

We now wish to differentiate an integral with a fluctuating upper limit, the following will beneeded.

Definition 2.4. If f ∈ R[a, b], then the function F : [a, b]→ R defined by

F (x) =

ˆ x

a

f for x ∈ [a, b],

is called the indefinite integral of f with base point a.

Definition 2.5. A function f : [a, b]→ R is said to be a Lipschitz function (or satisfy a Lipschitzcondition) on [a, b] if there exists a constant K > 0 such that

|f(x1)− f(x2)| ≤ K|x1 − x2| for all x1, x2 ∈ [a, b].

Every Lipschitz function is uniformly continuous.

Theorem 2.12. If f : [a, b] → R satisfies a Lipschitz condition, then f is uniformly continuouson [a, b].

Proof. Let f be as in Definition 2.5. For given ε > 0, we choose δ = ε/K. Now if |x1 − x2| < δ forall x1, x2 ∈ [a, b], then

|f(x1)− f(x2)| ≤ K|x1 − x2| < Kε

K= ε.

�17

The indefinite integral F of a Riemann integrable function f satisfies a Lipschitz condition andis therefore continuous.

Theorem 2.13. The indefinite integral F in Definition 2.4 is continuous on [a, b]. If |f(x)| ≤Mfor all x ∈ [a, b], then |F (x)− F (y)| ≤M |x− y| for all x, y ∈ [a, b].

Proof. Theorem 2.10 implies that if x, y ∈ [a, b] and y ≤ x, then

F (x) =

ˆ x

a

f =

ˆ y

a

f +

ˆ x

y

f = F (y) +

ˆ x

y

f,

from this we have

F (x)− F (y) =

ˆ x

y

f.

By hypothesis if −M ≤ f(x) ≤M for all x ∈ [a, b], then Theorem 2.3 implies that

−M(x− y) ≤ˆ x

y

f ≤M(x− y),

since y ≤ x it follows that

|F (x)− F (y)| =∣∣∣∣ˆ x

y

f

∣∣∣∣ ≤M |x− y|.Therefore F is uniformly continuous on [a, b] by Theorem 2.12. �

The derivative of the indefinite integral F exists at every point where f i continuous.

Theorem 2.14. Let f ∈ R[a, b] and let limx→c+ f(x) = f(c) for c ∈ [a, b). Then the indefiniteintegral

F (x) =

ˆ x

a

f,

has a right hand derivative at c equal to f(c).

Proof. Assume c ∈ [a, b). Since f is continuous at c, given ε > 0 there exists δε > 0 such that ifc ≤ x < c+ δε, then

f(c)− ε < f(x) < f(c) + ε. (2.2)

Let h satisfy 0 < h < δε. Since f ∈ R[a, b] Theorem 2.10 implies thatf ∈ (R[a, c],R[a, c+ h],R[c, c+ h]) and

F (c+ h)− F (c) =

ˆ c+h

c

f.

On the interval [c, c+ h] inequality 2.2 holds for f , thereforeˆ c+h

c

f(c)− ε <ˆ c+h

c

f(x) <

ˆ c+h

c

f(c) + ε.

Hence

(f(c)− ε) · h < F (c+ h)− F (c) =

ˆ c+h

c

f < (f(c) + ε) · h,

divide by h > 0 and subtract f(c) everywhere, we obtain

−ε < F (c+ h)− F (c)

h− f(c) < ε,

thus ∣∣∣∣F (c+ h)− F (c)

h− f(c)

∣∣∣∣ < ε,

18

since ε > 0 is arbitrary we conclude that

limh→0+

F (c+ h)− F (c)

h= f(c).

�

Approaching the limit from the left gives the following result.

Theorem 2.15. Let f ∈ R[a, b] and let limx→c− f(x) = f(c) for c ∈ [a, b). Then the indefiniteintegral

F (x) =

ˆ x

a

f,

has a left hand derivative at c equal to f(c).

Proof. Assume c ∈ (a, b]. Since f is continuous at c, given ε > 0 there exists δε > 0 such that ifc− δε < x ≤ c, then

f(c)− ε < f(x) < f(c) + ε. (2.3)

Let h satisfy 0 < h < δε. Since f ∈ R[a, b] Theorem 2.10 implies thatf ∈ (R[a, c− h],R[a, c],R[c− h, c]) and

F (c)− F (c− h) =

ˆ h

c−hf.

On the interval [c− h, c] inequality 2.3 holds for f , thereforeˆ c

c−hf(c)− ε <

ˆ c

c−hf(x) <

ˆ c

c−hf(c) + ε.

Hence

(f(c)− ε) · h < F (c)− F (c− h) =

ˆ c

c−hf < (f(c) + ε) · h,

divide by h > 0 and subtract f(c) everywhere, we obtain

−ε < F (c)− F (c− h)

h− f(c) < ε,

thus ∣∣∣∣F (c)− F (c− h)

h− f(c)

∣∣∣∣ < ε,

since ε > 0 is arbitrary we conclude that

limh→0−

F (c)− F (c− h)

h= f(c).

�

If f ∈ C[a, b], then we have a familiar result.

Corollary 2.1. If f ∈ C[a, b], then the indefinite integral F (x) =´ xaf is differentiable on the

whole interval [a, b] and F ′(x) = f(x) for all x ∈ [a, b].

Proof. Let c ∈ (a, b). Since f is continuous at c we have limx→c+ f(x) = limx→c− f(x) = f(c).Therefore according to Theorems 2.14 and 2.15 above we have

F ′(c) = limh→0+

F (c+ h)− F (c)

h= limh→0−

F (c)− F (c− h)

h= limh→0

F (c+ h)− F (c)

h= f(c).

If c = a or c = b we get

F ′+(a) = limh→0+

F (a+ h)− F (a)

h= f(a),

19

respectively

F ′−(b) limh→0−

F (b)− F (b− h)

h= f(b).

�

The indefinite integral of f need not to be an antiderivative of f , maybe the derivative does notexist at c or F ′(c) 6= f(c), however if f ∈ C[a, b], then its indefinite integral F is guaranteed to bean antiderivative of f by Corollary 2.1.

20

3. The gauge integral

In this section we shall generalize the Riemann integral. First in subsection 3.1 we introducesome notations, and then in subsection 3.2 we define the gauge integral. There we also provethat every Riemann integrable function is also gauge integrable, and give an example that thereis a function that is gauge integrable but not Riemann integrable. To be able to formulate theadvanced version of the fundamental theorem of calculus for the gauge integral we need some basicmeasure theory, this is done in subsection 3.3. Much in the same vein as for the Riemann integralwe prove in subsection 3.4 some basic properties. Finally in subsection 3.5 we obtain the aim ofthis essay ”the fundamental theorem of calculus for the gauge integral”.

3.1. Gauges and δ-fine partitions. In this subsection we shall introduce some necessary nota-tion and background for the gauge integral.

Definition 3.1. The closed neighborhood of x with radius r > 0 is the set

B[x, r] = {y ∈ R : |x− y| ≤ r},which is also called the closed ball with center x and radius r. The open neighborhood of x withradius r > 0 is the set

B(x, r) = {y ∈ R : |x− y| < r},which is also called the open ball with center x and radius r.

If I = [a, b], and a ≤ b, we define the length of I to be

l(I) = b− a.It is easy to see that l(I) ≥ 0, and that l(I) = 0 if and only if a = b. Similarly, the length of anyinterval of the forms (a, b), [a, b), (a, b], is also defined to be b− a. The length of the empty set isdefined as l(∅) = 0, this lengths coincide with the Lebesgue measure.

We say that an interval I ⊂ R is degenerate if it contains at most one point, and that it is nondegenerate if it contains at least two points, in which case it contains infinitely many points. Wesay that two intervals I1, I2 ⊂ R are disjoint if I1 ∩ I2 = ∅. We say that two intervals I1, I2 ⊂ Rare non overlapping if I1 ∩ I2 = ∅ or I1 ∩ I2 = {p}, where {p} is a singleton set, which is anendpoint of both intervals.

Example 3.1. If J ⊂ R is a bounded open interval and suppose that I1, ..., Im are nonoverlappingclosed and bounded intervals contained in J . Then

m∑i=1

l(Ii) ≤ l(J).

If I1, ..., Im are degenerate then∑mi=1 l(Ii) = 0 ≤ l(J). Suppose that the intervals Ii are

nondegenerate, so ai < bi. If necessary we relabel such that a1 ≤ a2 ≤ ... ≤ am. Since the intervalsare nonoverlapping, we have

α < a1 < b1 ≤ a2 < b2 ≤ ... ≤ am < bm < β,

where J = (α, β). It follows that bi − ai ≤ ai+1 − ai for i = 1, ...,m − 1 and bm − am < β − am.Adding the inequalities, we get

(b1 − a1) + (b2 − a2) + ...+ (bm − am) =

m∑i=1

(bi − ai) =

m∑i=1

l(Ii) < β − a1 < β − α = l(J).

�

21

Definition 3.2. If I = [a, b] ⊂ R, then the function δ : I → R is said to be a gauge on Iif δ(t) > 0 for all t ∈ I. The interval around t ∈ I controlled by the gauge δ is the intervalB[t, δ(t)] = [t− δ(t), t+ δ(t)].

Definition 3.3. Let I = [a, b] and let P = {Ii, ti}ni=1 be a tagged partition of I. If δ is a gauge

on I, then we say that P is δ-fine if

Ii ⊆ [ti − δ(ti), ti + δ(ti)] for all i = 1, ..., n.

Thus each subinterval Ii is a subset of the interval B[ti, δ(ti)] controlled by the point ti. When

a tagged partition P is δ-fine we sometimes write P � δ.

As always some examples will be needed to clarify the theory.

Example 3.2. If δ > 0 and δ ∈ R, define the gauge δ : I → R by letting δ(t) = δ for all x ∈ I.

This gauge is called a constant gauge, and was used in Definition 2.3. A partition P = {(Ii, ti)}ni=1

for this constant gauge is δ-fine if and only if

Ii ⊆ [ti − δ, ti + δ] = B[ti, δ] for all i = 1, ..., n.

Since P is δ-fine the length l(Ii) = xi − xi−1 ≤ 2δ for all i. �

Example 3.3. If δ1 and δ2 are gauges on I = [a, b]. Define

δ(t) = min{δ1, δ2} for t ∈ I,

then δ is a gauge on [a, b]. Every P = {Ii, ti)}ni=1 that is δ-fine will automatically be both δ1-fine

and δ2-fine. This is easy to realize since if P � δ then for each tagged point ti we have

ti − δ2(ti) ≤ ti − δ1(ti) ≤ xi−1 ≤ ti ≤ xi ≤ ti + δ1(ti) ≤ ti + δ2(ti) for all i = 1, ..., n.

assuming that δ(t) = min{δ1(t), δ2(t)} = δ1(t). Thus

Ii ⊆ B[ti, δ(ti)] ⊆ B[ti, δ1(ti)] ⊆ B[ti, δ2(ti)] ,

and similarly if δ(t) = min{δ1(t), δ2(t)} = δ2(t). This can be extended to any finite number ofgauges on I. �

Example 3.4. If δ1 and δ2 are gauges on I = [a, b] and if δ1(t) ≤ δ2(t) for all t ∈ I and if

P = {Ii, ti}ni=1 is δ1-fine, then it is also δ2-fine.

Since P � δ1, then for all i = 1, ..., n

ti − δ1(ti) ≤ xi−1 ≤ xi ≤ ti + δ1(ti),

orIi = [xi−1, xi] ⊆ B[ti, δ1(ti)] = [ti − δ1(ti), ti + δ1(ti)].

Using the fact that δ1(ti) ≤ δ2(ti) for i = 1, ..., n we get

Ii = [xi−1, xi] ⊆ B[ti, δ1(ti)] = [ti − δ1(ti), ti + δ1(ti)] ⊆ [ti − δ2(ti), ti + δ2(ti)] = B[ti, δ2(ti)].

Hence I ⊆ B[ti, δ2(ti)] so P � δ2.�

Example 3.5. If δ1, ..., δm are gauges on I = [a, b], then δ∗ defined by:

δ∗(t) = min{δ1(t), ..., δm(t)}, for t ∈ I,

is a gauge on I and each partition P of I is δ∗-fine if and only if P is δk-fine for each k = 1, ...,m.Assume P � δ∗, then δ∗ > 0 and

Ii ⊆ B[ti, δ∗(ti)] = [ti − δ∗(ti), ti + δ∗(ti)] ⊆ [ti − δk(ti), ti + δk(ti)]

mk=1 for all i = 1, ..., n.

