sequences of functions : different notions of convergence ...1218166/fulltext01.pdf · sequences of...

Sequences of Functions : Di�erent Notions of

Convergence and How They Are Related

Department of Mathematics, Linköping University

Erik Sätterqvist

LiTH-MAT-EX�2018/06�SE

Credits: 16 hp

Level: G2

Supervisor: Jana Björn,Department of Mathematics, Linköping University

Examiner: Anders Björn,Department of Mathematics, Linköping University

Linköping: June 2018

Abstract

In this thesis we examine di�erent types of convergence for sequences of functionsand how these are related. The functions considered are Lebesgue measurablefunctions f : A→ R, where A ⊂ R. We also devote a chapter to explore whencontinuity of a sequence of functions is preserved under pointwise convergence,and see that this happens precisely when the convergence is quasi uniform.

Keywords:

Sequences of functions, convergence, quasi uniform convergence, Lp-spaces

URL for electronic version:

http://urn.kb.se/resolve?urn=urn:nbn:se:liu:diva-148418

Sätterqvist, 2018. iii


Sammanfattning

I denna uppsats utforskar vi olika typer av konvergens för funktionsföljder för attse hur de är besläktade. Funktionerna i fråga är Lebesguemätbara funktioner f :A→ R, där A ⊂ R. Vi ägnar också ett kapitel åt att undersöka när kontinuitethos en följd av funktioner bevaras under punktvis konvergens och ser att dettahänder precis då konvergensen är kvasilikformig.

Nyckelord:

Funktionsföljder, konvergens, kvasilikformig konvergens, Lp-rum

URL för elektronisk version:


Sätterqvist, 2018. v


Acknowledgements

First of all I would like to thank my supervisor Jana Björn for suggesting thetopic of this thesis, and for all the help along the way. Your tips, both stylisticand mathematical, have elevated this thesis. I would also like to thank myexaminer Anders Björn. Under your keen eyes no mistakes, grammatical orstructural, can hope to survive. For the help with expelling minor �aws andinconsistencies Erik Landstedt deserves my thanks as well. Finally I wouldlike to thank my opponents Simon Calderon and Jakob Lundström for theirconstructive criticism and helpful comments.

Sätterqvist, 2018. vii

Nomenclature

Number Sets

C The complex numbers.

N The natural numbers, i.e. {0, 1, . . . }, N∗ := N \ {0}.

Q The rational numbers.

R The real numbers.

Z The integers.

Convergence

a.e.→ Almost everywhere convergence.

a

⇒ Almost uniform convergence.

⇒ Uniform convergence.

µ−→ Convergence in measure µ.

q

⇒ Quasi-uniform convergence.

→ Pointwise convergence or convergence in norm.

∗⇁ Weak-star convergence.

⇁ Weak convergence.

Set Theory

#A The cardinality of A.

χA The characteristic function on a set A.

Sätterqvist, 2018. ix

x

Ac The complement of A.

P(A) The power set of A, i.e. the set of all subsets of A.

A The closure of A.

A tB The union of two disjoint sets A and B.

A ⊂ B A is a subset of B (with A = B possible).

A ( B A ⊂ B and A 6= B i.e. A is a proper subset of B.

A◦ The interior of A.

Metric Spaces

(X, d) A metric space with metric d.

C(I) := {continuous f : I → R : µ(supp(f)) <∞}.

B(x, r) := {y : d(x, y) ≤ r}, the closed ball with radius r around x.

B(x, r) := {y : d(x, y) < r}, the open ball with radius r around x.

Lp The Lp-spaces, 1 ≤ p ≤ ∞.

p, q Conjugate exponents i.e. 1/p+ 1/q = 1.

Measure Theory

`(I) The length of an interval I.

B The Borel-sets on R.

M The Lebesgue measurable sets.

µ A measure.

µ∗ An outer measure.

f ∼ g The measurable functions f and g are equivalent i.e. f = g a.e.

Other Symbols

P The set of partitions of the interval [a, b].

∃ There exists.

∀ For every.

xi

dxe := min{n ∈ Z : n ≥ x}, the smallest integer not smaller than x.

bxc := max{n ∈ Z : n ≤ x}, the largest integer not bigger than x.

supp(f) := {x : f(x) 6= 0}, the support of f .

T A topology.

Contents

1 Introduction 1

2 Preliminaries 3

2.1 Numerical Sequences . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2.1 Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . . 162.2.2 Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.3.1 Topology and Metric Spaces . . . . . . . . . . . . . . . . . 22

2.4 Measure Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.4.1 The Lebesgue Measure . . . . . . . . . . . . . . . . . . . . 26

2.5 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . 282.6 The Lebesgue Integral . . . . . . . . . . . . . . . . . . . . . . . . 33

2.6.1 The Lp-spaces . . . . . . . . . . . . . . . . . . . . . . . . 362.7 List of Convergences . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.7.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.7.2 Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . . 392.7.3 Measure Theory . . . . . . . . . . . . . . . . . . . . . . . 41

3 Quasi uniform Convergence : Preserving Continuity 43

4 Relation Between Convergences 51

5 Further Research 67

A Convex Functions 69

Sätterqvist, 2018. xiii

List of Figures

2.1 Convergent numerical sequence . . . . . . . . . . . . . . . . . . . 52.2 Divergent numerical sequence . . . . . . . . . . . . . . . . . . . . 62.3 Numerical sequence that tends to +∞ . . . . . . . . . . . . . . . 72.4 Numerical sequence with no limit . . . . . . . . . . . . . . . . . . 92.5 Numerical sequence with no limit and no convergent subsequence 102.6 Unbounded numerical sequence with convergent subsequence . . 112.7 R \ {α} is not complete . . . . . . . . . . . . . . . . . . . . . . . 132.8 The Cantor set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.9 An example of a step function . . . . . . . . . . . . . . . . . . . . 292.10 Illustration of the Riemann integral . . . . . . . . . . . . . . . . . 302.11 A functions which is not Riemann integrable . . . . . . . . . . . 312.12 The space C([a, b]) is not complete . . . . . . . . . . . . . . . . . 332.13 Illustration of the Lebesgue integral . . . . . . . . . . . . . . . . 34

3.1 Continuous functions fn → f which is not continuous . . . . . . 443.2 Convergent fn ⇒ f which is not uniformly convergent . . . . . . 453.3 Continuous pointwise limit f with fn 6⇒ f . . . . . . . . . . . . . 463.4 Example of quasi uniform convergence . . . . . . . . . . . . . . . 49

4.1 Almost uniform but not L∞-convergent sequence . . . . . . . . . 554.2 Almost uniform but not Lp-convergent sequence . . . . . . . . . 564.3 Sequence which converges in Lp and measure but not a.e. . . . . 594.4 Sequence a.e. convergent but not convergent in measure . . . . . 604.5 Sequence weakly convergent but not convergent in Lp . . . . . . 644.6 Positive results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.7 Negative results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

A.1 Illustration of Lemma 4.9 . . . . . . . . . . . . . . . . . . . . . . 71

Sätterqvist, 2018. xv

Chapter 1

Introduction

Sätterqvist, 2018. 1

2 Chapter 1. Introduction

Sequences are essentially in�nite lists of objects. They are one of the funda-mental tools we have in analysis and are used everywhere. For example, wecan use sequences to determine whether a function is continuous, characterizeclosedness of sets (at least on metric spaces) and much more.

The limit of a sequence (whether or not it exists) describes the behavior at the�end� of the list. Speci�cally a sequence converges to a value if it, in some sense,�approaches� this value. For example, we can assign values to in�nite sums

∑∞1

by considering the limit of the partial sums Sn :=∑n

1 (This is the usual inter-pretation of what

∑∞1 means but not the only one). Similarly, one can assign

values to in�nite sums of the type∑∞−∞ or in�nite products.

Another natural application of convergence is approximating elements that are�far away� in a sequence. Consider a calculation using a data container of somesize N , with a calculation error eN that depends on N . If the limit of (eN ) iszero, then we can expect the error to be small for large N . How large N that isneeded for an acceptable error could then be determined numerically.

In this thesis we will consider sequences of functions. There are various notionsof convergence one could consider, from basic to more advanced, each highlight-ing di�erent properties of our functions. Quasi uniform convergence for examplewas developed to study the continuity of pointwise convergent sequences.

The convergences chosen for this thesis are introduced at the end of Chapter 2.In Chapter 3 we consider the less common quasi uniform convergence in moredetail, exploring the motivation for, and properties of, this convergence. Finallyin Chapter 4 the relations between the di�erent types of convergence are ex-amined. We formulate and prove theorems for the positive results, and providecounterexamples for the negative results.

Chapter 2

Preliminaries


4 Chapter 2. Preliminaries

Let us start by de�ning what a sequence is.

De�nition 2.1. (Abbott [1], De�nition 2.2.1) A sequence is a functionx : N→ X, where X is some set. We will usually write xn instead of x(n).