22

Conversely, assume that P = {Ii, ti}ni=1 � δk for all k = 1, ...,m then

Ii ⊆ [ti − δk(ti), ti + δk(ti)] for each k = 1, ...,m,

since δ∗(ti) ≤ δk(ti) for all k = 1, ...,m we get

Ii ⊆ [ti − δ∗(ti), ti + δ∗(ti)] ⊆ [ti − δk(ti), ti + δk(ti)]mk=1 ⊆ B[ti, δk(ti)]

mk=1,

thus P � δ∗. �

Example 3.6. Let I = [0, 1] and let

δ(t) =

{12 t, 0 < t ≤ 114 , t = 0 .

a) The partition P = {Ii, ti}3i=1 = {([0, 14 ], 0), ([ 14 ,12 ], 12 ), ([ 12 , 1], 34 )} is δ-fine, hence

I1 =

[0,

1

4

]⊆[0− 1

4, 0 +

1

4

]=

[−1

4,

1

4

]= B [t1, δ(t1)] = B

[0,

1

4

],

I2 =

[1

4,

1

2

]⊆[

1

4− 1

2· 1

2,

1

4+

1

2· 1

2

]= B [t2, δ(t2)] = B

[1

2,

1

2t2

],

I3 =

[1

2, 1

]⊆[

3

4− 1

2· 3

4,

3

4+

1

2· 3

4

]= B [t3, δ(t3)] = B

[3

4,

1

2t3

].

Therefore P � δ.b) The partition P = {Ii, ti}3i=1 = {([0, 14 ], 0), ([ 14 ,

12 ,

12 ), ([ 12 , 1], 6

10 )} is not δ-fine, since

I1 =

[0,

1

4

]⊆[−1

4,

1

4

]= B

[0,

1

4

]= B [t1, δ(t1)] ,

I2 =

[1

4,

1

2

]⊆[0,

1

2

]= B

[1

2,

1

4

]= B

[1

2,

1

2t2

]= B [t2, δ(t2)] .

But

B [t3, δ(t3)] = B

[6

10,

3

10

]=

[6

10− 1

2· 6

10,

6

10+

1

2· 6

10

]=

[3

10,

9

10

],

so I3 = [ 12 , 1] * [ 310 ,

910 ] = B[ 6

10 ,310 ] = B[t3, δ(t3)] and thus P 6� δ.

In this particular case every δ-fine partition must have tag t1 = 0.Since P is δ-fine [0, x1] ⊆ [t1 − δ(t1), t2 + δ(t2)] = B[t1, δ(t1)] must hold. This implies that

t1 − δ(t1) ≤ 0. Assume that t1 > 0, then δ(t1) = 12 t1 so t1 − δ(t1) = t1 − 1

2 t1 = 12 t1 > 0, this

contradicts the inequality t1 − δ(t1) ≤ 0. Therefore we must have t1 = 0 if P � δ. �

Example 3.7. Let a < c < b and let δ be a gauge on [a, b]. If P ′ is a δ-fine partition of [a, c] and

if P ′′ is a partition of [c, b] that is δ-fine, then P = P ′ ∪ P ′′ is a δ-fine partition of [a, b]. �

We can modify the gauge in Example 3.6 b) δ-fine for both partitions.

Example 3.8. Let I = [0, 1] and let δ1(t) =

{34 t, 0 < t ≤ 114 , t = 0 .

23

a) The partition will be δ1-fine according to Example 3.4 since δ(t) ≤ δ1(t).b) Since δ(t) ≤ δ1(t) the only interval thats need to be checked is I3,

I3 =

[1

2, 1

]⊆[

6

10− 3

4· 6

10,

6

10+

3

4· 6

10

]=

[3

20,

21

20

]= B

[6

10,

3

4t3

],

thus I3 =[12 , 1]⊆[

320 ,

2120

]= B

[610 ,

34 t3]

and therefore P � δ2. �

The question arises why do gauges work? Before the development of the integral we would liketo answer this question: Why do non constant gauges work better than constant gauges? In theRiemann integral, the measure of ”fineness” of a partition P is given by the maximum length ‖P‖of the subintervals Ii. In the approach of the gauge integral, with δ-fine partitions, more variationof the lengths of the subintervals Ii is allowed as long as the subintervals where the function quicklychanges have small length. There is no need to make the subintervals length small where a functionis roughly constant. Another important advantage with non constant gauges is that we can forcea particular point to be a tag, this can come in handy when a particular point is the origin ofdifficulty, by choosing the problem point as a tag we can control the difficulty at times. If I is acompact interval and δ is a gauge on I we can picture that every point t ∈ I restrain every pointin the interval B[t, δ(t)] = [t − δ(t), t + δ(t)]. Some points in I controls large intervals, and otherpoints controls intervals that are very small. This makes us inquisitive, for an arbitrary gauge δwhen can we find a tagged partition P = {(Ii, ti)}ni=1, where each tag ti controls its associatedsubinterval Ii. Theorem 3.3 will answer this question and guarantee that given any gauge δ wecan always find tagged partitions that are δ-fine. This result was first discovered by Pierre Cousin,in order to prove Cousin’s theorem we need the Archimedean property and the Nested intervalproperty.

Lemma 3.1. Let S ⊂ R be a non empty set, M = supS and y < M then there exist and x0 ∈ Swith x0 > y.

Proof. By contradiction, suppose that x ≤ y for every x ∈ S. Then y is upper bound for S, andy < M contradicts that M = supS. Hence there must exist an x0 ∈ S with x0 > y. �

Theorem 3.1 (Archimedean property). If α, β ∈ R with α > 0 and b > 0 then there exist ann ∈ N and nα > β.

Proof. The proof is by contradiction. Suppose that nα ≤ β for every n ∈ N. Let

S = {na : n ∈ N} .

Now S is bounded above by b, therefore M = supS exists. Now M − α < M , so by Lemma 3.1there exists an n0 ∈ N such that n0α > M − α. But then n0α + α > M , and so (n0 + 1)α > M .Since n0 + 1 ∈ N, we are led to the contradiction that (n0 + 1)α ∈ S and (n0 + 1)α > supS. Hencenα ≤ b cannot be true for every n ∈ N. Therefore there exist an n ∈ N with nα > b �

Corollary 3.1. Given arbitrary ε > 0 there exists an n ∈ N such that 1/n < ε.

Proof. Just put ε = α and b = 1 in Theorem 3.1. �

Theorem 3.2 (Nested interval property). Suppose that I1 = [a1, b1], ..., In = [an, bn] with In ⊆... ⊆ I2 ⊆ I1 and limn→∞(bn − an) = 0. Then there is exactly one number A ∈ R common to allthe intervals In.

Proof. The sequence {an}∞n=1 is monotone increasing and bounded above by b1. Therefore

limn→∞

an = A

24

exists and A = supn an ≤ bk for each k ∈ N. Thus ak ≤ A ≤ bk for every natural number k. Thatis A is contained in each interval Ik. Suppose B ∈ In for every n ∈ N. Then an ≤ B ≤ bn for alln, and so 0 ≤ B − an ≤ bn − an. Since limn→∞(bn − an) = 0 we also have limn→∞(B − an) = 0.Hence B = limn→∞ an = A, so A ∈ R is the only common number to all the intervals In. �

Theorem 3.3. If I = [a, b] ∈ R is a non degenerate compact interval and δ is a gauge on I, then

there exists a tagged partition P of I such that P � δ.

Proof. By contradiction we suppose that I does not have a δ-fine partition. Let c = 12 (a + b)

and bisect I to [a, c] and [c, b]. Since we assumed that I had no δ-fine partition at least one ofthese subintervals cannot have a δ-fine partition. Because if they both were δ-fine then their unionwould a be δ-fine partition as we saw in Example 3.7. Let I1 = [a, c] if this is the subinterval withno δ-fine partition, otherwise let I1 = [c, b]. Rename I1 as [a1, b1], let c1 = 1

2 and bisect I1 to[a1, c1] and [c1, b1]. As above, at least one of these subintervals does not have a δ-fine partition.Let I2 = [a1, c1] if it does not have a δ-fine partition, otherwise let I2 = [c1, b1]. Rename I2 as[a2, b2] and bisect again. Continue this bisection on and on gives us a sequence {In}∞n=1 of compactsubintervals of I = [a, b] that is nested in the meaning that

[a, b] = I ⊃ I1 ⊃ ... ⊃ In ⊃ In+1 ⊃ ...Theorem 3.2 implies that there is exactly one number ξ common to all of the intervals In. Sinceδ(ξ) > 0, Theorem 3.1 implies that there exist k ∈ N such that

l(Ik) = (b− a)/2k < δ(ξ),

where lp ⊂ [ξ − δ(ξ), ξ + δ(ξ)]. Thus the interval around ξ is controlled by the gauge δ, thereforethe pair (lp, ξ) is a δ-fine partition of lp. But we have constructed subintervals In of I that haveno δ-fine partitions. Hence for every gauge δ on I, there exists a δ-fine partition of I. �

3.2. The gauge integral and the relation to the Riemann integral. In Definition 3.6, weshall define the concept of gauge integral. Then in Theorem 3.6 it is proved that every Riemannintegrable function is gauge integrable, but as shown in Example 3.13 there are functions that aregauge integrable but not Riemann integrable.

Lets restate the definition for a function f to be Riemann integrable. The purpose in doing sois comparison that a little modification to the definition will give us huge advantages and widelybroaden the class of integrable functions.

Definition 3.4. A function f : I → R is said to be Riemann-integrable on I if there exists anumber A ∈ R such that for every ε > 0 there exists a number δε such that if P = {Ii, ti}ni=1 isany tagged partition of I such that l(Ii) ≤ δε for i = 1, ..., n then

|S(f, P )−A| < ε,

and we write lim‖P‖→0 S(f, P ) = A.

Definition 3.5. A function f : I → R is said to be gauge integrable on I if there exists a numberB ∈ R such that for every ε > 0 there exists a gauge γε on I such that if P = {Ii, ti}ni=1 is anytagged partition of I such that l(Ii) ≤ γε(ti) for i = 1, ..., n then

|S(f, P )−B| < ε, (3.1)

and we write lim‖P‖→0 S(f, P ) = B.

The only difference from Definition 3.4 is that the constant δε is raplaced by a gauge on I, thatis, by a function δε : I → (0,∞). The following definition is more practical thou.

25

Definition 3.6. A function f : I → R is said to be gauge-integrable on I if there exists a numberC ∈ R such that for every ε > 0 there exists a gauge δε on I such that if P = {Ii, ti}ni=1 is anytagged partition of I that is δε-fine, then

|S(f, P )− C| < ε, (3.2)

and we write lim‖P‖→0 S(f, P ) = C.

The collection of all gauge integrable functions on I = [a, b] will be denoted by G[a, b] or G(I).

Theorem 3.4. Definitions 3.5 and 3.6 lead to equivalent integrals.

Proof. Let ε > 0 and f ∈ G(I) be given in the sense of Definition 3.5. Then there exists a

number B ∈ R and a gauge γε(t) on I such that for any tagged partition P = {Ii, ti}ni=1 we havel(Ii) ≤ γε(ti) for i = 1, ..., n. Define δε(t) = 1

2γε(t) for t ∈ I, so that δε is a gauge on I, then forany δε-fine partition of I we have

Ii ⊆ [ti − δε(ti), ti + δε(ti)] =

[ti −

1

2γε(ti), ti +

1

2γε(ti)

].

Everything in Definition 3.5 is satisfied so inequality 3.1 holds. Hence if P is any δε-fine partitionof I, then

|S(f, P )−B| < ε.

Since ε > 0 is arbitrary f ∈ G(I) in the sense of Definition 3.6 with C = B. Conversely, supposethat f ∈ G(I) in the sense of Definition 3.6. Then there exists a number C ∈ R such thatgiven ε > 0 there exists a gauge δε. Define γε(t) = δε, so γε is a gauge on I. If the partition

P = {(Ii, ti)}ni=1 is δε-fine then Ii ⊆ [ti − δε, ti + δε] for all i = 1, ..., n and so

l(Ii) ≤ γε(ti) = δε(ti).

Hence Definition 3.6 is satisfied so inequality 3.2 holds. Thus if P is any partition of I withl(Ii) ≤ γε(ti) for all i = 1, ..., n then |S(f, P )− C| < ε. Since ε > 0 is arbitrary, then f ∈ G(I) inthe sense of defintion 3.5 with B = C. �

The number B from Definition 3.5 and the number C from Definition 3.6 is unique.

Theorem 3.5. There is only one number C that satisfies Definition 3.6.

Proof. The proof is by contradiction. Suppose C 6= C and let ε = 14 |C − C| > 0. If C satisfies

Definition 3.6, then there exists a gauge δε on I such that if P is any δε-fine partition of I then

|S(f, P )| < ε.

Similarly if C satisfies Definition 3.6, there exists a gauge δε on I such that if P is any δε-finepartition of I then

|S(f, P )| < ε.

Let δε = min{δε, δε}, then δε is a gauge on I(see Example 3.5) and let P � δε. Then P � δε and

P � δε. By the triangle inequality we get

|C − C| = |C − S(f, P ) + S(f, P )− C|

≤ |C − S(f, P )|+ |S(f, P )− C| ≤ ε+ ε =1

2|C − C| < |C − C|.

Ofcourse |C − C| ≮ |C − C|, therefore the assumption C 6= C is false and C = C = C. �

We will now see that the Riemann integral is a subset of the gauge integral.

Theorem 3.6. A function f that is Riemann integrable is also gauge integrable.

26

Proof. Let ε > 0 be given and assume f ∈ R(I), then there exists a number A ∈ R and γε > 0

such that if P = {(Ii, ti)}nt=1 is any tagged partition of I such that l(Ii) ≤ γε for i = 1, ..., n then

|S(f, P )−A| < ε.

Define a (see Example 3.2) gauge γε : I → R by γε(t) = γε for all x ∈ I. Hence the conditionin Definition 3.5 is satisfied so inequality 3.1 holds, since ε > 0 is arbitrary, then f ∈ G(I) withB = A and thus {f : f ∈ R[a, b]} ⊂ {f : f ∈ G[a, b]}. �

The strict inequality 3.2 can be allowed to be equal to ε.

Theorem 3.7. If f : I → R is integrable in the sense of Definition 3.6 if and only if for eachε > 0 there exists a gauge ηε on I such that if Q� ηε then

|S(f, Q)− C| ≤ ε.

Proof. If Definition 3.6 is satisfied then for each partition Q� ηε we have |S(f, Q)−C| < ε, then

|S(f, Q)− C| ≤ ε. Conversely, if P � δε/2, then |S(f, P )− C| ≤ ε/2 < ε. �

Lets have a look at some examples that illustrates the concept of this new integral.

Example 3.9. Let f : I → R be defined by f(x) = c, where c=constant. Then f ∈ G[a, b] and´ baf = c(b− a). If P = {[xi−1, xi], ti}ni=1 is any tagged partition of I then

S(f, P ) =

n∑i=1

f(ti)(xi − xi−1) =

n∑i=1

c(xi − xi−1) = c(b− a).

Hence all Riemann sums equals c(b − a) and therefore we can choose δε with no effort. For

example choose δε(t) = δ. If P � δε ( δ-fine partitions always exists according to Theorem 3.3)

then |S(f, P )− c(b−a)| = 0 < ε, thus f ∈ G[a, b] with´ baf = c(b−a). Of course we already knew

that f ∈ G[a, b], since f ∈ R[a, b]. �

Example 3.10. Let I = [0, 1] and define

h(1/n) =

{n, n = 1, 2, 3, 4, 5

0, all other x ∈ [0, 1] .

Then h ∈ G[0, 1] and h ∈ R[0, 1].

Let δε(t) = Kε for all t ∈ [0, 1] then for each P � δε(t) (an δ-fine partition is guaranteed tobe found by Theorem 3.3) we have that t = 1 is tag for at most one subinterval (xn − xn−1) =(1 − xn−1) ≤ Kε and each of the points 1

2 ,13 ,

14 ,

15 can be tags for at most two subintervals with

length ≤ Kε because [xi−1, xi] ⊆ B[ti,Kε] for i = 1, 2, 3, 4, 5 and all other tags will not contributeto the Riemann sum because h(ti) = 0 there, thus

|S(h, P )− 0| =n∑i=1

h(x)(xi − xi−1) =

5∑i=1

h

(1

n

)(xi − xi−1)

= h(1)(x1 − x0) + h(1/2)(x2 − x1) + h(1/3)(x3 − x2) + h(1/4)(x4 − x3) + h(1/5)(x5 − x4)

≤ Kε+ 2Kε+ 3Kε+ 4Kε+ 5Kε = 15Kε.

Choose K = 115 then S(h, P ) ≤ ε. We got away with a constant gauge, hence f ∈ G[a, b] and

f ∈ R[a, b] with´ 10h(1/n) = 0. �

What will happen if we let n ∈ N.27

Example 3.11. Let I = [0, 1] and define

k(1/i) =

{i, i ∈ N0, all other x ∈ [0, 1] .

Then k ∈ G[0, 1] but k /∈ R[0, 1]. A constant gauge cannot be used here since S(k, P ) will not be

arbitrary small as n→∞ and therefore k /∈ R[0, 1]. Define

δε(1/i) =

{εi2i , t ∈ N1, otherwise .

If P � δε, then the point 1/i can be the tag of atmost two subintervals with lenght ≤ ε/i2i so

|S(k, P )− 0| =n∑i=1

k(ti)(xi − xi−1) =

n∑i=1

i(xi − xi−1)

≤n∑i=1

iε

i2i=

n∑i=1

ε

2i≤ limn→∞

n∑i=1

ε

2i=

∞∑i=1

ε

2i= ε.

Since ε > 0 is arbitrary it follows that k ∈ G[0, 1], and´ 10k = 0. �

Although the next example is Riemann integrable it will be instructive to see how the nonconstant gauges can be chosen.

Example 3.12. Let f : [a, b]→ R be defined by

f(x) =

{α, a ≤ x ≤ cβ, c ≤ x ≤ b .

Then´ baf = α(c − a) + β(b − c). Note that f is continuous at all points except at x = c, so the

”trouble” is concentrated at that point. Let δε : [a, b]→ R be defined by

δε(t) =

{12 |t− c|, t 6= c

δ, t = c .

This gauge forces c to be the tag of a subinterval, later δ will be chosen as needed. LetP = {[xi−1, xi]}ni=1 be a δε-fine partition of [a, b]. If P � δε then δε forces c to be the tag ofany subinterval that contains c. Assume that c is the tag for the two subintervals [xk−1, xk] and[xk, xk+1], where xk = c. Then we have

S(f, P ) =

n∑i=1

f(ti)(xi − xi−1) = α(xk+1 − a) + β(b− xk+1),

thus

|S(f, P )− (α(c− a) + β(b− c)) | = |α(xk+1 − a) + β(b− xk+1)− (α(c− a) + β(b− c)) |= |α(xk+1 − c)− β(xk+1 − c)| = |α− β|(xk+1 − c) ≤ |α− β|δ.

We see that δε(c) = ε/|α − β| will do. Since ε > 0 is arbitrary we conclude that f ∈ G[a, b] and´ baf = α(c− a) + β(b− c). �

Lets consider some functions that are not Riemann integrable and for which nonconstant gaugesis of the essence.

28


f(x) =

{1 if x ∈ Q ∩ [0, 1]

0 if x ∈ (R\Q) ∩ [0, 1] .

Then f ∈ G[0, 1] and´ 10f = 0. We saw in Example 2.4 that f /∈ R[0, 1]. Since the rational

numbers Q in [0, 1] is a countable set, let {rk : rk ∈ Q, k ∈ N} be an enumeration of all rationalnumbers in [0, 1] and let ε > 0 be given. Define δε : [0, 1]→ R by

δε(t) =

{ε

2k+1 , t = rk ∈ (Q ∩ [0, 1])

1, t ∈ ([0, 1] \Q) .

Now let P = {Ii, ti}ni=1 be a δε-fine partition of [0, 1]. If ti ∈ Ii is irrational then f(ti) = 0 so

there is no contribution to the Riemann sum. If ti is rational then f(ti) = 1 and the lenght

l(Ii) = xi − xi−1 is small since P � δε. That is if the k:th rational number rk is tag for theinterval Ii then Ii ⊆ [rk − δε, rk + δε], and therefore l(Ii) ≤ 2δε(rk) = ε/2k+1 + ε/2k+1 = ε/2k(see

Example 3.2). If rk is tag for two subintervals in P , the sum of the lenghts of these two intervalscannot exceed ε/2k, therefore

|S(f, P )| ≤ limn→∞

n∑k=1

ε

2k= ε.

Since ε > 0 is arbitrary this shows that f ∈ G[0, 1] with´ 10f = 0.

�

This makes us suspicious that every ”small” set which could be covered with ε ”balls” behaveslike this? How do we compare ”smallness” and what is small enough? The answers is given below.

3.3. A little taste of exceptional sets, null sets and null functions. To formulate a generalversion of the fundamental theorem of calculus for the gauge integral we need the concept of thata set has Lebesgue measure zero. Measure theory is unfortunately way beyond the scope of thisthesis, and therefore we introduce only the basics in this subsection and refer to [4] for more details.

Definition 3.7. a) A set Z ⊂ R is said to have (Lebesgue) measure zero or be a null set iffor every ε > 0 there exists a countable collection {Jk}∞k=1 of open sets such that

Z ⊆∞⋃k=1

Jk and

∞∑k=1

l(Jk) ≤ ε.

b) If A ⊆ R, then a function f : A→ R is said to be null function if the set {x ∈ A : f(x) 6= 0}has measure zero.

c) If P (x) is a statement about the point x ∈ A and if E ⊂ A, we say that E is an exceptionalset for P if the statement P (x) holds for all x ∈ A− E.

d) In part c), If E ⊂ A is a set of measure zero, we say that P (x) holds almost everywhere onA, and we write

P (x) holds a.e. on A.

29

e) In part c), If E is a finite (or countable) set, we say that P (x) holds with finitely manyexceptions(or with countability). In these cases we write

P (x) holds f.e. [respectively, c.e.]on A.

Some examples of sets of measure zero(or null sets) to clarify the concept.

Example 3.14. a) If A ⊂ R is a set of measure zero then any E ⊆ A is also a set of measurezero. Readily seen since A is a set of measure zero there exists a countable union {Jk}∞k=1

of open sets such that

A ⊆∞⋃k=1

Jk and

∞∑k=1

l(Jk) ≤ ε,

since E ⊆ A we can use exactly the same countable union {Jk}∞k=1 to establish that

E ⊆ A ⊆∞⋃k=1

Jk and

∞∑k=1

l(Jk) ≤ ε.

Hence E is a set of measure zero.

b) Any singleton set {p} is a set of measure zero. Easily seen, given ε > 0 we can cover withJ1 = (p− ε

2 , p+ ε2 ) and J2 = J3 = ...... = ∅.

c) Any countable set in R is a set of measure zero. For if Z = {p1, p2, ...} is an enumerationof Z and if ε > 0, then for k = 1, 2, ... we let

Jk = (pk − ε/2k+1, pk + ε/2k+1) .

Since l(Jk) = ε/2k, it follows that∑∞k=1 l(Jk) =

∑∞k=1

ε2k

= ε.

d) A countable union of sets of measure zero is a set of measure zero. Let Z = {Zm}∞m=1 bea countable collection of sets of measure zero in R. If ε > 0 and m ∈ N, let {Jm,k}∞k=1 bea countable collection of open sets such that

Zm ⊆∞⋃k=1

Jm,k and

∞∑k=1

l(Jm,k) ≤ ε

2m,

since Z is a countable collection of sets of measure zero, then the collection {Jm,k}∞m,k=1

will be countable, hence

Z ⊆∞⋃m=1

∞⋃k=1

Jm,k =

∞⋃m,k=1

Jm,k and

∞∑m=1

∞∑k=1

l(Jm,k) ≤∞∑m=1

ε

2m= ε.

So Definition 3.7 is satisfied.

e) The function in Example 3.13 is a null function since {x ∈ [0, 1] : f(x) 6= 0} = Q ∩ [0, 1] iscountable and thus have measure zero. Or f(x) = 0 a.e.

f) If P1(x) holds f.e. on A, then P1(x) also holds c.e. Since any finite set is countable. Also,if P2(x) holds c.e. on A, then P2(x) holds a.e. on A, since the set E where P2(x) does nothold is countable and therefore have measure zero. �

Looking back at Example 3.13 one might suspect that every null function belongs to G(I) and´If = 0. This will be proven to be true, but before we investigate the general case lets look at a

simpler situation where the function is constant.30

Example 3.15. Let Z ⊂ I = [a, b] be any set of measure zero. Define ϕ : I → R by

ϕ(x) =

{1 if x ∈ Z0 if x ∈ (I \ Z) .

Then ϕ ∈ G(I) withÍϕ = 0.

Let ε > 0 be given and let {Jk}∞k=1 be a countable collection of open intervals as in Definition 3.7.Define a gauge δε : I → R as follows. If t ∈ (I \ Z), let δε(t) = 1 and if t ∈ Z, let k(t) be thesmallest index k such that t ∈ Jk and choose δε(t) > 0 such that [t− δε(t), t+ δε(t)] ⊂ Jk(t).

Now let P = {(Ii, ti)}ni=1 be a δε-fine partition of I. If ti ∈ (I \Z), then ϕ(ti) = 0 so the sum of

the terms in S(ϕ, P ) with tags in (I \Z) equals 0. For each k ∈ N, the (nonoverlapping) intervalsIi with tags in Z ∩ Jk have total lenght ≤ l(Jk). (See Example 3.1). Consequently, the sum of the

terms in S(ϕ, P ) with tags in Jk is ≤ l(Jk). Therefore i

0 ≤ S(ϕ, P ) =

n∑k=1

ϕ(ti)l(Ii) ≤n∑k=1

l(Jk) <

∞∑k=1

l(Jk) ≤ ε.

Since ε > 0 is arbitrary, we conclude that ϕ ∈ G(I) and thatÍϕ = 0. �

And the general case.

Theorem 3.8. Let φ be any null function on I. Then φ ∈ G(I) andÍφ = 0.

Proof. Let Z = {x ∈ I : φ(x) 6= 0} so Z has measure zero. For each n ∈ N letZn = {x ∈ Z : n − 1 ≤ |φ(x)| < n}. Then Zn has measure zero since Zn ⊆ Z. Given ε > 0 andn ∈ N, let {Jn,k}∞k=1 that exist a collection of open intervals such that

Zn ⊆∞⋃k=1

Jn,k and

∞∑k=1

l(Jn,k) ≤ ε

n2n.