This means that a sequence is essentially an ordered (countable) list of values.We will denote sequences by this list x0, x1, . . . or simply (xn).

Example 2.2. An easy example of a (numerical) sequence is 1,−1, 1, . . ., whichcan also be written ((−1)n).

De�nition 2.3. Given a sequence (xn) a subsequence is the composed functionxσ(n), where σ : N→ N is strictly increasing.

A trivial example of a subsequence would be xk, xk+1, . . . induced by a right-shift σ(n) = n + k, with k ∈ N. This does not change the properties we areinterested in (namely limits) and we will therefore not distinguish between theseshifted subsequences and the sequence from whence they came, denoting themas we have above or by (xn)

∞n=k. Another less trivial subsequence is acquired

by letting σ(n) = 2n. Doing this in Example 2.2 yields the constant sequence(1n).

In this thesis we will be studying the convergence of sequences, i.e. if a sequence(xn) �approaches� an x as n → ∞. It seems clear that any constant sequence(x) should converge to x. To move beyond constant sequences however, we needto de�ne what �approach� means, and to do this we need to know more aboutX. The starting point for our journey will be the (hopefully) familiar settingX = R.

2.1 Numerical Sequences

De�nition 2.4. A numerical sequence is a sequence x : N → R. Such asequence (xn) converges to y ∈ R if

∀ε > 0 ∃N ∈ N ∀n ≥ N : |xn − y| < ε,

and we write xn → y. If there is such a y the sequence is convergent, otherwiseit is divergent.

Note that using the above de�nition any constant numerical sequence (α), α ∈ Rconverges to α (simply let N = 0). Let us give a more involved example of aconvergent sequence.

2.1. Numerical Sequences 5

Example 2.5. Let xn = 1/n for n ∈ N∗. We will show, using the de�nition,that this sequence is convergent and converges to 0. Let ε > 0 and take N ∈ Nsuch that N > 1/ε. Then 1/N < ε, so

∀n ≥ N |xn − 0| = 1

n≤ 1

N< ε.

Hence |xn − 0| < ε so xn → 0 and we are done.

n

1 2 3 5 6N

y

0· · ·

y = xn

y = ε

Figure 2.1: The sequence (xn) = (1/n) converges to 0.

Using De�nition 2.4, what do divergent sequences look like?

Example 2.6. Let (xn) = (n). Assume that xn converges to some x ∈ R. Byde�nition there then exists an N ∈ N such that |xn−α| < 1/2 whenever n ≥ N .Letting n ≥ N we have that

|xn+1 − xn| = |(xn+1 − α)− (xn − α)| ≤ |xn+1 − α|+ |xn − α|< 1/2 + 1/2 = 1.

However, we also know that

|xn+1 − xn| = n+ 1− n ≥ 1 6< 1.

Thus we have a contradiction, so (xn) is divergent.


n

1 2 3 4

y

0· · ·

y = xn

Figure 2.2: The sequence (xn) = (n) is divergent.

We have established that the sequence (n) is divergent, that is it does notapproach any number. However, it is strictly increasing, so one could say thatit approaches +∞. This motivates the next de�nition.

De�nition 2.7. If (xn) converges to x ∈ R, then x is the limit of (xn). We saythat +∞ is the limit of (xn) if

∀ω > 0 ∃N ∈ N ∀n ≥ N : xn ≥ ω.

Finally −∞ is the limit of (xn) if +∞ is the limit of (−xn). If x ∈ R∪{±∞} isthe limit of (xn) we say that (xn) tends to x, and write xn → x or x = limxn.

This essentially extends De�nition 2.4 to include the values ±∞. With De�ni-tion 2.7 in mind let us return to Example 2.6. Let ω > 0, set N := dωe and letn ≥ N . Then xn = n ≥ N ≥ ω, implying xn → +∞.

With our de�nitions �rmly established we consider some properties of numericalsequences.


n

1 2 4

y

0· · ·N

y = xn

y = ω

Figure 2.3: The sequence (xn) = (n) tends to +∞.

Proposition 2.8. The limit of a numerical sequence (xn) is unique.

Proof: We divide the proof into three cases.

(i) Assume that xn → x ∈ R and xn → y ∈ R. Let ε > 0, then there existN1, N2 ∈ N such that

∀n ≥ N1 : |xn − x| < ε/2 and ∀n ≥ N2 : |xn − y| < ε/2.

Now let n ≥ N := max{N1, N2} be arbitrary, then

|x− y| = |(x− xn)− (y − xn)| ≤ |x− xn|+ |y − xn| < ε/2 + ε/2 = ε.

Hence |x− y| = 0 since ε was arbitrary, but this implies x = y and we aredone.

(ii) Assume that xn → x ∈ R and xn → +∞. Let ε = 1/2 and ω = |x| + 1,then there exists an N ∈ N such that xn ≥ |x| + 1, for any n ≥ N . Butthen |xn − x| ≥ 1 ≥ ε for every n ≥ N , implying xn 6→ x. This is acontradiction and thus xn converges to at most one of +∞ and x. Thecase xn → −∞ is similar.

(iii) Assume that xn → +∞ and xn → −∞. Let ω > 0, then there exists anN ∈ N such that xn ≥ ω whenever n ≥ N . This means that xn 6→ −∞which is a contradiction. Thus xn converges to at most one of ±∞, whichconcludes the proof.


Proposition 2.9. If a numerical sequence (xn) tends to some value x ∈ R∪ {±∞},then all subsequences tend to x as well. Speci�cally if (xn) is convergent, thenall of its subsequences are convergent as well, and converge to the same numberas (xn).

Proof: First assume that xn → x ∈ R. Let xσ(n) be a subsequence of x andtake ε > 0. Then by de�nition there exists an N ∈ N such that |xn − x| < ε,whenever n ≥ N . Note that σ is strictly increasing on N, so σ(n) ≥ n for anyn ∈ N. Thus letting n ≥ N we have that |xσ(n) − x| < ε as desired.

Next assume that xn → +∞. Let xσ(n) be a subsequence of x and take ω > 0.Then by de�nition there exists an N ∈ N such that xn ≥ ω, whenever n ≥ N .Again letting n ≥ N we acquire xσ(n) ≥ ω as desired. The proof for xn → −∞is similar.

Combining Proposition 2.8 with Proposition 2.9 we have the following usefulresult.

Corollary 2.10. If a numerical sequence (xn) has two subsequences with dif-ferent limits, then it is divergent.

Next let us have a look at a simple divergent sequence.

Example 2.11. Let (xn) = ((−1)n), we know from before that (1n) is a sub-sequence which converges to 1. Another subsequence is (x2n+1) = (−1), whichobviously converges to −1. Thus we have two subsequences converging to dif-ferent values, so by Corollary 2.10 (xn) is divergent.

In this example the divergent sequence at least had convergent subsequences.However, applying Proposition 2.8 to Example 2.6, we see that the sequence(n) does not have any convergent subsequences (they all tend to +∞). If wemodify Example 2.11 a little bit we can even get a sequence with no limit andno convergent subsequence.

Example 2.12. Consider the sequence (xn) = (n(−1)n), the subsequences(x2n) = (2n) and (x2n+1) = (−2n − 1) have limits +∞ and −∞ respectively.Note that since we have two subsequences with di�erent limits the limit of (xn)does not exist by Corollary 2.10. Next we will show that there is no convergentsubsequence.


n

1 2 3 4 5 6 7

y

0· · ·

y = x2n

y = x2n+1

Figure 2.4: The sequence (xn) = ((−1)n) has no limit.

Assume that there is a convergent subsequence (xσ(n)), then any subsequenceof this subsequence converges. There are either in�nitely many odd or evenelements in (σ(n)). Assume that there are in�nitely many even elements andlet (xσ′(n)) be a subsequence of (xσ(n)) such that σ′(n) is always even. Then(xσ′(n)) is also a subsequence of (x2n), but this sequence has limit +∞, andtherefore no convergent subsequence, leading to a contradiction. Similarly weacquire a contradiction if there are in�nitely many odd elements in (σ(n)). Thus(xn) can have no convergent subsequence.

The two examples of sequences without convergent subsequences had one thingin common. The numbers of the sequence got very big or very negatively big.To be more precise we introduce the following de�nition.

De�nition 2.13. A sequence is bounded if

∃M > 0 ∀n : |xn| ≤M.

Otherwise it is unbounded.

Clearly the sequences in Example 2.6 and Example 2.12 where unbounded, butdo all unbounded sequences lack convergent subsequences? The answer is nowhich the next, quite simple, example will show.


n

1 2 3 4 5 6 7

y

0· · ·

y = xny = xσ(n)y = xσ′(n)

Figure 2.5: The sequence (xn) = (n(−1)n) has no limit and no convergentsubsequence.