Define a gauge δε : I → R as follows. If t ∈ (I \ Z), let δε(t) = 1 and if t ∈ Z, then since {Zn}∞n=1

are pairwise disjoint, that is Zn ∩ Zm = ∅ for n,m ∈ N, there exists a unique n(t) ∈ N such thatt ∈ Zn(t). Let k(t) be the smallest index k such that t ∈ Jn(t),k. And then choose δε > 0 such that

[t− δε, t+ δε] ⊂ Jn(t),k(t). Now let P = {(Ii, ti)}ni=1 be a δε-fine partition of I. If ti ∈ (I \Z), then

φ(ti) = 0 so the sum of the terms in S(φ, P ) with tags in (I \ Z) equals 0. Fix n ∈ N then foreach k ∈ N, the (non overlapping) intervals Ii with tags in Zn ∩ Jn,k are contained in Jn,k, andtherefore have total lenght ≤ l(Jn,k). Since |φ(ti)| < n for each ti ∈ Zn, the sum of the terms in

S(φ, P ) with tags in Zn will be

S(φ, P ) =∑ti∈Zn

φ(ti)l(Ii) ≤∑ti∈Zn

φ(ti)l(Jn,k) <∑ti∈Zn

nl(Jn,k) ≤ n ε

n22=

ε

2n.

Since the sets {Zn}∞n=1 are pairwise disjoint and Z =⋃∞n=1 Zn the sum of the terms with tags

in Z will be ≤∑∞n=1 ε/2

n = ε. Therefore |S(φ, P )| ≤ ε and since ε > 0 is arbitrary it follows thatφ ∈ G(I) and

Íφ = 0. �

This result is very useful, for instance lets apply Theorem 3.8 on the function in Example 3.13.Let Z = {x ∈ I : f(x) = 1} which is all rational numbers Q in [0, 1], since this set (Q

⋂[0, 1]) is

a set of measure 0 it follows that f ∈ G[0, 1] and´ 10f = 0.

31

3.4. Properties of the gauge integral. Now we will establish some elementary properties ofthe gauge integral. This will be familiar since the properties is only slight different from those forthe classical Riemann integral, even the proofs a quite alike.

Theorem 3.9. Suppose f, g ∈ G[a, b] and k ∈ R. Then

a) kf ∈ G[a, b] and´ bakf = k

´ baf ,

b) (f + g) ∈ G[a, b] and´ ba

(f + g) =´ baf +´ bag.

Proof. a) For k = 0 there is nothing to prove so we assume k 6= 0. Give ε > 0, let P ={([xi−1, xi], ti)}ni=1 be a δε-fine partition, then∣∣∣∣∣S(f, P )−

ˆ b

a

f

∣∣∣∣∣ < ε

|k|whenever P � δε.

Thus, if P is δε-fine then∣∣∣∣∣S(kf, P )− kˆ b

a

f

∣∣∣∣∣ =

∣∣∣∣∣kS(f, P )− kˆ b

a

f

∣∣∣∣∣ = |k|

∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣ < |k| ε|k| = ε.

Since ε > 0 is arbitrary, it follows that kf ∈ G[a, b] andˆ b

a

kf = k

ˆ b

a

f.

b) Given ε > 0, let δ1ε , δ2ε be gauges on I such that if the partition P = {([xi−1, xi], ti)}ni=1 is

δ1ε -fine, then ∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣ < ε

2whenever P � δ1ε ,

and if P is δ2ε -fine, then ∣∣∣∣∣S(g, P )−ˆ b

a

g

∣∣∣∣∣ < ε

2whenever P � δ2ε .

Now let δε(t) = min{δ1ε(t), δ2ε(t)} so that if P � δε, then it is both δ1ε -fine and δ2ε -fine. Since

S(f + g, P ) =

n∑i=1

(f + g)(ti)(xi − xi−1)

=

n∑i=1

f(ti)(xi − xi−1) +

n∑i=1

g(ti)(xi − xi−1) = S(f, P ) + S(g, P ),

we have that∣∣∣∣∣S(f + g, P )−

(ˆ b

a

f +

ˆ b

a

g

)∣∣∣∣∣ =

∣∣∣∣∣S(f, P ) + S(g, P )−ˆ b

a

f −ˆ b

a

g

∣∣∣∣∣≤

∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣+

∣∣∣∣∣S(g, P )−ˆ b

a

g

∣∣∣∣∣ < ε

2+ε

2= ε.

Since ε > 0 is arbitrary it follows that f + g ∈ G[a, b] andˆ b

a

f + g =

ˆ b

a

f +

ˆ b

a

g.

�32

Example 3.16. Suppose that f ∈ G(I) and g : I → R such that g(x) = f(x) a.e on I. Theng ∈ G(I) and

Íg =

Íf . To realize this define h : I → R by h = g − f , and let Z = {x ∈

I : g(x) 6= f(x)}, then Z is a set of measure zero since {g(x) 6= f(x)} have measure zero. Also{x ∈ I : h(x) 6= 0} is a set of measure zero and therefore h is a null function (or function of measurezero) by definition. By Theorem 3.8 h ∈ G(I) and

Íh = 0, According to Theorem 3.9 above we

have that since f ∈ G(I) and h ∈ G(I) then g = (f +h) ∈ G(I) andÍg =Íf +Íh =Íf . �

By using mathematical induction, we can extend Theorem 3.9 to a linear combination of func-tions in G[a, b].

Theorem 3.10. Let f1, ..., fn ∈ G(I), I = [a, b] and c1, ..., cn ∈ R then the linear combinationf =

∑nk=1 ckfk belongs to G(I) and ˆ

I

f =

n∑k=1

ck

Î

fk.

Proof. Given ε > 0, let δ1ε , ..., δnε be gauges on I such that if the partition P = {([xi−1, xi], ti)}ni=1

is δ1ε -fine, then ∣∣∣∣∣S(f1, P )−ˆ b

a

f1

∣∣∣∣∣ < ε

nwhenever P � δ1ε ,

and if P is δnε -fine, then∣∣∣∣∣S(fn, P )−ˆ b

a

g

∣∣∣∣∣ < ε

nwhenever P � δnε .

Now let δε(t) = min{δ1ε(t), ..., δnε (t)} so that if P � δε, then it is δ1ε -fine,..., δnε -fine. Since

S(f1 + ...+ fn, P ) =

n∑i=1

(f1 + ...+ fn)(ti)(xi − xi−1)

=

n∑i=1

f1(ti)(xi − xi−1) + ...+

n∑i=1

fn(ti)(xi − xi−1) = S(f1, P ) + ...+ S(fn, P ),

we have that∣∣∣∣∣S(f1 + ...+ fn, P )−

(ˆ b

a

f1 + ...+

ˆ b

a

fn

)∣∣∣∣∣=

∣∣∣∣∣S(f1, P ) + ...+ S(fn, P )−ˆ b

a

f1 − ...−ˆ b

a

fn

∣∣∣∣∣≤

∣∣∣∣∣S(f1, P )−ˆ b

a

f1

∣∣∣∣∣+ ...+

∣∣∣∣∣S(fn, P )−ˆ b

a

fn

∣∣∣∣∣ < ε

n+ ...+

ε

n= ε.

Since ε > 0 is arbitrary it follows that´ ba

(f1 + ...+ fn) =´ baf1 + ...+

´ bafn. Now apply Theorem

3.9 part a) to´ bac1f1, ...,

´ bacnfn and add the terms to obtain

c1

ˆ b

a

f1 + ...+ cn

ˆ b

a

fn =

n∑k=1

ck

ˆ b

a

fk.

Let p(2) be the statement that´ baf1 + f2 =

´ baf1 +

´ baf2 is true and let p(n) be the statement

that∑nk=1 ck

´ bafk is true. Now

∑n+1k=1 ckfk = cn+1fn+1 +

∑nk=1 ckfk and since p(n) is true and

p(n+ 1) = cn+1fn+1 + p(n) another application of Theorem 3.9 implies that p(n+ 1) is true. �33

Theorem 3.11. If f ∈ G(I) and f(x) ≥ 0 for all x ∈ I, thenÎ

f ≥ 0.

Proof. Let δε be a gauge on I such that for any partition P � δε. Since f ∈ G(I), we have

|S(f, P )−Íf | ≤ ε. Since f(x) ≥ 0 for all x ∈ I, the Riemann sum becomes

S(f, P ) =

n∑i=1

f(ti)(xi − xi−1) ≥ 0.

Therefore we have 0 ≤ S(f, P ) ≤Íf + ε, and since ε > 0 is arbitrary, it follows that

Íf ≥ 0. �

Corollary 3.2. If f, g ∈ G(I) and f(x) ≤ g(x) for all x ∈ I, then

Î

f ≤Î

g.

Proof. Define h = g − f so h is integrable by Theorem 3.9. Given that h(x) ≥ 0 for all x ∈ I,it follows from Theorem 3.11 that

Íh ≥ 0. Since

Íg −Íf =

Íh the conclusion is that´

If ≤Íg. �

Corollary 3.3. If f ∈ G(I), I = [a, b] and if m,M ∈ R are such that m ≤ f(x) ≤ M for allx ∈ I, then

m(b− a) ≤ˆ b

a

f ≤M(b− a).

Proof. Since f(x)−m ≥ 0 for all x ∈ I, it follows from Theorems 3.9 and 3.11 thatˆ b

a

f −ˆ b

a

m =

ˆ b

a

(f −m) ≥ 0

Hence´ baf ≥´ bam = m(b−a). Similarly since M−f(x) ≥ 0 for all x ∈ I, it follows from Theorems

3.9 and 3.11 that ˆ b

a

M −ˆ b

a

f =

ˆ b

a

(M − f) ≥ 0.

Thus´ baM = M(b − a) ≥

´ baf combining these inequalities yields the desired result m(b − a) ≤´ b

af ≤M(b− a). �

There is a Cauchy criterion for the gauge integral as well, which is useful when there is nodistinct value that can be foreseen to be the integral of a function.

Theorem 3.12. A function f : I → R belongs to G[a, b] if and only if for every ε > 0 there exists

a gauge ηε > 0 such that if P and Q are any partitions of I with P � ηε and Q� ηε, then

|S(f, P )− S(f, Q)| ≤ ε.

Proof. Let f ∈ G(I) withÍf = A, let ηε = δε/2 > 0 be a gauge on I such that if P , Q� ηε, then

|S(f, P )−A| ≤ ε/2 and |S(f, Q)−A| ≤ ε/2,

hence for such partitions P and Q we have

|S(f, P )− S(f, Q)| = |S(f, P )−A+A− S(f, Q)|

≤ |S(f, P )−A|+ |A− S(f, P )| ≤ ε/2 + ε/2 = ε.

34

Conversely for each n ∈ N, let δn > 0 be a gauge on I such that if P , Q� δn, then

|S(f, P )− S(f, Q)| < 1/n.

We can assume that δn ≥ δn+1 for each n ∈ N, if not just substitute δn with δ′n = min{δ1, ..., δn}.For each n ∈ N, let Pn � δn. Hence, if m > n then both Pn � δn and ˙Pm � δn, therefore

|S(f, Pn)− S(f, ˙Pm)| ≤ 1/n if m > n,

consequently {S(f, ˙Pm)}∞m=1 is a Cauchy sequence in R and therefore this sequence converges in

R to say A, thus sending m to the limit gives us limm→∞ S(f, ˙Pm) = A, and

|S(f, Pn)−A| ≤ 1/n for all n ∈ N.

To be convinced that´ baf = A, given ε > 0, let K ∈ N satisfy K > 2/ε. If Q is an arbitrary

partition of I with Q� δK , then

|S(f, Q)−A| = |S(f, Q)− S(f, ˙PK) + S(f, ˙PK)−A|

≤ |S(f, Q)− S(f, ˙PK)|+ |S(f, PK)−A| ≤ 1/K + 1/K = 2/K < ε.

Since ε > 0 is arbitrary f ∈ G(I) and´ baf = A. �

The gauge integral is additive in the same sense as the Riemann integral.

Theorem 3.13. Let f : [a, b]→ R and let c ∈ (a, b) then f ∈ G[a, b] if and only if f ∈ G[a, c] andf ∈ G[c, b] and we have ˆ b

a

f =

ˆ c

a

f +

ˆ b

c

f.

Proof. Suppose that f1 ∈ G[a, c] with´ caf1 = A1 and f2 ∈ G[c, b] with

´ bcf1 = A2. Then for given

ε > 0 there exists gauges δ′ on [a, c] and δ′′ on [c, b] such that if P1 � δ′ and P2 � δ′′, then

|S(f1, P1)−A1| ≤ ε/2 and |S(f2, P2)−A2| ≤ ε/2.Define a gauge δε : I → R by

δε(t) =

min{δ′ε(t), 12 (c− t)} if t ∈ [a, c)

min{δ′ε(c), ε′′ε (c)} if t = c

min{δ′′ε (t), 12 (t− c)} if t ∈ (c, b] .

Let P be a partition of I = [a, b] such that P � δε, then the point c must be tag of at least one

subinterval in P . Let P1 be the partition of I1 consisting of the points P1 ∩ I1, and let P2 be thepartition of I2 consisting of the points P2 ∩ I2, so that

S(f, P ) = S(f1, P1) + S(f2, P2),

since P1 � δ′ε and P2 � δ′′ε , the conclusion is that

|S(f, P )− (A1 +A2)| = |S(f, P1) + S(f, P2)− (A1 +A2)|

≤ |S(f, P1)−A1|+ |S(f, P2)−A2| ≤ε

2+ε

2= ε.

Since ε > 0 is arbitrary, we conclude that f ∈ G[a, b] and´ baf =´ caf +´ bcf .

Conversely, suppose that f ∈ G[a, b]. For given ε > 0, let ηε > 0 be a gauge, by Theorem 3.12we have

|S(f, P )− S(f, Q| ≤ ε.35

Let f1 be the restriction of f to I1 = [a, c]. Let η′η be the restriction of ηε to I1, and let P1, Q1 bepartitions of I1 that are η′ε-fine. By adding additional partition points and tags from I2 = [c, b],

we can extend P1 and Q1 to partitions P and Q of I = [a, b] that are ηε-fine. If we add the same

additional points and tags in I2 = [c, b] for both P and Q, then

S(f1, P1)− S(f1, Q1) = S(f, P )− S(f, Q),

since P , Q� ηε, we have|S(f1, P1)− S(f1, Q1)| ≤ ε.

Therefore the restriction f1 of f to [a, c] belongs to G[a, c] by Theorem 3.12. A repetition of theargument above shows that the restriction f2 of f belongs to G[c, d], only difference is that thepartition we are ”borrowing” points from is I1 = [a, c], instead of I2 = [c, b]. Now since since

f ∈ G[a, c] and f ∈ G[c, b] has been established the equality´ caf +´ bcf =

´ baf follows from the

first part of the theorem. �

Corollary 3.4. If f ∈ G[a, b] and if a = co < c1 < ... < cn = b, then the restrictions of f to eachof the subintervals [ci−1, ci] are integrable andˆ b

a

f =

n∑i=1

ˆ ci

ci−1

f.

Proof. The case for n = 2 was proved above in Theorem 3.13. The general case follows by usingmathematical induction. �

3.5. The fundamental theorems of calculus for the gauge integral. Now to the main eventof this essay, the one and only, the fundamental theorem of calculus. There are two aspects ofwhat is commonly called the ”fundamental theorem of calculus”. One part concerns with theintegration of derivatives, and the other part with differentiation of integrals. One first encounterthe fundamental theorem in early courses and learn to evaluate the integral of a function f over[a, b] (provided that f ∈ R[a, b]) by finding a function F on [a, b] with F ′(x) = f(x) for all x ∈ [a, b],and then evaluate F (b)− F (a).

We rarely evaluate the integrals by using the definition directly or by calculating Riemann sums.One might supsect that the usual Leibniz-Newton formula that enables us to evaluate Riemann-integrals, then a more advanced method will be required to compute gauge-integrals. As a matter offact the guidelines for evaluating gauge integrals is the same as for Rimann integrals. Furthermore,since the derivative of a function is always gauge integrable (but not always Riemann integrable oreven Lebesgue integrable), the evaluation is indeed easier for the G-integral than for the R-integral.

Definition 3.8. Let I = [a, b] ⊂ R and let F, f : I → R.

a) We say that F is a primitive (or an antiderivative) of f on I if the derivative F ′(x) existsand F ′(x) = f(x) for all x ∈ I.

b) We say that F is a-primitive of f on I if F is continuous on I, and there exists a set E ⊂ Iof measure zero such that for x ∈ E the derivative F ′(x) does not exist or F ′(x) 6= f(x).The set E is called the exceptional set of f .

c) We say that F is c-primitive of f on I if F is continuous on I, and there exists a countableset E ⊂ I such that for x ∈ E the derivative F ′(x) does not exist or F ′(x) 6= f(x). Theset E is called the exceptional set of f .

d) We say that F is f-primitive of f on I if F is continuous on I, and there exists a finite setE ⊂ I such that for x ∈ E the derivative F ′(x) does not exist or F ′(x) 6= f(x). The set Eis called the exceptional set of f .

36

e) If f ∈ G(I) and u ∈ I, then the function Fu : I → R defined by

Fu(x) =

ˆ x

u

f

is called the indefinite integral of f with base point u. If the base point equals the left(u = a) end point of I, the subscript is not written. Any function that differs from Fa bya constant is called an indefinite integral of f.

Notation 3.1. a) In Definition 3.8 a) F is automatically continuous on I since F is differentiableon I, however in Definition 3.8b) it is important to assume that F is continuous on I.

b) For t in the exceptional set E the derivative F ′(t) may not exist or F ′(t) exist but F ′(t) 6= f(t).Sometimes the function f is not even defined at certain points in E, when this is the case we setf = 0 at such points.

c) We always have Fu(u) = 0. If Fu is the indefinite integral of f with base point u, then sinceFu(x) =

´ xaf −´ uaf = Fa(x) −

´ uaf for x ∈ I, it differs from Fa with constant −

´ uaf and thus

Fu is an indefinite integral of f . �

We will return to primitives and indefinite integrals later in this section. We will need thefollowing lemma for the first part of the theorems.

Lemma 3.2. Let F : I → R be differentiable at t ∈ I. Given ε > 0 there exists δε(t) > 0 suchthat if u, v ∈ I satisfy

t− δε(t) ≤ u ≤ t ≤ v ≤ t+ δε(t),

then

|F (v)− F (u)− F ′(t)(v − u)| ≤ ε(v − u). (3.3)

Proof. Since F (t) is differentiable at the point t ∈ I, then for given ε > 0 there exists δε(t) > 0such that if 0 < |z − t| ≤ δε(t) for z ∈ I, then∣∣∣∣F (z)− F (t)

z − t− F ′(t)

∣∣∣∣ ≤ ε,it follows that

|F (z)− F (t)− F ′(t)(z − t)| ≤ ε |z − t| .For all z ∈ I such that |z − t| ≤ δε(t). Now choose u ≤ t and v ≥ t in B[t, δε(t)] and note thatv − t ≥ 0 and t− u ≥ 0. Adding and subtracting the term F (t)− F ′(t)t gives

|F (v)− F (u)− F ′(t)(v − u)|= |F (t)− F ′(t)t− F (t) + F ′(t)t+ F (v)− F (u)− F ′(t)(v − u)|= |[F (v)− F (t)− F ′(t)(v − t)]− [F (u)− F (t)− F ′(t)(u− t)]|≤ |F (v)− F (t)− F ′(t)(v − t)|+ |F (u)− F (t)− F ′(t)(u− t)|

≤ ε|v − t|+ ε|u− t| = ε(v − t) + ε(t− u) = ε(v − u).

�

Notation 3.2. The existence of a derivative on I automatically provides the existence of a gaugeδε on I.

37

The next result is the first edition of the fundamental theorems that makes sure that thederivative of any function on an interval I = [a, b] will always be G-integrable on [a, b]. Furtheron stronger results will be obtain in Theorem 3.15.

Theorem 3.14. If f : I → R has a primitive F on I = [a, b] then f ∈ G(I) andˆ b

a

f = F (b)− F (a).

Proof. Let ε > 0 be given, since F ′(t) = f(t) exists on I let the gauge δε be as in Lemma 3.2.

Choose P = {([xi−1, xi], ti)}ni=1 such that P � δε since xi−1 and xi ”straddle” the tag ti, it followsfrom equation 3.3 that

|F (xi)− F (xi−1)− f(ti)(xi − xi−1)| ≤ ε(xi − xi−1).

We have to make the quantity F (b)−F (a)−S(f, P ) arbitrary small. In order to do so we use thefact that it is a telescoping sum and therefore F (b)− F (a) =

∑ni=1 [F (xi)− F (xi−1)] and hence

F (b)− F (a)− S(f, P ) = F (b)− F (a)−n∑i=1


=

n∑i=1

[F (xi)− F (xi−1)− f(ti)(xi − xi−1)] .

By the triangle inequality we obtain∣∣∣F (b)− F (a)− S(f, P )∣∣∣ =

∣∣∣∣∣n∑i=1

[F (xi)− F (xi−1)− f(ti)(xi − xi−1)]

∣∣∣∣∣≤

n∑i=1

|F (xi)− F (xi−1)− f(ti)(xi − xi−1)| ≤n∑i=1

ε(xi − xi−1) = ε(b− a).

Since ε > 0 is arbitrary, the conclusion is that f ∈ G[a, b] and´ baf = F (b)− F (a). �

In the next example it will be shown that we can allow one point where F ′ does not exist orF ′ 6= f .

Example 3.17. Define

g(x) =

{1/√x if x ∈ (0, 1]

0 if x = 0 .

g is not bounded and therefore g /∈ R[0, 1]. Let G(x) = 2√x for x ∈ [0, 1], then G is continuous on

[0, 1] and G′(x) = g(x) for all x ∈ (0, 1], but G′(0) does not exist. Therefore G is not a primitiveof g but a f -primitive of g on [0, 1] with the exceptional set E = {0}. Given ε/2 > 0 let

δε(t) =

{δε if t ∈ (0, 1]

ε2/16 if t = 0 ,

where δε will be chosen such that inequality (3.3) holds for G. Now if 0 ≤ v ≤ δε(0), then

G(v)−G(0) = 2√v ≤ ε/2. Let P = {([xi, xi−1], ti)}ni=1 be a δε-fine partition of [0, 1]. If all of the

tags ti belongs to (0, 1], then Theorem 3.14 applies with the only change that F = G. But if thefirst tag t1 = 0, then t1 = 0 ≤ x1 ≤ δε(0) so G(x1) − G(0) = 2

√x1 ≤ ε/2. Also the first term in

S(f, P ) = g(0)(x1 − x0) = 0, hence38

∣∣∣∣∣G(1)−G(0)−n∑i=1


∣∣∣∣∣ =

∣∣∣∣∣n∑i=1

G(xi)−G(xi−1)− g(ti)(xi − xi−1)]

∣∣∣∣∣=

∣∣∣∣∣G(x1)−G(x0)− g(0)(x1 − x0) +

n∑i=2

[G(xi)−G(xi−1)− g(ti)(xi − xi−1)]

∣∣∣∣∣≤ |G(x1)−G(0)|+

∣∣∣∣∣n∑i=2

[G(xi)−G(xi−1)− g(ti)(xi − xi−1)]

∣∣∣∣∣≤ |2√x1|+

n∑i=2

| [G(xi)−G(xi−1)− g(ti)(xi − xi−1)] |

≤ ε

2+

n∑i=2

ε

2(xi − xi−1) =

ε

2+ε

2(xn − x1) =

ε

2+ε

2(1− x1) ≤ ε

2+ε

2= ε.

Since ε/2 > 0 is arbitrary, we conclude that g ∈ G[0, 1] and´ 10

(1/√x) dx = G(1) − G(0) =

2√

1 − 2√

0 = 2. It is understood that the integrand is given the value 0 at the point where it isnot defined. �

This argument in Example 3.17 can be extended to any integrand f such that there exists acontinuous function F on [a, b] such that F ′(x) = f(x) for all but a finite number of points. Belowwe shall show that it will work even if we allow a countable number of points where F ′ does notexist or F ′(x) 6= f(x) at those points. This is a significant improvement of Theorem 3.14.

Theorem 3.15. If f : [a, b]→ R has a c-primitive F on [a, b], then f ∈ G[a, b] and

ˆ b

a

f = F (b)− F (a).

Proof. Let E = {ck}∞k=1 be the exceptional countable set where F ′ does not exist or F ′(x) 6= f(x).Since E is countable it has measure zero. We can without a lost of generality assume that f(ck) = 0.

If f(ck) 6= 0 just define h = g− f , with g = f a.e, then h = 0 a.e, so g ∈ G[a, b] and´ bag =´ baf as

we saw in Example 3.16. Lets define a gauge on I = [a, b], for given ε/2(b− a) > 0 and t ∈ (I \E)we use the fact that F is differentiable at those points so just let δε(t) be as in Lemma 3.2. If t ∈ E,then t ∈

⋃∞k=1 ck(t = ck for some k ∈ N) and since F is continuous on I = [a, b] we can choose

δε(ck) > 0 such that for all z ∈ I that satisfies |z − ck| ≤ δε(ck) we have |F (z)− F (ck)| ≤ ε/2k+3.

Let P = {([xi, xi−1], ti)}ni=1 be a δε-fine partition of I = [a, b]. If t ∈ (I \ E) then the proof givenin Theorem 3.14 applies without change. If ck ∈ E is the tag of a subinterval [xi−1, xi], then

|F (xi)− F (xi−1 − f(ck)(xi − xi−1)| = |F (xi)− F (ck) + F (ck)− F (xi−1 − f(ck)(xi − xi−1)|

≤ |F (xi)− F (ck)|+ |F (ck)− F (xi−1|+ |f(ck)(xi − xi−1)| ≤ ε/2k+3 + ε/2k+3 + 0 = ε/2k+2.