Example 2.14. Let (xn) be the numerical sequence given by x2n = 2n andx2n+1 = 1 for n ∈ N. Clearly (xn) cannot be bounded since it has an unboundedsubsequence (x2n). However, it also has the obviously convergent subsequence(x2n+1) = (1n) as seen in Figure 2.6.

What if we have a convergent sequence? Does it have to be bounded? This isanswered by our next proposition.

Proposition 2.15. Any convergent numerical sequence is bounded.

Proof: Let (xn) be a numerical sequence converging to x ∈ R. By de�nitionthere then exists an N ∈ N such that |xn − x| < 1, whenever n ≥ N . Lettingn ≥ N and applying the triangle inequality we have that |xn| ≤ |x| + 1. Nextset M := maxm≤N |xm|. We then have that |xm| ≤ max{|x| + 1,M}, for anym ∈ N. Thus (xn) is bounded and we are done.

2.2. Metric Spaces 11

n

1 2 3 4 5 6 7

y

0· · ·

y = x2n

y = x2n+1

Figure 2.6: The sequence (xn) with x2n = n, x2n+1 = 1 is unbounded but hasa convergent subsequence.

Recall that our sequence with no convergent subsequence in Example 2.12 wasunbounded. In fact, it was the very reason we introduced bounded and un-bounded sequences. This begs the question, are there any bounded sequenceswho lack convergent subsequences? The answer is no as stated by the followingtheorem.

Theorem 2.16. Any bounded numerical sequence has a convergent subsequence.

This does not hold in more general settings, but it is a special case of the moregeneral Theorem 2.60 discussed in Section 2.3.

2.2 Metric Spaces

Feeling comfortable with numerical sequences we move on to the more generalsetting of metric spaces. Remember that what we need in order to de�ne con-vergence is a notion of what it means for a sequence to �approach� an element.In the case of our numerical sequences, a sequence approached a real number ifit got arbitrarily close for big n, i.e. if the Euclidean distance was small. We willnow replace this Euclidean distance with other notions of distance, and allowother sets than R, to generalize our notion of convergence.


De�nition 2.17. (Kreyzig [7], De�nition 1.1-1) A metric is a functiond : X ×X → R such that for all x, y, z ∈ X the following hold

(i) d(x, y) = d(y, x).

(ii) d(x, y) ≥ 0.

(iii) d(x, y) = 0 ⇐⇒ x = y.

(iv) d(x, z) ≤ d(x, y) + d(y, z).

A metric space is a pair (X, d ), where X is a set and d is a metric.

Remark 2.18. Notice that if Y ⊂ X, then d|Y is a metric on Y , so (Y, d|Y ) is ametric space.

Example 2.19. Let X be a set and de�ne a metric by

d(x, y) :=

{1, if x 6= y,

0, if x = y.

Clearly (X, d ) is a metric space. Moreover B(x, 1) = X for any x ∈ X.

Now that we have distance, we introduce a notion of convergence.

De�nition 2.20. (Kreyzig [7], De�nition 1.4-1) Let (X, d ) be a metric space.A sequence (xn) ⊂ X converges to x if

∀ε > 0 ∃N ∈ N ∀n ≥ N : d(xn, x) < ε.

We say that x is the limit of x and write xn → x or x = limxn. If there is nosuch x, then we say that the sequence is divergent.

Before we move on let us consider a very familiar metric space.

Example 2.21. The setX = R equipped with the Euclidean distance d(x, y) :=|x − y|, x, y ∈ R is a metric space. It is easily seen that d is a metric, indeedmetrics are a way of generalizing this notion of distance that we have on R toother sets.

Having established that R is a metric space we realize that what was callednumerical sequences in the previous section are actually sequences in a metricspace. This makes us wonder if the properties established for convergence ofnumerical sequences can be generalized to sequences in metric spaces. This isindeed the case and the results are summarized in the following proposition.


Proposition 2.22. Let (xn) be a sequence in some metric space X, then

(i) The limit of (xn) is unique.

(ii) If (xn) converges to x ∈ X, then all subsequences converge to x as well.

(iii) If (xn) has two subsequences with di�erent limits, then it is divergent.

The proofs of (i)-(ii) are completely analogous to the proofs of Proposition2.8 and Proposition 2.9. Simply replace the Euclidean distance with a generalmetric d. Part (iii) then follows directly from (i)-(ii).

Since R is a metric space Remark 2.18 tells us that taking subsets we acquirea whole family of metric spaces {A : A ⊂ R}. One of these is the topic of ournext example.

Example 2.23. Consider the metric spaceX = R\{α}, where α ∈ R , equippedwith the Euclidean distance. De�ne a sequence (xn) in X by xn = α + 1/n,n ∈ N∗. In R this sequence converges to α. In X, however, we have removedα, and since limits are unique by Proposition 2.22, the sequence (xn) cannotconverge to any β ∈ X. Therefore the sequence is divergent in X.

α α+ 1

X = R \ {α} xn = α+ 1/n

Figure 2.7: The sequence xn is divergent.

Evidently for a metric space X one could have sequences that are divergent inX, but convergent if we replace X with some �larger� set. Our next de�nitioncaptures precisely these sequences.

De�nition 2.24. (Kreyzig [7], De�nition 1.4-3) Given a metric space (X, d ), aCauchy sequence is a sequence (xn) such that

∀ε > 0 ∃N ∈ N ∀n,m ≥ N : d(xn, xm) < ε.

The space is said to be complete if all Cauchy sequences converge in X.

Cauchy sequences have the following important properties.


Proposition 2.25. If a Cachy sequence (xn) in a metric space (X, d ) has asubsequence which converges to some x ∈ X, then (xn) converges to x as well.

Proposition 2.26. All Cauchy sequences are bounded.

Proof of Proposition 2.25: Let (X, d ) be a metric space and let (xn) be a Cauchysequence in this space. Assume that (xσ(n)) is a subsequence converging tox ∈ X. Let ε > 0. By de�nition there exists an N1 ∈ N such that

∀n ≥ N1 : |xσ(n) − x| < ε/2.

Also since xn is a Cauchy sequence there exists an N2 ∈ N such that

∀n,m ≥ N2 : |xn − xm| < ε/2.

Letting n ≥ max{N1, N2}, we have that

|xn − x| = |(xn − xσ(n))− (xσ(n) − x)| ≤ |xn − xσ(n)|+ |xσ(n) − x| < ε.

This proves that xn → x.

Proof of Proposition 2.26: Let (X, d ) be a metric space and let (xn) be a Cauchysequence in this space. By de�nition there exists an N ∈ N such that

∀m,n ≥ N : |xm − xn| < 1. (2.1)

Fix n ≥ N , then∀m ≥ N : |xm| < |xn|+ 1.

Finally set C := maxm<N |xm|. We have that

∀m ∈ N : |xm| ≤ max{C, |xn|+ 1}.

Thus (xn) is bounded.

Remark 2.27. From these two results it follows that R is complete. Let (xn) bea numerical Cauchy sequence, then by Proposition 2.26 (xn) is bounded. NextProposition 2.15 implies that (xn) has a convergent subsequence. Thus (xn)converges by Proposition 2.25.

Completeness is often a desirable trait for metric spaces, which begs the ques-tion, does there exist a way to �complete� noncomplete metric spaces? In Ex-ample 2.23, we saw that the space X could be �completed� simply by adding α.Indeed this method of �lling in the limits of nonconvergent Cauchy sequenceswill always yield a complete space. Before we move on to the theorem though,let us consider a more complicated noncomplete metric space.


Example 2.28. Consider the set c00 of numerical sequences x = (x(k))∞k=1, forwhich #{x(k) 6= 0 : k ∈ N∗} <∞. Equip this set with the metric

d(x, y) =

∞∑k=1

|x(k)− y(k)|.

The resulting metric space is not complete. Consider the sequence of sequences(xn) de�ned by

xn(k) :=

{1/k2, if k < n,

0, if k ≥ n,n ∈ N∗.

Obviously the sequence (xn) lies in c00. Next we will show that it is Cauchy.Let ε > 0. Since

∞∑k=1

1

k2=π2

6<∞,

there exists an N ∈ N∗ such that∑∞k=N

1k2 < ε. Let m > n ≥ N , then

d(xm, xn) =

∞∑k=1

|xm(k)− xn(k)| =m−1∑k=n

1

k2≤∞∑k=N

1

k2< ε.

Thus the sequence (xn) is Cauchy. However, it converges to the sequence(1/k2)∞k=1 which does not belong to c00. Thus c00 is not complete.

De�nition 2.29. (Kreyzig [7], De�nition 1.6-1) Let (X, d ) and (X̃, d̃) be met-ric spaces. A function T : X → X̃ is an isometry if it preserves distance,i.e.

∀x, y ∈ X : d(x, y) = d̃(Tx, Ty).

If the isometry T is bijective, then the spaces X and X̃ are isometric, and wewrite X ∼= X̃.