Now each point of E can be the tag of at most two subintervals in P and the lenght of thosesubintervals cannot exceed ε/2k+2 + ε/2k+2 = ε/2k+1. Thus if ti ∈ E we have

∑ti∈E|F (xi)− F (xi−1)− f(ti)(xi − xi−1)| ≤

∞∑k=1

ε/2k+1 = ε/2.

39

Also by Lemma 3.2, the sum of the terms with ti ∈ (I \ E) satisfies∑ti∈(I\E)

|F (xi)− F (xi−1)− f(ti)(xi − xi−1)|

≤∑

ti∈(I\E)

ε

2(b− a)(xi − xi−1) =

ε

2(b− a)

m∑i=1

(xi − xi−1)

≤ ε

2(b− a)(b− a) = ε/2.

Now if ti ∈ (E⋃I \ E) = I we have

∣∣∣F (b)− F (a)− S(f, P )∣∣∣ =

∣∣∣∣∣n∑i=1

F (xi)− F (xi−1)− f(ti)(xi − xi−1)

∣∣∣∣∣=

∣∣∣∣∣∣∑

ti∈(I\E)

F (xi)− F (xi−1)− f(ti)(xi − xi−1) +∑ti∈E

F (xi)− F (xi−1)− f(ti)(xi − xi−1)

∣∣∣∣∣∣≤

∑ti∈(I\E)

|F (xi)− F (xi−1)− f(ti)(xi − xi−1)|+∑ti∈E|F (xi)− F (xi−1)− f(ti)(xi − xi−1)|

≤ ε/2 + ε/2 = ε.

Since ε/2(b− a) > 0 is arbitrary it follows that f ∈ G[a, b] and´ baf = F (b)− F (a). �

Example 3.18. Let F : [0, 1]→ R be defined by

F (x) =

{x2 cos(1/x2) for x ∈ (0, 1]

0 for x = 0 .

We saw in Example 2.7 that F ′ /∈ R[0, 1] since the derivative limx→0+ F′ = ∞, hence F ′ fails to

have a limit at x = 0. However that is no longer a problem, let E = {0} be the exceptional set.Therefore F ′ has a f -primitive and also a c-primitive on [0, 1] by Theorem 3.15, thus

ˆ 1

0

F ′ = F (1)− F (0) = 12 cos(1/12)− 0 = cos(1).

�

Example 3.19. Let A(x) = arcsin(x) for x ∈ [−1, 1] so that A is continuous on [−1, 1] and A′(x) =

1/√

1− x2 for x ∈ (−1, 1). We define a(x) = A′(x) for x ∈ (−1, 1) and let a(−1) = a(1) = 0.

Then Theorem 3.15 with E = {−1, 1} implies that a ∈ G[−1, 1] and that´ 1−1 a = A(1)−A(−1) =

π2 −

−π2 = π, which we write with as

ˆ 1

−1

1√1− x2

= arcsin(1)− arcsin(−1) = π.

�

Lets now turn our focus to the part of the fundamental theorem that discusses the differentiationof an indefinite integral. For the moment we will focus our attention on the differentiation of anindefinite integral at a specific point c ∈ [a, b]. We will see that one-sided continuity of f at a pointc will imply one-sided differentiability of F at c.

40

Theorem 3.16. Let f ∈ G[a, b] and let limx→c+ f(x) = A for c ∈ [a, b). Then the indefiniteintegral

Fu(x) =

ˆ x

u

f

has a right hand derivative at c equal to A.

Proof. We can without loss of generality assume a = u (since any other indefinite integral onlydiffers from Fa by a constant function) and denote Fa = F . Let h satisfy 0 < h < δ. Since

f ∈ (G[a, c],G[a, c+ h],G[c, c+ h]) according to Corollary 3.4 we have F (c+h)−F (c) =´ c+ha

f−´ caf =´ c+hc

f and since limx→c+ f(x) = A for x ∈ (c, c+ h] ⊂ (c, c+ δ) we have |f(x)−A| ≤ ε or

equivalent A−ε ≤ f(x) ≤ A+ε. Now by Corollary 3.3 we have´ c+hc

A−ε ≤´ c+hc

f ≤´ c+hc

A+ε,

considering A ± ε are constants we obtain (A − ε)h ≤´ c+hc

f ≤ (A + ε)h. Since h > 0 we have

A− ε ≤ 1h

´ c+hc

f ≤ A+ ε or A− ε ≤ F (c+h)−F (c)h ≤ A+ ε, hence∣∣∣∣F (c+ h)− F (c)

h−A

∣∣∣∣ ≤ ε.Since ε > 0 is arbitrary we conclude that

limh→0+

F (c+ h)− F (c)

h= A.

That is F ′+(c) = A. �

Approaching the limit from the left side yields.

Theorem 3.17. Let f ∈ G[a, b] and let limx→c− f(x) = B for c ∈ (a, b]. Then the indefiniteintegral

Fu(x) =

ˆ x

u

f

has a left hand derivative at c equal to B.

Proof. We can without loss of generality assume a = u (since any other indefinite integral onlydiffers from Fa by a constant function) and denote Fa = F . Let h satisfy 0 < h < δ. Sincef ∈ (G[a, c− h],G[a, c],G[c− h, c]) according to Corollary 3.4 we have F (c)− F (c− h) =

´ caf −´ c−h

af =´ cc−h f and since limx→c− f(x) = B for x ∈ [c−h, c) ⊂ (c−δ, c) we have |f(x)−B| ≤ ε or

equivalent B− ε ≤ f(x) ≤ B+ ε. Now by Corollary 3.3 we have´ cc−hB− ε ≤

´ cc−h f ≤

´ cc−hB+ ε,

considering B ± ε are constants we obtain (B − ε)h ≤´ cc−h f ≤ (B + ε)h. Since h > 0 we have

B − ε ≤ 1h

´ cc−h f ≤ B + ε or B − ε ≤ F (c)−F (c−h)

h ≤ B + ε, hence∣∣∣∣F (c)− F (c− h)

h−B

∣∣∣∣ ≤ ε.Since ε > 0 is arbitrary we conclude that

limh→0−

F (c)− F (c− h)

h= B.

That is F ′−(c) = B. �

Corollary 3.5. If f is continuous at c ∈ [a, b] and f ∈ G[a, b]. Then the indefinite integral Fu off is differentiable at c and F ′u(c) = f(c).

41

Proof. Let c ∈ (a, b), since f is continuous at c we have limx→c+ f(x) = limx→c− f(x) = f(c).Therefore according to Theorems 3.16 and 3.17 above we have

F ′(c) = limh→0+

F (c+ h)− F (c)

h= limh→0−

F (c)− F (c− h)

h= limh→0

F (c+ h)− F (c)

h= f(c).

If c = a or c = b we get

F ′+(a) = limh→0+

F (a+ h)− F (a)

h= f(a),

respectively

F ′−(b) limh→0−

F (b)− F (b− h)

h= f(b).

�

Earlier we said in Definition 3.8 that a function Fu is said to be an indefinite integral of f ∈ G(I)in case Fu−Fa is a constant function. Corollary 3.5 above implies that if f is continuous at c ∈ I,

then F ′u(c) =(´ caf −´ uaf)

= F ′a(c)−(´ uaf)′

= f(c), since´ uaf is constant. This points us to the

familiar result below.

Corollary 3.6. Let f be continuous on I = [a, b]. Then any indefinite integral F of f is differen-tiable on I and F ′(x) = f(x) for all x ∈ I.

Proof. Take x0 ∈ {x : x ∈ I} arbitrary and apply Corollary 3.5, hence F ′(x0) = f(x0) and sincex0 ∈ {x : x ∈ I} was arbitrary it follows that F ′(x) = f(x) for all x ∈ I. �

In other words if f is continuous on I = [a, b], then any indefinite integral F of f is a primitiveof f on [a, b]. Before we get to the much stronger theorem about the differentiation of the integralwe need some terminology and preparation.

Definition 3.9. Let I = [a, b] be a non degenerate closed interval

a) A subpartition of I is a collection {Jj}sj=1 of non overlapping closed intervals in I.

b) A tagged subpartition of I is a collection P0 = {(Jj , tj)}sj=1 of ordered pairs, consisting ofintevals {Jj}sj=1 that form a subpartition of I, and tags tj ∈ Jj for j = 1, ..., s.

c) If δ is a gauge on I, we say that the tagged subpartition P0 is δ-fine if Jj ⊆ B[tj , δ(tj)] =[tj − δ(tj), tj + δ(tj)] for j = 1, ..., s.

d) If δ is a gauge on a subset E ⊆ I, we say that the tagged subpartition P0 is (δ, E)-fine if all

tags tj ∈ E and Jj ⊆ [tj−δ(tj), tj +δ(tj)] for j = 1, ..., s. Sometimes we write P0 � (δ, E).

Notation 3.3. a) Any subset of a partition of I is a subpartition of I. Also if Π1 = {Jj}sj=1 is asubpartition of I, then there exists a subpartition Π2 of I such that Π1 ∪ Π2 is a partition of I.To see this suppose that the intervals Π1 = {Jj}sj=1 = {[aj , bj ]}sj=1 are pairwise disjoint. We order

these intervals so that bj < aj+1 for j = 1, ..., s− 1. Let Π2 = {[bj , aj+1]}s−1j=1 ∪ [a, a1]∪ [bs, b] whenthese intervals are nondegenerate. Thus

Π1 ∪Π2 = [a, a1] ∪ [a1, b1] ∪ ... ∪ [as, bs] ∪ [bs, b] = I.

b) Definition 3.9 (d) only requires δ to be defined on E. However we can always set δ(t) = 1 fort ∈ I \ E and obtain a gauge on I.

Definition 3.10. If P0 = {(Jj , tj) : j = 1, ..., s} is a tagged partition of I, then we let U(P0) =⋃sj=1 Jj . If f ∈ G(I), we define

S(f, P0) =

s∑j=1

f(tj)l(Jj) and

ˆU(P0)

f =

s∑j=1

ˆJj

f

42

where l(Jj) denotes the lenght of the subinterval Jj .

This definition makes sense by Corollary 3.4. The next lemma is an important step in ourjourney to prove the deeper properties of the gauge integral. It will be used in order to prove thecontinuity and the almost everywhere differentiability of the indefinite integral. It is called theSaks-Henstock Lemma, but for its use for the general integral is is certainly due to Henstock.

Lemma 3.3. Let f ∈ G[a, b] and for given ε > 0 let δε be a gauge on I such that if P � δε, then∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣ ≤ ε.If P0 = {(Jj , tj) : J = 1, ..., s} is any δε-fine subpartition of I, then∣∣∣∣∣∣

s∑j=1

(f(tj)l(Jj)−

ˆJj

f

)∣∣∣∣∣∣ =

∣∣∣∣∣S(f, P0)−Û(P0)

f

∣∣∣∣∣ ≤ ε.Proof. Let K1, ...,Kk denote closed subintervals in I such that {Jj}sj=1∪{Kk}mk=1 form a partitionof I. We show that since f ∈ G(Kk) and obtain partitions of these intervals that are so fine thatthe contribution to the sum is arbitrarily small. Now let α > 0 be arbitrary. By Corollary 3.4 therestriction of f to each subinterval Kk where k = 1, ...,m is integrable. Since f ∈ G(Kk), there

exits a gauge δα,k on Kk such that if Qk is a δα,k-fine partition of Kk, then∣∣∣∣S(f, Qk)−ˆKk

f

∣∣∣∣ < α

m.

We may assume that δα,k(x) ≤ δε(x) for all x ∈ Kk. Otherwise P cannot be δε-fine. Let P ∗ denote

the tagged partition P ∗ = P0 ∪ Q1 ∪ ... ∪ Qm of I. Clearlyg P ∗ is δε-fine. so∣∣∣S(f, P )−

´ baf∣∣∣ ≤ ε

holds for P ∗. Also since

S(f, P ∗) = S(f, P0) + S(f, Q1) + ...+ S(f, Qm),

therefore Î

f =

Û(P0)

f +

ˆK1

f + ...+

ˆKm

f,

consequently, we obtain∣∣∣∣∣S(f, P0)−Û(P0)

f

∣∣∣∣∣=

∣∣∣∣∣(S(f, P ∗)−

m∑k=1

S(f, Qk)

)−

(Î

f −m∑k=1

ˆKk

f

)∣∣∣∣∣=

∣∣∣∣∣(S(f, P ∗)−

Î

f

)−

(m∑k=1

S(f, Qk)−ˆKk

f

)∣∣∣∣∣≤∣∣∣∣S(f, P ∗)−

Î

f

∣∣∣∣+

∣∣∣∣∣m∑k=1

S(f, Qk)−ˆKk

f

∣∣∣∣∣ ≤ ε+

m∑k=1

α

m= ε+ α.

Since α > 0 is arbitrary we have |S(f, P0)−Ú(P0)

f | ≤ ε as asserted. �

If the possible error doubles, we can interchange the absolute value and the sum.43

Corollary 3.7. Let the conditions be as in Lemma 3.3, then

s∑j=1

∣∣∣∣∣f(tj)l(Jj)−ˆJj

f

∣∣∣∣∣ ≤ 2ε.