One should think of isometric spaces as being essentially the same.

Theorem 2.30. (Kreyzig [7], Theorem 1.6-2) Let (X, d ) be a metric space.

Then there exists a complete metric space (X̂, d̂) with a dense subspace (X̃, d̃),

such that X ∼= X̃. The space (X̂, d̂) is unique up to isometries.


2.2.1 Normed Spaces

We now impose a bit more structure on our metric spaces.

De�nition 2.31. (Kreyzig [7], De�nition 2.1-1) A vector space over R is a setX together with two binary operations, called addition + : X × X → X andmultiplication by scalar ∗ : R × X → X, such that for all x, y, z ∈ X andα, β ∈ R

A1 x+ y = y + x.

A2 x+ (y + z) = (x+ y) + z.

S1 α ∗ (β ∗ x) = (αβ) ∗ x.

S2 1 ∗ x = x.

S3 α ∗ (x+ y) = α ∗ x+ α ∗ y.

S4 (α+ β) ∗ x = α ∗ x+ β ∗ x.

Moreover there exists a unique neutral element 0 ∈ X called the zero vectorsuch that for all x ∈ X.

A3 x+ 0 = x.

A4 There exists an additive inverse (−x) ∈ X i.e. x+ (−x) = 0.

The symbol ∗ will be omitted, writing α ∗ x as simply αx.

De�nition 2.32. (Kreyzig [7], De�nition 2.2-1) Let X be a vector space overR, then ‖ · ‖ : X → R is a norm on X if for all x, y ∈ X and α ∈ R

(i) ‖x‖ ≥ 0.

(ii) ‖αx‖ = |α| ‖x‖.

(iii) ‖x+ y‖ ≤ ‖x‖+ ‖y‖.

(iv) ‖x‖ = 0 ⇐⇒ x = 0.

The pair (X, ‖ ·‖) is called a normed space. For x, y ∈ X, the function d(x, y) :=‖x−y‖ is a metric X. Thus (X, d ) is a metric space. If we drop the requirement(iv), then ‖·‖ is a seminorm, and it no longer induces a metric. If X is completeregarded as a metric space, then it is a Banach space.


Example 2.33. Perhaps the easiest example of a normed space is Rn endowedwith the Euclidean norm

‖x‖ :=( n∑j=1

x2j

)1/2

.

Example 2.34. The space

B([a, b]) :=

{f : [a, b]→ R : sup

x∈[a,b]|f(x)| <∞

},

with norm ‖f‖ = supx∈[a,b] |f(x)| is a normed space. We will show that it isalso Banach. Let (fn) be a Cauchy sequence in B([a, b]). Take ε > 0, then

∃N ∈ N ∀n,m ≥ N : ‖fn − fm‖ ≤ ε/2.

Fix x ∈ [a, b], then

∀m,n ≥ N : |fn(x)− fm(x)| ≤ ‖fn − fm‖ ≤ ε/2. (2.2)

Thus (fn(x)) is a Cauchy sequence in R. Since R is complete the limit existsand we set f(x) := limn→∞ fn(x). Let n,m ≥ N , letting m → ∞ in (2.2) weacquire

|fn(x)− f(x)| ≤ ε/2,

and since this holds for all x ∈ R we have that

‖fn − f‖ ≤ ε/2 < ε.

Thus fn → f , and by the triangle inequality ‖f‖ < ‖f‖n + ε, so f ∈ B([a, b]).Hence B([a, b]) is Banach.

Having introduced the extra structure of a vector space we wonder if any normedspace X can be completed to obtain a Banach space. That is, can we applyTheorem 2.30 in a way that preserves the norm structure? The answer is yes.

Theorem 2.35. (Kreyzig [7], Theorem 1.6-2) Let (X, ‖ · ‖) be a normed space.

Then there exists a Banach space X̂ with a dense subspace (X̃, d̃) , such that

X ∼= X̃. The space (X̂, d̂) is unique up to isometries.


2.2.2 Dual Spaces

We will now consider functions T : X → Y , where X and Y are normed. Thesefunctions will be referred to as operators, and a class of these be used to formnormed spaces. In particular a certain type of these operators will always giveus Banach spaces. But let us start from the beginning.

De�nition 2.36. (Kreyzig [7], De�nitions 2.6-1 and 2.8-1) Let X and Y bevector spaces over R. A linear operator is a function T : X → Y which is lineari.e.

∀x, y ∈ X, α ∈ R : T (x+ y) = T (x) + T (y) and T (αx) = αT (x).

If Y = R, then T is a linear functional.

De�nition 2.37. (Kreyzig [7], De�nitions 2.7-1 and 2.8-2) Let X and Y benormed spaces. A linear operator T : X → Y is bounded if

∃c > 0 ∀x ∈ X : ‖T (x)‖ ≤ c‖x‖.

If Y = R, then T is a bounded linear functional. Furthermore the set of boundedlinear operators from X to Y forms a vector space which we denote by B(X,Y ).

Note that this di�ers from the way bounded functions are de�ned on R. Bound-edness here instead means that bounded sets are mapped to bounded sets. Wealso have the following characterization.

Theorem 2.38. (Kreyzig [7], Theorem 2.8-3) A linear functional is continuousif and only if it is bounded.

Next we use our bounded operators to make a normed space.

Theorem 2.39. (Kreyzig [7], Theorem 2.10-1) Let X and Y be normed spaces.Then the vector space B(X,Y ) of all bounded linear functions from X to Y isa normed space, with norm de�ned by

‖T‖ := supx∈Xx6=0

‖Tx‖‖x‖

= supx∈X‖x‖=1

‖Tx‖‖x‖

.

Theorem 2.40. (Kreyzig [7], Theorem 2.10-2) Let X and Y be normed spaces.If Y is a Banach space, then B(X,Y ) is a Banach space.

Since R is complete the set of linear functionals on a normed space X is alwayscomplete. In fact, this space has its own name and notation.


De�nition 2.41. (Kreyzig [7], De�nition 2.10-3) Let X be a normed space.The Banach space B(X,R) is called the dual of X, and is denoted X∗. Thedual of the dual is called the bidual, denoted X∗∗.

In preparation for our next example we have the following de�nition.

De�nition 2.42. 1 ≤ p, q ≤ ∞ are conjugate exponents if 1/p+1/q = 1, where1/∞ := 0.

Note that for 1 < p, q <∞

1/p+ 1/q = 1 ⇐⇒ p =q

q − 1⇐⇒ q = p(q − 1).

For the rest of this thesis p and q will denote conjugate exponents.

Example 2.43. (Kreyzig [7], Example 2.10-6 and 2.10-7) Let 1 ≤ p <∞. De-�ne the spaces

lp := {x = (x(k))∞k=1 :

∞∑k=1

|x(k)|p <∞},

l∞ := {x = (x(k))∞k=1 : supk∈N∗

|x(k)| <∞},

with norms

‖x‖p :=( ∞∑k=1

|x(k)|p)1/p

and ‖x‖∞ := supk∈N∗

|x(k)|.

Then (lp)∗ ∼= lq, speci�cally (l1)∗ ∼= l∞.

De�nition 2.44. (Kreyzig [7], Section 4.6) Let X be a normed space. Thecanonical mapping J : X → X∗∗ is de�ned by Jx(f) := f(x), x ∈ X. This isan injective isometry and if J is surjective then X∗∗ is re�exive.

Example 2.45. (Kreyzig [7], Section 4.6) Let 1 < p < ∞. We know that(lp)∗ ∼= lq, wherefore (lp)∗∗ ∼= lp. This on its own does not show that lp isre�exive, since the isometry may not be given by the canonical mapping. Itdoes however turn out that lp is re�exive. The spaces l1 and l∞ on the otherhand, are not re�exive.


2.3 Topology

The aim of this section is to start with a general introduction to topology, andthen work our way to the usual topology on metric spaces (X, d ). Let us startby de�ning a topology.

De�nition 2.46. (Bass [2], De�nition 20.1) Let X be a set. A topology on Xis a family T of subsets of X such that

(i) ∅, X ∈ T .

(ii) If {Uα : α ∈ I} ⊂ T , where I is any index set (possibly uncountable),then

⋃α∈I

Uα ∈ T .

(iii) If U1, . . . , Un ∈ T , thenn⋂j=1

Uj ∈ T .

The pair (X, T ) is called a topological space. If a subset A of X is in T , then Ais open and if the complement Ac is in T , then A is closed.

Example 2.47. Let X be a set, then T1 := {∅, X} and T2 := P(X) are topolo-gies (called the trivial and discrete topology respectively). This is the smallestrespectively largest topology on X.

Remark 2.48. If (X, T ) is a topological space and Y ⊂ X, then the restrictionTY := {U ∩ Y : U ∈ T } is a topology on Y . This is called the relative topologyinduced on Y .

Remark 2.49. Let Ai be closed sets, then Aci are open. By De Morgan's laws

and De�nition 2.46 we therefore have that

(i) (⋂i∈I Ai)

c=⋃i∈I A

ci ∈ T , where I is an arbitrary index set.

(ii) (⋃ni=1Ai)

c=⋂ni=1A

ci ∈ T , n ∈ N∗.

i.e. arbitrary intersections and �nite unions of closed sets are closed.

This notion of open and closed sets may seem very di�erent from the way openand closed sets are usually de�ned on R, using interior and limit points. Theseconcepts will now instead be de�ned using De�nition 2.46.

2.3. Topology 21

De�nition 2.50. (Bass [2], Section 20.1) Let X = (X, T ) be a topologicalspace and let A ⊂ X.

(i) The union of all open subsets of A is the interior of A, denoted A◦. Apoint x ∈ A◦ is called an interior point.

(ii) The intersection of all closed sets containing A, denoted A, is the closureof A. If A = X then A is dense in X.

(iii) A neighborhood of A is an open set V ⊂ X such that A ⊂ V . Similarly aneighborhood of a point x ∈ X is an open set U ⊂ X such that x ∈ U .

Remark 2.51. Notice that the set A◦ is open since it is a union of open sets,thus it is the largest open set contained in A. But this means that A is open ifand only if A = A◦ i.e. when every point in A is an interior point.

Remark 2.52. The set A is closed since it is an intersection of closed sets. ThusA is the smallest closed set that contains A, so A is closed if and only if A = A.

So far we know what a topology is but have no way of describing it other thansimply listing the whole collection T . The following de�nition provides us witheasier building blocks.

De�nition 2.53. (Bass [2], Section 20.1) Let T be a topology and let σ be acollection of sets such that every element in T is a union of elements in σ. Thenσ is a basis for T .

Finally we consider continuity and compactness as topological properties.

De�nition 2.54. (Folland [5], Section 4.2) Let X,Y be topological spaces andlet f : X → Y . Then f is called continuous if the preimage f−1(V ) is open inX for every open V ⊂ Y . If x ∈ X then f is called continuous at x if f−1(V )contains a neighborhood of x for every neighborhood V of f(x).

Proposition 2.55. (Folland [5], Proposition 4.8) A function f : X → Y iscontinuous if and only if it is continuous at every point x ∈ X.

De�nition 2.56. (Kreyzig [7], Section A1.5) Let X be a set, an open cover iscollection of open sets {Gi : i ∈ I}, such that X ⊂

⋃i∈I G. If any open cover of

X has a �nite subcover, then X is compact.


2.3.1 Topology and Metric Spaces

How does this all look in the context of metric spaces? To begin with a basis forthe topology of a metric space is given by T := {B(x, r) : x ∈ X, r > 0}, thatis, all open sets can be written as unions of open balls. Thus a neighborhood ofa point x ∈ X is a set U such that B(x, r) ⊂ U for some r > 0. Next we have a(more familiar) characterization of continuity.

Theorem 2.57. Let (X, dX) and (Y, dY ) be metric spaces. A function f :X → Y is continuous at x0 ∈ X if and only if

∀ε > 0 ∃δ > 0 : dX(x, x0) < δ =⇒ dY (f(x), f(x0)) < ε.

Proof: ⇐= Let V ⊂ X be a neighborhood of f(x0) and take ε > 0 such thatB(f(x0), ε) ⊂ V . We know that

∃δ > 0 : dX(x, x0) < δ =⇒ dY (f(x), f(x0)) < ε.

This means that f(B(x0, δ)) ⊂ B(f(x0), ε) so B(x0, δ) ⊂ f−1(B(f(x0), ε)) ⊂f−1(V ). Thus f is continuous at x0 since B(x0, δ) is a neighborhood of x0.

=⇒ Take ε > 0, then f−1(B(f(x0), ε)) is a neighborhood of x0, thus itcontains B(x0, δ) for some δ > 0 and we are done.

We also have characterizations of closedness and compactness.

Theorem 2.58. (Kreyzig [7], Theorem 1.4-6) Let Y be a subset of a metricspace X. The set Y is closed if and only Y 3 xn → x ∈ X implies x ∈ Y .

De�nition 2.59. (Kreyzig [7], Section A1.5) Let Y be a subset of a metricspace X. If every sequence in Y has a convergent subsequence, then Y issequentially compact.

Theorem 2.60. (Kreyzig [7], Section A1.5) On metric spaces, compactness andsequential compactness are equivalent.

This is a very useful tool for characterizing compact sets in metric spaces. Dowe even have that the compact sets are the closed and bounded sets as on R?Our �nishing example answers this question with a clear no.

2.4. Measure Theory 23

Example 2.61. Consider the sequence (en)∞n=1 in l∞ given by

en(k) =

{1, if k = n,

0, if k 6= n,n ∈ N∗.

Note that ‖en‖ = 1 for each n ∈ N∗. Thus the sequence lies in the closed unitball B(0, 1), a set which is by de�nition closed and bounded. However, since

∀n,m ∈ N∗ : ‖xn − xm‖ = 1,

the sequence has no convergent subsequence.

2.4 Measure Theory

Measure theory is the theory of determining the �sizes� of sets. This will beneeded to construct the Lebesgue integral and it will be used to de�ne conver-gence in measure later on.

De�nition 2.62. (Bass [2], De�nition 2.1) Let X be a set. An algebra A is acollection of subsets of X such that

(i) ∅ ∈ A and X ∈ A.

(ii) A ∈ A implies Ac ∈ X.

(iii) If A1, . . . , An ∈ A then⋃nk=1Ak ∈ A and

⋂nk=lAk ∈ A.

A is a σ-algebra if in addition

(iv) If A1, A2, · · · ∈ A, then⋃∞k=1Ak ∈ A and

⋂∞k=1Ak ∈ A.

Lemma 2.63. (Bass [2], De�nition 2.7) Let {Aα}α∈I be a family of σ-algebras,where I is some (possibly uncountable) index set. Then A :=

⋂α∈I Aα is a σ-

algebra.

We will need this lemma for the following de�nition.

De�nition 2.64. (Bass [2], p. 8) Let C be a collection of subsets of X and set

σ(C) :=⋂{Aα : Aα is a σ-algebra, C ⊂ Aα}.

Then σ(C) is the σ-algebra generated by C and it is the smallest σ-algebra thatcontains the collection C.


Remark 2.65. By Lemma 2.63, σ(C) is a σ-algebra and the intersection is neverempty since C ⊂ P(X), which is a σ-algebra, so σ(C) is well de�ned.

An example of a generated σ-algebra � which will be important for this thesis� is the Borel σ-algebra on the real line. This is generated by the open sets onR, and will be denoted B. A natural question to ask is whether we need all ofthe open sets on R to generate B.

Proposition 2.66. (Bass [2], Proposition 2.8) The Borel σ-algebra B is gener-ated by each of the following collections of sets

(i) C1 = {(a, b) : a, b ∈ R}.

(ii) C2 = {[a, b] : a, b ∈ R}.

(iii) C3 = {(a, b] : a, b ∈ R}.

(iv) C4 = {(a,∞) : a ∈ R}.

We will prove (i) since it will be used later. However, �rst we need the followinglemma.

Lemma 2.67. Let A ⊂ R be an open set. Then A is a countable union ofdisjoint open intervals.

Proof. De�ne the following equivalence relation on A. If x, y ∈ A then x ∼ y if(min{x, y},max{x, y}) ⊂ A. This is an equivalence relation and the equivalenceclasses are open disjoint intervals. Let A be the set of equivalence classes, then

A =⊔I∈A

I. (2.3)

We are done if we can show that A is countable. Since Q is dense in R we havethat

∀I ∈ A ∃q ∈ Q such that q ∈ I.

But this means that #A ≤ #Q and thus A is countable.

Proof of Proposition 2.66: Notice that the set C1 is made up of open sets.Thus C1 ⊂ T , where T is the collection of open sets on R, and it followsthat σ(C1) ⊂ B. For the reverse inclusion we know that any countable unionof open intervals is in σ(C1), but by the previous lemma all elements of Tare countable unions of open intervals. Therefore we have T ⊂ σ(C1), henceB ⊂ σ(σ(C1)) = σ(C1) and the proof is complete.


De�nition 2.68. (Bass [2], De�nition 3.1) Let X be a set and let A be a σ-algebra consisting of subsets of X. A measure on (X,A) is a function µ : A →[0,∞] such that

(i) µ(∅) = 0.

(ii) If A1, A2, · · · ∈ A are pairwise disjoint then

µ

( ∞⊔j=1

Aj

)=

∞∑j=1

µ(Aj).

If in addition µ(X) < ∞, the measure is called a �nite measure. The triple(X,A, µ) is called a measure space and the sets in A are called µ-measurable orsimply measurable.