Proof. Let ˙P+0 be those parts of P0 such that f(tj)l(Jj)−

´Jjf ≥ 0, and let ˙P−0 be the parts where

f(tj)l(Jj)−´Jjf < 0. If we now apply Lemma 3.3 to ˙P+

0 and ˙P−0 . We obtain

∑˙P+

0


f

∣∣∣∣∣ =∑˙P+

0

(f(tj)l(Jj)−

ˆJj

f

)≤ ε

and ∑˙P−0


f

∣∣∣∣∣ = −∑˙P−0

(f(tj)l(Jj)−

ˆJj

f

)≤ ε,

therefore

s∑j=1


f

∣∣∣∣∣=∑˙P+

0


f

∣∣∣∣∣+∑˙P−0


f

∣∣∣∣∣=∑˙P+

0

(f(tj)l(Jj)−

ˆJj

f

)−∑˙P−0

(f(tj)l(Jj)−

ˆJj

f

)≤ ε+ ε = 2ε.

�

An important application of Lemma 3.3 is that it can be used to establish the continuity ofthe indefinite integral of an integrable function. As always we only consider the indefinite integralwith the left endpoint a as base point, since any other indefinite integral differs from this one bya constant and if F is continuous so is F + c, where c is a constant.

Theorem 3.18. Let f ∈ G[a, b], then the indefinite integral F (x) =´ xaf is continuous from the

right on [a, b].

Proof. Let c ∈ [a, b) and let ε2 > 0 be given. Since f ∈ G[a, b] let δε/2 be a gauge as in Lemma 3.3.

Define δ′ : [a, b]→ R by

δ′ε/2(t) =

{min{δε/2(t), 12 |t− c|} if t ∈ I, t 6= c

min{δε/2(c), ε/2(|f(c)|+ 1)} if t = c .

This forces c to be a tag of any subinterval in P that contains c. Now let 0 < h < δ′ε/2(c) and let

P0 be the δ′ε/2-fine partition consisting of the pair ([c, c+ h], c). Lemma 3.3 implies that∣∣∣∣∣∣s∑j=1

f(tj)l(Jj)−ˆJj

f

∣∣∣∣∣∣ =

∣∣∣∣∣f(c)h−ˆ c+h

c

f

∣∣∣∣∣ ≤ ε

2.

44

It follows from the fact that since h ≤ ε/2(|f(c)|+ 1) that h < ε/2|f(c)| and hence h|f(c)| < ε/2,therefore

|F (c+ h)− F (c)| =

∣∣∣∣∣ˆ c+h

c

f − f(c)h+ f(c)h

∣∣∣∣∣≤

∣∣∣∣∣ˆ c+h

c

f − f(c)h

∣∣∣∣∣+ |f(c)h| =

∣∣∣∣∣ˆ c+h

c

f − f(c)h

∣∣∣∣∣+ |f(c)|h < ε

2+ε

2= ε.

Since ε/2 > 0 it follows that limx→c+ F (x) = F (c). �

The other direction is handed similarly.

Theorem 3.19. Let f ∈ G[a, b], then the indefinite integral F (x) =´ xaf is continuous from the

left on [a, b].

Proof. Let d ∈ (a, b] and let ε2 > 0 be given. Since f ∈ G[a, b] let δε/2 be a gauge as in Lemma

3.3. Define δ′ : [a, b]→ R by

δ′ε/2(t) =

{min{δε/2(t), 12 |t− d|} if t ∈ I, t 6= d

min{δε/2(d), ε/2(|f(d)|+ 1)} if t = d

This forces d to be a tag of any subinterval in P that contains d. Now let 0 < h < δ′ε/2(d) and let

P0 be the δ′ε/2-fine partition consisting of the pair ([d− h, d], d). Lemma 3.3 implies that∣∣∣∣∣∣s∑j=1

f(tj)l(Jj)−ˆJj

f

∣∣∣∣∣∣ =

∣∣∣∣∣f(d)h−ˆ d

d−hf

∣∣∣∣∣ ≤ ε

2.

It follows from the fact that since h ≤ ε/2(|f(d)|+ 1) that h < ε/2|f(d)| and hence h|f(d)| < ε/2,therefore

|F (d)− F (d− h)| =

∣∣∣∣∣ˆ d

d−hf − f(d)h+ f(d)h

∣∣∣∣∣≤

∣∣∣∣∣ˆ d

d−hf − f(d)h

∣∣∣∣∣+ |f(d)h| =

∣∣∣∣∣ˆ d

d−hf − f(d)h

∣∣∣∣∣+ |f(d)|h < ε

2+ε

2= ε.

Since ε/2 > 0 it follows that limx→d− F (x) = F (d). �

Corollary 3.8. If f ∈ G[a, b], then the indefinite integral F (x) =´ xaf is continuous on [a, b].

Proof. Since F is right continuous at every c ∈ [a, b) by Theorem 3.18, F will be right continuousat every c′ ∈ (a, b). Also F is left continuous at every d ∈ (a, b] by Theorem 3.19, F will be leftcontinuous at every d′ ∈ (a, b). Let c′ = d′, hence

limx→c′+

F (x) = limx→c′−

F (x) = F (c′).

�

Before we state the main differential theorem we need a detour about coverings.

Definition 3.11. Let E ⊆ [a, b] and let F be a collection of nondegenerate closed intervals in[a − 1, b + 1]. We say that F is a Vitali covering for E if for every x ∈ E and every s > 0 thereexists an interval J ∈ F such that x ∈ J and 0 < l(J) < s.

As an example of a countable Vitali covering for the inteval I = [0, 1], consider the collection ofclosed balls B[r, 1/n] = [r − 1/n, r + 1/n], where r ∈ I ∩Q and n ∈ N.

45

Theorem 3.20. Let E ⊆ [a, b] and let F be a Vitali covering fo E. Then, given ε > 0 there existsdisjoint intervals I1, ..., Ip from F and a countable collection of closed intervals {Ji : i = p+ 1, ...}in R with

E \p⋃i=1

Ii ⊆∞⋃

i=p+1

Ji and

∞∑i=p+1

l(Ji) ≤ ε.

Therefore, it follows that

E ⊆p⋃i=1

Ii ∪∞⋃

i=p+1

Ji.

Proof. See e.g. Theorem 6.3.1 in [4]. �

Lastly we are equipped to prove the more general differential theorem.

Theorem 3.21. If f ∈ G[a, b], then any indefinite integral F is continuous on [a, b] and is ana-primitive of f on [a, b]. Hence, there exists a set Z ⊂ I of measure zero such that

F ′(x) = f(x) for all x ∈ I \ Z.

Proof. As before, it is sufficient to handle F (x) =´ xaf . Let E be the set of points x ∈ [a, b)

such that the right hand derivative F ′+(x) 6= f(x) or F ′+(x) does not exist. The strategy is to

prove that E is a set of measure zero, and then likewise prove that the set E of points in (a, b],where F ′−(x) 6= f(x) or F ′−(x) does not exist also is a set of measure zero. Then the set Z whereF ′(x) 6= f(x) or F ′(x) does not exist is the union of these two sets. The union of two sets ofmeasure zero is itself a set of measure zero. If F has a right hand derivative F ′+(x) = f(x) at thepoint x ∈ [a, b], then for any α > 0 there exists an s > 0 such that if v ∈ (x, x+ s) ⊂ [a, b], then∣∣∣∣F (v)− F (x)

v − x− f(x)

∣∣∣∣ ≤ α. (3.4)

On the other hand if 3.4 would not hold and if x ∈ E, then there exists α(x) > 0 such that forevery s > 0 there exists a point vx,s ∈ (x, x+ s) ⊂ [a, b] such that∣∣∣∣F (vx,s)− F (x)

vx,s − x− f(x)

∣∣∣∣ > α(x),

which give us|(F (vx,s)− F (x))− f(x)(vx,s − x)| > α(x)(vx,s − x). (3.5)

Now fix n ∈ N and let En = {x ∈ E : α(x) ≥ 1/n}. Given ε > 0, since f ∈ G[a, b], there exists a

gauge δε on [a, b] such that if P is a δε-fine partition of [a, b], then∣∣∣∣∣S(f, P )−ˆ b

a

f

∣∣∣∣∣ ≤ ε

n. (3.6)

Also let Fn = {[x, vx,s] : x ∈ En, 0 < s ≤ δε(x)}, then Fn is a Vitali covering for En. By Theorem3.20 there exists intervals Ii = [x1, v1], ..., Ip = [xp, vp] in F and a sequence {Ji}∞i=p+1 of closedintervals such that

En ⊆p⋃i=1

Ii ∪∞⋃

i=p+1

Ji and

∞∑i=p+1

l(Ji) ≤ ε. (3.7)

Regard the sump∑i=1

∣∣∣∣f(xi)(vi − xi)−ˆ vi

xi

f

∣∣∣∣ =

p∑i=1

|f(xi)(vi − xi)− (F (vi)− F (xi))| . (3.8)

46

It follows from 3.5 with α(xi) ≥ 1/n that the sum on the right in 3.8 is greater than

1

n

p∑i=1

(vi − xi). (3.9)

Since xi ≤ vi ≤ xi + δε(xi) for i = 1, ..., p the ordered pairs {(Ii, xi)}pi=1 form a subpartition of aδε-fine partition of I for which 3.6 holds. Therefore Corollary 3.7 implies that the sum in 3.8 is lessthan or equal to 2ε/n. This estimation in combination with 3.9 and multiplying with n we have

p∑i=1

(vi − xi) ≤ np∑i=1

|f(xi)(vi − xi)− (F (vi)− F (xi))|

= n

p∑i=1

∣∣∣∣f(xi)(vi − xi)−ˆ vi

xi

f

∣∣∣∣ ≤ n2ε

n= 2ε.

Considering 3.7 we conclude that En is contained in a countable union of intervals with total length≤ 3ε. Since ε > 0 is arbitrary, it follows that the set En have measure zero (or is a null set). SinceEn is a set of measure zero then by Example 3.14 the countable union E = ∪∞n=1En is a set ofmeasure zero. Therefore the set E where F ′+(x) 6= f(x) or F ′+(x) does not exist is a set of measure

zero. A repetition of the above argument shows that the set E where F ′−(x) 6= f(x) or F ′−(x) does

not exist is a set of measure zero, hence we conclude that the set Z = E ∪ E where F ′(x) 6= f(x)or F ′(x) does not exist is a set of measure zero.

�

Let us end this essay with the nowhere continuous function we encountered earlier.


f(x) =

{1 if x ∈ Q ∩ [0, 1]

0 if x ∈ (R\Q) ∩ [0, 1] .

a) Then f has a c-primitive function F (x) = 0 for x ∈ [0, 1] and F is the indefinite integral of fwith base point 0. Indeed F ′(x) = 0 for all x ∈ [0, 1], so F ′(x) = f(x) for x ∈ (R\Q)∩[0, 1].Therefore F (x) = 0 is a a-primitive of f . Since the restriction of f to [0, x] is a functionof measure zero({x : x ∈ Q ∩ [0, 1]} is a countable set), we have

´ x0f = 0 = F (x) for

all x ∈ [0, 1]. This example shows the broaden generality of Theorem 3.21, thus f is notcontinuous at any point c ∈ [0, 1], yet has an indefinite integral F that is differentiable atevery point c ∈ [0, 1].

b) f has a c-primitive function G(x) = 1 for x ∈ [0, 1]. While G is an indefinite integral of f , ithas no base point. Since G′(x) = 0 for all x ∈ [0, 1], so G′(x) = f(x) for x ∈ (R\Q)∩ [0, 1].Thus G(x) is a a-primitive of f . Since f is a null function on every interval, then G(x) =1 6= 0 =

´ xuf for all u, x ∈ [0, 1]. Therefore G has no base point in [0, 1] but G(x) differs

from F (x) by a constant and hence is an indefinite integral of f .

�

47

Appendix A. An open letter to authors of calculus books

The following open letter being distributed to publishers’ representatives at the Joint Mathe-matics Meetings in San Diego, California, in January 1997. It was signed by: Robert Bartle, RalphHenstock, Jaroslav Kurzweil, Eric Schechter, Stefan Schwabik, and Rudolf Vyborny.

To: The authors of calculus textbooks

From: Several authors of more advanced books and articles

Robert Bartle, USARalph Henstock, IrelandJaroslav Kurzweil, Czech RepublicEric Schechter, USAStefan Schwabik, Czech RepublicRudolf Vyborny, Australia

Subject: Replacing the Riemann integral with the gauge integral

It is only an accident of history that the Riemann integral is the one used in all calculus bookstoday. The gauge integral (also known as the generalized Riemann integral, the Henstock integral,the Kurzweil integral, the Riemann complete integral, etc.) was discovered later, but it is a ”better”integral in nearly all respects. Therefore, we would like to suggest that in the next edition of yourcalculus textbook, you present both the Riemann and gauge integrals, and then state theoremsmainly for the gauge integral.