Example 2.69. Let X = N and set m(A) := #A. We see that m(∅) = 0, and

m

∞⊔j=1

Aj

=∑

a∈Aj , j∈N∗1 =

∞∑j=1

∑a∈Aj

1 =

∞∑j=1

m(Aj).

Thus m is a measure. However, it is not �nite since m(N) = #N = ∞. Notealso that only the empty set has measure zero.

Let us list some basic properties of a measure.

Proposition 2.70. (Bass [2], Proposition 3.5) Let µ be a measure on (X,A).Then the following hold.

(i) If A,B ∈ A : A ⊂ B then µ(A) ≤ µ(B).

(ii) If A1, A2, · · · ∈ A then µ(⋃∞

j=1Aj

)≤∑∞j=1 µ(Aj).

How does one construct a measure? The next de�nition gives us a startingpoint.


De�nition 2.71. (Bass [2], De�nition 4.1) Let X be a set. An outer measureµ∗ on X is a function µ∗ : P(X)→ R such that

(i) µ∗(∅) = 0.

(ii) If A ⊂ B, then µ∗(A) ≤ µ∗(B).

(iii) If A1, A2, · · · ⊂ X, then µ∗ (⋃∞n=1An) ≤

∞∑n=1

µ∗(An).

If µ∗(A) = 0, then A is a null set (with respect to µ∗).

From an outer measure we can then construct a measure.

De�nition 2.72. (Bass [2], De�nition 4.5) Let µ∗ be an outer measure on X.A set A ⊂ X is measurable if

∀E ⊂ X : µ∗(E ∩A) + µ∗(E ∩Ac) = µ∗(A).

Theorem 2.73. (Bass [2], Theorem 4.6) If µ∗ is an outer measure on X, thenthe collection A of all µ∗-measurable sets is a σ-algebra. If µ is the restrictionof µ∗ to A then µ is a measure. Furthermore A contains all the null sets.

2.4.1 The Lebesgue Measure

Consider the real line R. The length of an interval I = [a, b], (a, b], [a, b) or(a, b) is given by `(I) = b− a. We wish to extend `, to obtain a measure whichcaptures our natural understanding of length.

De�nition 2.74. (Bass [2], Section 4.2) De�ne µ∗ by

µ∗(A) := inf

{ ∞∑k=1

`(ak, bk) : (ak, bk) ⊂ R and A ⊂∞⋃k=1

(ak, bk)

}.

This is the outer Lebesgue measure on R.

Let us check De�nition 2.71 to verify that µ∗ is an outer measure on R. The�rst condition (i) follows from the fact that

∀n ∈ N∗ : ∅ ⊂(− 1

n,1

n

)and µ∗

(− 1

n,1

n

)=

2

n→ 0 as n→∞.


Next (ii) follows from the fact that if A ⊂ B, then any covering of B is also acovering of A. Finally we consider (iii). Let A1, A2, · · · ⊂ R. Let n ∈ N∗ andtake ε > 0. There exist open intervals In,m, m ∈ N∗ such that

An ⊂∞⋃k=1

In,k and µ∗( ∞⋃k=1

In,k

)≤ µ∗(An) + ε2−n.

Thus

µ∗( ∞⋃n=1

An

)= µ∗

( ∞⋃n=1

∞⋃k=1

In,k

)≤∞∑n=1

µ∗( ∞⋃k=1

In,k

)≤∞∑n=1

(ε2−n + µ∗(An))

= ε+

∞∑n=1

µ∗(An).

Hence letting ε→ 0 we have that µ( ∞⋃n=1

An)≤∑∞n=1 µ

∗(An), as desired.

Since µ∗ is an outer measure on R, the collection of µ∗-measurable sets forms aσ-algebraM.

De�nition 2.75. (Bass [2], Section 4.2) The Lebesgue measure µ is the restric-tion of the outer Lebesgue measure µ∗ toM. The sets inM are called Lebesguemeasurable or simply measurable.

Note that µ(R) =∞ so µ is not �nite. However, if we restrict of µ to a boundedset A ∈M (e.g. a bounded interval), then the resulting measure µ|A is �nite.

Remark 2.76. The collectionM contains all the open intervals on R, but sincethese generate B � as we saw in Proposition 2.66 � we have that B ⊂M. Thusall sets in B are measurable.

From here on µ will denote the Lebesgue measure unless stated otherwise, andnull sets with respect to the Lebesgue measure will be referred to as simply nullsets. Speaking of this, which sets are null sets? The next proposition providesus with a few.

Proposition 2.77. All countable sets A ⊂ R have Lebesgue measure zero.

Proof: Let A ⊂ R be countable, we may then order the set A = {a1, a2, . . . }.Take ε > 0, then {B(aj , ε2

−1−j) : j ∈ N∗} is an open cover of A so

µ∗(A) ≤ µ∗( ∞⋃j=1

B(aj , ε2−1−j)

)≤∞∑j=1

µ∗(B(aj , ε2−1−j)) =

∞∑j=1

ε2−j = ε.


Thus letting ε→ 0 we see that µ∗(A) = 0.

The countable sets however, do not make up all the null sets. We �nish thissection by showing that the uncountable Cantor set (Abbott [1], Section 3.1) isalso a null set.

Example 2.78. De�ne

Cn :=

3n−1−1⋃k=0

(3k + 1

3n,3k + 2

3n

), n ∈ N∗,

and set

C := [0, 1] \∞⋃n=1

Cn.

This is the Cantor set. Since the sets Cn are open they and the Cantor set aremeasurable. Notice that

1 = µ([0, 1]) = µ(([0, 1] \ C) t C) = µ([0, 1] \ C) + µ(C),

soµ(C) = 1− µ([0, 1] \ C). (2.4)

Next letD1 = C1 and Dn+1 = Cn ∩Dc

n, n ∈ N∗.The sets Dn are disjoint with µ(Dn) = 2n−13−n and

⋃∞n=1 Cn =

⊔∞n=1Dn.

Thus

µ([0, 1] \ C) = µ

( ∞⋃n=1

Cn

)= µ

( ∞⊔n=1

Dn

)=

∞∑n=1

µ(Dn) =

∞∑n=1

2n−1

3n

=1

3

∞∑n=0

(2

3

)n=

1

3

1

1− 2/3= 1.

Hence by (2.4) we have that µ(C) = 0.

2.5 The Riemann Integral

In this section we consider the Riemann integral. We want to use this integralto de�ne a norm on some space of functions, and then from this normed spaceobtain a Banach space (using Theorem 2.35 if necessary). However, we will seethat the Riemann integral is insu�cient for this purpose �indeed the Lebesgueintegral will be a natural �x for the problems that arise� but we are gettingahead of ourselves, for now let us construct the Riemann integral.

2.5. The Riemann Integral 29

...

C

( )( ) ( )

( ) ( ) ( ) ( )

Figure 2.8: The Cantor set.

De�nition 2.79. (Abbott [1], Chapter 7) A partition of an interval [a, b] is a�nite set P = {x0, . . . , xn} where a = x0 < x1 < · · · < xn = b. We denote theset of all partitions by P .

A step function is a function ξ : [a, b)→ R such that ξ = αk on each subinterval[xk−1, xk) of a partition P = {x0, . . . , xn}, where α1, . . . , αk are real constants.We de�ne the integral of such a function by∫ b

a

ξ(x)dx :=

n∑k=1

αk(xk − xk−1).

x

a b

y

0

y = ξ(x)

Figure 2.9: An example of a step function.


De�nition 2.80. (Abbott [1], Chapter 7) For a partition P ∈ P , de�ne thenumbers

mk = inf{f(x) : x ∈ [xk−1, xk]}, Mk = sup{f(x) : x ∈ [xk−1, xk]},

and use these to create the step functions

fP(x) = mk, x ∈ [xk−1, xk), and fP (x) =Mk, x ∈ [xk−1, xk).

Next de�ne

L(f) := sup

{∫ b

a

fP(x)dx : P ∈ P

}and U(f) := inf

{∫ b

a

fP (x)dx : P ∈ P

}.

We say that the function f is Riemann integrable on [a, b] if L(f) = U(f), andin this case we de�ne ∫ b

a

f(x)dx := L(f) = U(f).

x

y

a b

y = f(x)

y = fP (x)

y = fP(x)

P

Figure 2.10: Illustration of the Riemann integral.

This is quite a chunky de�nition but it essentially means that a function isRiemann integrable if we can approximate it from above and below using stepfunctions. Because of this any Riemann integrable function needs to be bounded.A natural question to ask is which functions are Riemann integrable on a giveninterval [a, b]. Clearly any step function is Riemann integrable and the followingtheorem due to Lebesgue gives us a complete classi�cation.

Theorem 2.81. (Abbott [1], Theorem 7.6.5) Let f be a bounded function on[a, b], then f is Riemann integrable on [a, b] if and only if its set of discontinuitieson the interval has Lebesgue measure zero.