This switch would only require altering a few pages in your calculus book. Any freshmancalculus book is devoted almost entirely to derivatives and antiderivatives – how to compute themand how to use them; that material would not change at all. The only changes would be in themore theoretical sections of the book – i.e., the definitions and theorems – which take up only afew pages.

The reasons for making this change are twofold:

(1) It would actually make some parts of your book more readable. Some definitions andtheorems can be stated more simply (and more strongly) if the gauge integral is usedinstead of the Riemann integral. This is particularly true of the Second FundamentalTheorem of Calculus, discussed below.

(2) It would be a better preparation for the handful of calculus students who will go on tohigher math courses. The gauge integral is far more useful than the Riemann integral, asa bridge to more advanced analysis.

The idea of introducing the gauge integral to college freshmen is not entirely new; it was pro-moted, for instance, in the paper ”The Teaching of the Integral” by Bullen and Vyborny, Journalof Mathematical Education in Science and Technology, vol. 21 (1990). However, we feel that theidea deserves wider promotion; hence this letter.

Introduction to the integral. If you are not already familiar with the gauge integral, wewould recommend the article by Bartle in the American Mathematical Monthly, October 1996;it provides a quick introduction to the subject and gives most of the main references. Perhapsa second introduction would be the book of DePree and Swartz (Introduction to Real Analysis,Wiley, 1988); it is written for advanced undergraduates and is probably the most elementaryamong the currently available introductions to this subject. Professors Bartle and Schechter have

49

volunteered to make themselves available, at least to some extent, to answer further questions youmay have about this subject.

The definition. The gauge integral is a very slight generalization of the Riemann integral. Insteadof a constant ε and a constant δ, it uses a constant ε and a function δ. The two definitions can beformulated so that they are nearly identical, and then placed side by side. This would be helpfulto the teachers who are using the book and learning about the gauge integral for the first time.

We think that the slight increase in the complexity of the definition will make little difference tothe students learning from the book. It has been our experience that, for the most part, freshmancalculus students do not fully grasp the definition of the Riemann integral anyway; it is just toocomplicated for students at that level. The definition of the Riemann integral is included in acalculus book more for completeness and integrity than for teaching. The few students who haveenough mathematical maturity to grasp the definition of the Riemann integral will probably haveno greater difficulty with the gauge integral, and they will benefit from being exposed to thisconcept in their calculus book.

Existence and nonexistence of integrals. In recent calculus books it is customary to statewithout proof that continuous functions and monotone functions on a compact interval are Riemannintegrable. The omission of the proofs is unavoidable – a proof would involve the completenessof the reals, uniform continuity on compact intervals, and other notions that are far beyond thereach of freshmen.

Any Riemann integrable function is also gauge integrable. Many more functions are gaugeintegrable, as will be evident later in this letter; see especially our remarks about the DominatedConvergence Theorem.

It is customary to give the characteristic function of the rationals as an example of a functionthat is not Riemann integrable. Most students don’t fully understand this example, because thecharacteristic function of the rationals is too unlike familiar examples to appeal to the intuitionof most students. Still, it is useful to include such an example in the book, because the studentcomes away convinced (perhaps through trust rather than through understanding) that not everybounded function on a closed bounded interval is Riemann integrable.

A book covering the gauge integral might still include the characteristic function of the rationals,as an example of a bounded function that is gauge integrable but not Riemann integrable on aclosed bounded interval.

Ideally, one would like to also include a concrete example of a bounded function that is notgauge integrable on a closed bounded interval. Even if it is complicated, the presence of suchan example would convince the student (again, through trust) that not every function is gaugeintegrable. Unfortunately, there are no concrete examples, as explained later in this letter. Thatis both a strength of the theory (essentially all functions are gauge integrable) and a weakness ofthe theory (it is hard to convey the concept of the few functions that are not gauge integrable).

The book might include a statement something like this: ”Heuristically, every bounded functionthat one is likely to encounter in applications is gauge integrable on any closed bounded interval.Although it is possible to prove mathematically that there exist bounded functions that are NOTgauge integrable on closed bounded intervals, such functions do not arise in practical applications.We shall discuss this further, briefly, in the appendix.” Some possibilities for the appendix arediscussed later in this letter.

The first fundamental theorem of calculus: derivatives of integrals. There are twoversions of this theorem – an elementary version and a more advanced version. The elementaryversion is this:

50

Suppose F (t) =´ taf(s)ds exists. Then F is continuous. Moreover, at every t where f(t) is

continuous, the derivative F ′(t) exists and equals f(t).

We don’t have to assume f is continuous everywhere. The theorem is not hard to prove. Aremark might be made that ”a more advanced and stronger version of this theorem can be foundin the appendix.” The more advanced version says:

Suppose F (t) =´ taf(s)ds exists. Then F is continuous. Moreover, at almost every t, the

derivative F ′(t) exists and equals f(t).

(”Almost every” would have to be defined in the appendix.) We don’t have to make anyassumption at all about continuity of the integrand, f . However, this more advanced version isharder to prove; you’d have to just state it without proof.

The second fundamental theorem of calculus: integrals of derivatives. Here the gaugeintegral offers a big advantage over the Riemann integral. Again there are two versions. Theelementary version is:

If f(t) = F ′(t) exists for all t in [a, b], then´ taf(s)ds exists and equals F (t)− F (a).

We don’t have to make any assumption of continuity of f . The proof of this theorem is easyenough to include in a freshman calculus book. In an appendix, after introducing countable sets,you could state and prove this advanced version:

Suppose F is continuous. Also suppose f(t) = F ′(t) exists for all but countably many points t in

[a, b]; define f arbitrarily at those points. Then´ taf(s)ds exists and equals F (t)− F (a).

The proof is not very hard. (By the way, in this theorem, ”for all but countably many points”cannot be replaced by ”except on a set of measure 0”; that is shown by the Cantor Function.)

Convergence theorems. Calculus books traditionally have included the theorem that uniformlimits preserve integrability and integrals. However, for the gauge integral we can state this muchnicer and stronger Dominated Convergence Theorem:

Suppose f1, f2, f3, . . . are gauge integrable functions on [a,b], converging pointwise to a limitfunction f . Suppose that e ≤ fn ≤ g for all n, where e and g are gauge integrable functions.

Then f is gauge integrable, and´ baf(t)dt = limn→∞

´ bafn(t)dt .

Thus, under rather simple and mild assumptions, pointwise limits preserve integrability andintegrals. This theorem can be used to demonstrate the integrability of a very wide class offunctions.

The Monotone Convergence Theorem and Dominated Convergence Theorem could be statedwithout proof, and then examples and exercises can be given. Applications can be found in anybook on the Lebesgue integral.

(Actually, the Monotone and Dominated Convergence Theorems are also true, in appropriateforms, for Riemann integrals. Some references can be found in a paper by Simons, Math. Intelli-gencer 17 (1995), 67-70. However, the proofs in that setting are not much shorter than the proofsin the more general setting of gauge or Lebesgue integrals, and so there is not much reason tobother with them in an introductory course.)

51

Improper integrals. A separate definition is not required. A statement such asˆ 1

0

1√tdt = lim

a↓0

ˆ 1

a

1√tdt

becomes a theorem, rather than a definition of the left side of the equation. Admittedly, we stillend up using the right side of the equation as a method for computing numerical values, but it isconceptually advantageous to have this equation as a theorem instead of a definition.

A separate definition can also be avoided for integrals over unbounded intervals. A statementsuch as ˆ ∞

1

t−2dt = limb↑∞

ˆ b

1

t−2dt

becomes a theorem, rather than a definition of the left side of the equation, if the definition of theintegral is formulated as in McLeod’s book – i.e., unbounded subintervals make 0 contribution tothe approximating Riemann sum. (Admittedly, this may be counterintuitive to beginners, so it isperhaps debatable whether this more general definition is desirable in a calculus textbook.)

Integration formulas. Any calculus book must include the substitution rule (i.e., change ofvariables or reparametrization in integrals) and integration by parts. Usually these are stated ascomputational formulas, with the regularity hypotheses glossed over. If one does wish to statethe regularity hypotheses explicitly, those hypotheses can be weakened slightly by switching fromRiemann integrals to gauge integrals.

Appendices. At the end of your book you may wish to add a few pages of material which has notcustomarily appeared in calculus books. This is more for the benefit of the teachers and occasionalgifted students than for the typical student. Here is an outline of some suggested material:

Countability. It would be helpful to discuss the difference between countable and uncountable sets.Examples would include the fact that the integers and the even integers are equipollent; Cantor’sproofs that the rationals are countable but the reals are not. The union of countably manycountable sets is countable. The gauge integral of the characteristic function of any countable setis 0. These results are not hard to prove.

Anticipating results below about measure, you might also prove that the Cantor set is uncount-able. (To simplify the exposition, remove the middle 8/10 instead of the middle 1/3; then you canuse decimal expansions instead of ternary expansions.)

A theorem of Cantor states that any set has smaller cardinality than its power set. The proofis is actually very short and easy, except that it is rather abstract.

Measure. A set is measurable if and only if its characteristic function is gauge integrable; then themeasure of the set is equal to the value of that integral. Easy examples show that this extends thenotion of length, and gives us a notion of ”size” for a very large class of subsets of the real line.Admittedly, defining size in terms of integrals is conceptually backward – many mathematiciansfind it more intuitive to define integrals in terms of measures – but for purposes of your calculusbook, this approach saves several pages.

Some basic properties can be stated, with or without proof. The union of countably manydisjoint measurable sets is measurable, and its measure is the sum of their measures; etc. If afunction vanishes outside a set of measure 0, then that function is gauge integrable and its integralis 0. Any countable set has measure 0. The Cantor set is an uncountable set with measure 0, sinceit is contained in simpler sets with arbitrarily small positive measure.

Unmeasurable sets. Vitali’s classical proof of the existence of an unmeasurable set is fairly shortand elementary. It can be found in any book on Lebesgue integration. A sketch of the proof might

52

be useful in the appendix to a calculus book. The typical calculus student would not understandthe proof, but a few students might understand a few parts of the proof. In any case, the studentwould go away believing (perhaps on trust rather than on understanding) that there do existunmeasurable sets.

Vitali’s proof uses the Axiom of Choice in a fairly simple and intuitive way. The expositor canpoint out the Axiom of Choice when it is used, and emphasize that the proof is not constructive.It can be stated that there are no constructive ways of obtaining unmeasurable sets. (The proof ofthe fact that there are no constructive ways is rather difficult; it involves much set theory and logic.It is discussed in part of Schechter’s book, Handbook of Analysis and its Foundations, AcademicPress, 1996.) Thus, we know that there exist bounded functions that are not gauge integrable ona closed bounded interval, but we cannot produce explicit examples of such functions.

In summary, we feel that the changes in the calculus book would not be major but would improvethe teaching of calculus. We invite your questions on these matters.

Sincerely,

Robert Bartle, Ralph Henstock, Jaroslav Kurzweil,Eric Schechter, Stefan Schwabik, and Rudolf Vyborny

53

Acknowledgements

First and foremost I would like to thank my supervisor Per Ahag, without him this essay wouldnot be what it is. His guidelines and enthusiasm has been priceless. Secondly I would like to thankmy examiner Lisa Hed, for reading my essay and giving me constructive feedback.

55

References

[1] Anderson M., Katz V., Wilson W., Sherlock Holmes in Babylon and other tales of mathematical history. Edited

by Marlow Anderson, Victor Katz and Robin Wilson. MAA Spectrum. Mathematical Association of America,Washington, DC, 2004.

[2] Bartle R. G., A modern theory of integration. Graduate Studies in Mathematics, 32. American MathematicalSociety, Providence, RI, 2001.

[3] Bartle R. G., Sherbert D. R., Introduction to real analysis. Second edition. John Wiley & Sons, Inc., New York,

1992.[4] Cohn D. L., Measure theory. Second edition. Birkhauser Advanced Texts: Basler Lehrbucher. Birkhauser

/Springer, New York, 2013.

[5] Denjoy A., Une extension de l’integrale de M. Lebesgue. C. R. Acad Sci. Paris 154 (1912), 859-862.[6] Gordon R. A., The integrals of Lebesgue, Denjoy, Perron, and Henstock. Graduate Studies in Mathematics, 4.

American Mathematical Society, Providence, RI, 1994.

[7] Hawkins T., Lebesgue’s theory of integration. Its origins and development. AMS Chelsea Publishing, 1975.Second AMS printing 2002

[8] Henstock R. The efficiency of convergence factors for functions of a continuous real variable. J. London Math.

Soc. 30 (1955), 273-286.[9] Kurzweil J. Generalized ordinary differential equations and continuous dependence on a parameter. Czechoslo-

vak Math. J. 7 (82) (1957) 418-449.

[10] Parzynski W. R., Zipse P. W., Introduction to mathematical analysis. International Series in Pure and AppliedMathematics. McGraw-Hill Book Co., New York, 1982.

[11] Perron O., Uber den inIntegralbegriff. Stizber. Heidelberg Akad. Wiss., Math.-Naturw. Klasse Abt. A 16 (1914),1-16.

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on the fundamental theorem of calculus - diva...

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