2.5. The Riemann Integral 31

It seems that lots of functions are Riemann integrable. There are howeverquite simple examples of functions that are not Riemann integrable. The nextexample is quite standard.

Example 2.82. For a set A ⊂ R the characteristic function on A is given by

χA =

{1 if x ∈ A,0 if x /∈ A.

Let f = χQ|[a,b], let P = {x0, . . . , xn} be a partition of [a, b] and let k ∈ {1, . . . , n}.Since Q is dense in R there exists yk ∈ [xk−1, xk)∩Q, which implies thatMk = 1.Since also R \ Q is dense in R there exists zk ∈ [xk−1, xk) \ Q, which impliesthat mk = 0. Thus fP (x) = 1 and f

P(x) = 0, independently of the partition

P . Hence U(f) = 1 and L(f) = 0, so U(f) 6= L(f) and f is not Riemannintegrable.

x

a b

y

y = f(x)

y = fP (x)

y = fP(x)

Figure 2.11: Upper and lower approximation of f = χQ|[a,b] with step functions.

This might not seem like a problem, when would we ever want to integrate sucha function anyway? Let us forget about it for now and instead move on with ourmission. Consider the set C([a, b]) of continuous functions f : [a, b] → R. Sincecontinuous functions are Riemann integrable we may use the Riemann integralto de�ne the norm

‖f − g‖R :=

∫ b

a

|f − g|dx,

on C([a, b]). Does this turn C([a, b]) into a Banach space?


Example 2.83. The space C([a, b]) equipped with the norm ‖ · ‖R is not com-plete. Let c = (a+ b)/2 and de�ne fn by

fn(x) :=

0, if a ≤ x ≤ c,n(x− c), if c < x < c+ 1/n,

1, if c+ 1/n ≤ x ≤ b,n ≥ 2/(b− a).

Obviously these functions are continuous, moreover they make up a Cauchysequence in C([a, b]) with respect to ‖ · ‖R. To see this let m > n ≥ 2/(b − a).We know that 1 ≥ fm ≥ fn ≥ 0, and that fm = fn on [0, c] ∪ [c+ 1/n, b] so

‖fm − fn‖R =

∫ b

a

|fm(x)− fn(x)|dx =

∫ c+1/n

c

(fm(x)− fn(x))dx

≤∫ c+1/n

c

fm(x)dx ≤∫ c+1/n

c

dx =1

n→ 0 as m,n→∞.

This Cauchy sequence does not have a limit in C([a, b]). Assume that thereexists an f ∈ C([a, b]) such that fn → f , then

‖fn − f‖R =

∫ b

c

|fn(x)− f(x)|dx

=

∫ c

a

|f(x)|dx+

∫ c+1/n

c

|n(x− c)|dx+

∫ b

c+1/n

|1− f(x)| → 0,

as n → ∞. Since f is continuous we would then have that f = 0 on [a, c] andf = 1 on [c+ 1/n, b] for every n, i.e. f = 0 on [a, c] and f = 1 on (c, b]. But nosuch continuous function exists. Thus C([0, 1]) is not complete.

We know by Theorem 2.35 that the completion of C([a, b]) with respect to thenorm ‖·‖R exists. Let us try and �nd it. The problem in our previous example isthat the Cauchy sequence converges to a function with at least one discontinuity.Thus if we expand our set C([a, b]) to the set D([a, b]) of all functions for whichthe set of discontinuities is a null set, then the Cauchy sequence will converge.A problem is that ‖ · ‖R is no longer a norm (only a seminorm).

Example 2.84. De�ne f : [a, b]→ R by

f :=

{1, if x = a,

0, if x 6= a,

then ‖f‖R = 0.

2.6. The Lebesgue Integral 33

x

a c b

y

1

· · ·

y = fn(x)

y = f(x)

Figure 2.12: A Cauchy sequence that does not converge in C([a, b]).

To make ‖ · ‖R become a norm we have to consider equivalence classes wheref ∼ g if f = g except for on a null set. Then however we run into a muchworse problem, namely that the Riemann integrable function 0 is equivalentwith χQ|[a,b]. A function which is not Riemann integrable as we saw in Example2.82. Clearly we need a new integral which can handle these sorts of things. Letus get to work constructing it.

2.6 The Lebesgue Integral

The idea for this new integral is to integrate along the y-axis instead of thex-axis. Given a function f : [a, b] → R, partition f([a, b]) into small disjointintervals Ij = [αj , βj) and let Aj = f−1(Ij). Then

g(x) ≈∞∑j=1

αjχAj, (2.5)

and the approximation should get better and better as we re�ne our partition.This should take care of the function in Example 2.82 since in this case f([a, b]) ={0, 1}. The functions on the right hand side of (2.5) will be the building blocksof our new integral, as the step functions were for the Riemann integral.


De�nition 2.85. (Beals [3], Section 11C) A simple function is a function f :R→ R of the form

f =

n∑j=1

αjχAj, αj ≥ 0,

where Aj are pairwise disjoint measurable subsets of R and n ∈ N∗. TheLebesgue integral of a such a function is∫

f :=

n∑j=1

αjµ(Aj).

x

y

a b0

y = g(x)

y = α3χA3

y = α2χA2

y = α1χA1

α3

α2

Figure 2.13: Illustration of the Lebesgue integral.

Example 2.86. Returning to Example 2.82 we see that we can view f as asimple function by setting f(x) := 0 for x /∈ [a, b]. We then have that∫

fdx =

∫ b

a

χQdx =

∫Q∩[a,b]

dx = µ(Q ∩ [a, b]) = 0,

since Q ∩ [a, b] is countable and therefore a null set.

Thus we have taken care of the function which caused the Riemann integral tobreak down. Now we need to expand our set of Lebesgue integrable functionsto at least include C([a, b]). In light of De�nition 2.85 it seems natural to re-quire that the sets Aj in (2.5) be measurable. Otherwise the right hand sideapproximations of our function f may not be simple functions. This motivatesthe next de�nition.


De�nition 2.87. (Beals [3], Section 11A) A function f : D → R, D ⊂ R, ismeasurable if for every open A ⊂ D, the set f−1(A) is measurable.

Note that R is open, so in particular f−1(R) = D is measurable. Notice alsothat since all open sets are measurable, continuous functions are measurable.What other functions are measurable? Since the whole point of introducing thisnew integral is to expand the number of integrable functions, one would hopethat there are a lot more. And indeed there is, as we shall see in Proposition2.89. First however we need the following de�nition.

De�nition 2.88. Let f, g : D → R, D ⊂ R, then

(f ∧ g)(x) := min{f(x), g(x)}, and (f ∨ g)(x) := max{f(x), g(x)}.

Proposition 2.89. (Beals [3], Section 11A) Let f and g be measurable func-tions, then the following functions are also measurable

(i) cf, c ∈ R, (ii) f + g, (iii) fg, (iv) |f |, (v) f ∧ g, (vi) f ∨ g.

Having an abundance of measurable functions, we now move on with the con-struction of the Lebesgue integral.

De�nition 2.90. (Beals [3], Section 11C) If f : R → R is a nonnegative mea-surable function, then we de�ne the Lebesgue integral of f by∫

f := sup

{∫g : g is a simple function and 0 ≤ g ≤ f

}.

Note that∫f =∞ is possible.

De�nition 2.91. (Beals [3], Section 11C) Let f : R → R, we de�ne the func-tions f+ and f− by

f+(x) :=

{f(x), if f(x) > 0,

0, if f(x) ≤ 0,f−(x) :=

{−f(x), if f(x) < 0,

0, if f(x) ≥ 0.

Note that f = f+ − f− and |f | = f+ + f−.

De�nition 2.92. (Beals [3], Section 11C) A function f : R→ R is integrable if∫|f | <∞. In this case the Lebesgue integral of f is∫

f :=

∫f+ −

∫f−.


De�nition 2.93. (Beals [3], Section 11C) Let D ⊂ R and suppose that f :

D → R is a measurable function. De�ne the extension f̃ : R→ R of f by

f̃ :=

{f(x), if x ∈ D,0, if x /∈ D.

The function f is integrable if f̃ is and∫f :=

∫f̃ .

Suppose that g : R → R is integrable and A ⊂ R is measurable. Then theintegral of g on A is de�ned by

∫Ag :=

∫gχA.

Having acquired this new integral, we wonder how it compares to our old Rie-mann integral.

Proposition 2.94. (Beals [3], Section 12B*) Let f : [a, b] → R be Riemannintegrable, then it is also Lebesgue integrable, and the integrals of f are equal.

So what is di�erent? For example we have a powerful result concerning theLebesgue integral, which will be used a lot later in the thesis.

Theorem 2.95. The dominated convergence theorem (Beals [3], Theorem11.6) Let (fn) be a sequence of measurable functions dominated by an integrablefunction g, i.e. |fn| ≤ g. Then

limn→∞

∫fn(x)dx =

∫limn→∞

fn(x)dx.

2.6.1 The Lp-spaces

Recalling the end of section 2.5 we prepare for equivalence classes.

De�nition 2.96. (Beals [3], Section 12A) Two functions f and g are equalalmost everywhere (a.e. for short) if they are equal except for on a null set, andwe write f ∼ g.

Proposition 2.97. (Beals [3], Section 12A) Let f be a measurable function andsuppose that f = g a.e., then g is also measurable.

This proposition ensures that ∼ is well de�ned on the set of measurable func-tions, in fact, on this set ∼ de�nes an equivalence relation.


De�nition 2.98. (Folland [5], Section 6.1) Let 1 ≤ p <∞ and let A ⊂ R. Letf : A → R be a measurable function. The Lp-norm of f , denoted ‖f‖p, isde�ned by

‖f‖p :=(∫

A

|f |pdx)1/p

.

Using this norm we de�ne

Lp(A) := {f : A→ R : f is measurable and ‖f‖p <∞},

where f, g ∈ Lp(A) are considered equal if f ∼ g. Endowed with the Lp-norm,this becomes a normed space. We will usually write Lp(A) as simply Lp.

Proposition 2.99. (Folland [5], Section 6.1) For 1 ≤ p <∞ the space Lp is aBanach space.

Theorem 2.100. (Pedersen [9], Section 2.1.14) The completion of the normedspace C([a, b]), with norm

‖x‖p :=(∫|x(t)|pdt

)1/p

,

using the Riemann integral, is the space Lp([a, b]).

Thus we have accomplished our goal. However, the attentive reader will noticethat this leaves out the spaces Lp(I), for unbounded intervals I (i.e. intervalsof the type [a,+∞), (−∞, b] and R).

De�nition 2.101. (Bass [2], Section 26.1) Let f : [a, b] → R. The support off in, denoted supp(f), is de�ned by

supp(f) := {x : f(x) 6= 0}.

Let I be a closed interval (possibly unbounded), and set

C(I) := {f : I → R : f continuous and supp(f) is bounded}.

This coincides with our previous de�nition for bounded closed intervals. En-dowing this vector space with the norm ‖ · ‖R it becomes a normed space, andTheorem 2.100 still holds.

Recalling Example 2.34, we wonder what happens if we consider functions f ∼ gequal on the Banach space B(A), where A is measurable. What does this spacelook like and is it also Banach?


De�nition 2.102. (Folland [5], Section 6.1) Let A ⊂ R and let f : A → R bea measurable function. The L∞-norm, denoted ‖ · ‖∞, is de�ned by

‖f‖∞ = ess supA|f | := inf

g∼fsupA|g|.

Using this we de�ne

L∞(A) := {f : A→ R : f is measurable and ‖f‖∞ <∞},

where again f, g ∈ L∞(A) are considered equal if f ∼ g. Endowed with the L∞-norm, this becomes a normed space. We will usually write L∞(A) as simplyL∞.

Proposition 2.103. (Folland [5], section 6.1) The space L∞ is a Banach space.

Note that the ‖·‖∞ norm is in fact the norm in B(A) of the �smallest� function inthe equivalence class. We now �nish this chapter by considering some importantproperties of Lp-spaces.

Proposition 2.104. The Hölder inequality (Folland [5], section 6.1)Suppose that f and g are measurable functions. Then

‖fg‖1 ≤ ‖f‖p‖g‖q.

Proposition 2.105. (Folland [5], Theorem 6.15 and Corollary 6.16) If1 ≤ p <∞, then (Lp)∗ ∼= Lq, and for 1 < p <∞ the space Lp is re�exive.

2.7 List of Convergences

This section provides a quick overview of the convergences studied in this thesis,and the notation used for them.

2.7.1 Functions

As the main title suggests, this thesis is mainly concerned with convergence offunctions. More precisely real functions de�ned on measurable subsets of R. Westart with two elementary notions of convergence.

2.7. List of Convergences 39

De�nition 2.106. (Abbott [1], De�nition 6.2.1) A sequence of functions (fn),where fn : A→ R, converges pointwise to a function f : A→ R if

∀x ∈ A : fn(x)→ f(x),

and we write fn → f .

De�nition 2.107. (Abbott [1], De�nition 6.2.3) A sequence of functions (fn),where fn : A→ R, converges uniformly to f : A→ R if

∀ε > 0 ∃N ∈ N ∀x ∈ A ∀n ≥ N : d(fn(x), f(x)) < ε,

and we write fn ⇒ f .

Remark 2.108. Note that fn ⇒ f ⇐⇒ supx∈A |fn(x)− f(x)| → 0.

Next we have a more exotic kind of convergence due to Arzelà, which will bestudied in detail in Chapter 3.

De�nition 2.109. (Caserta et al. [4], De�nition 2.1) Let (fn) be a sequence offunctions fn : [a, b]→ R where [a, b] ⊂ R is compact, which converges pointwiseto f . The sequence (fn) converges quasi uniformly to f if for all ε > 0 andn0 ∈ N there exist natural numbers n1, . . . , nr ≥ n0 and corresponding openintervals In1 , . . . , Inr which cover [a, b], and

∀x ∈ Inj∩ [a, b] |fnj

(x)− f(x)| < ε, j = 1, . . . r.

We then write fnq

⇒ f .

2.7.2 Normed Spaces

Functions can also be seen as elements of normed spaces, where we have thefollowing notions of convergence:

De�nition 2.110. (Kreyzig [7], De�nition 4.8-1) Let X be a normed space. Asequence of elements xn ∈ X converges to x in norm if

‖xn − x‖ → 0.

This convergence is the same as in De�nition 2.20 using the induced metricd(x, y) := ‖x− y‖, x, y ∈ X.


De�nition 2.111. (Kreyzig [7], De�nition 4.8-2) A sequence (xn) in a normedspace X converges weakly to x if

∀f ∈ X∗ limn→∞

f(xn) = f(x),

and we write xn ⇁ x.

De�nition 2.112. (Kreyzig [7], De�nition 4.9-4) Let X be a normed space andlet (fn) be a sequence of bounded functionals on X, then

(i) fn converges to f ∈ X∗ in norm if

‖fn − f‖ → 0,

and we write fn → f .

(ii) fn converges weak* to f ∈ X∗ if

∀x ∈ X : fn(x)→ f(x),

and we write fn∗⇁ f .

Remark 2.113. The normed spaces considered in this thesis will mainly be thefollowing.

(i) Lp(A), 1 ≤ p <∞, with the norm

‖f‖p :=(∫

A

|f(t)|pdt)1/p

.

(ii) L∞(A) with the norm

‖f‖∞ := ess supx∈A

|f(x)|.

Convergence in norm in these spaces will be referred to as convergence in Lp

and L∞ respectively.

2.7. List of Convergences 41

2.7.3 Measure Theory

Finally we can also de�ne convergence for functions using measure theory.

De�nition 2.114. (Bass [2], p. 78) Let (X,µ) be a measure space and let fn :D → R be measurable functions. The sequence (fn) converges almost uniformlyto f : D → R if for every ε > 0 there is a measurable set Eε ⊂ D such that

(i) µ(Eε) < ε.

(ii) fn ⇒ f on D \ Eε.

and we write fna

⇒ f .

De�nition 2.115. (Bass [2], De�nition 10.1) Let X = (X,µ) be a measurespace and let (fn) be a sequence of measurable functions.

(i) fn converges almost everywhere to f if there is a set A ∈ X with measure0 such that

∀x /∈ A : fn(x)→ f(x),

and we write fna.e.→ f .

(ii) fn converges to x in measure if

∀ε > 0 µ({x : |fn(x)− f(x)| > ε})→ 0,

and we write fnµ−→ f .

Chapter 3

Quasi uniform Convergence :

Preserving Continuity


44 Chapter 3. Quasi uniform Convergence : Preserving Continuity

In this chapter we will study when continuity of a sequence of functions carriesover to its limit. To be more precise, consider a sequence (fn) of continuousfunctions which converge pointwise to a function f . When is the function fcontinuous? We will see that quasi uniform convergence arises quite naturallyto deal with this exact question. To begin with we ask, is continuity alwayspreserved? The answer is no as we shall see in the next example.

Example 3.1. Let the functions fn : [0, 1] → R, n ∈ N∗, be de�ned byfn(x) := xn. Take x ∈ [0, 1), then xn → 0 as n → ∞. However, we havethat 1n → 1 as n→∞. Thus fn → f where

f =

{1, if x = 1,

0, if 0 ≤ x < 1.

This is obviously a discontinuous function.

x

y

0 1

y = fn(x)

y = f(x)

Figure 3.1: Continuous functions fn → f which is not continuous.

Clearly we need something stronger than simply pointwise convergence. Mayberequiring uniform convergence will do the trick.

Proposition 3.2. Uniform convergence preserves continuity and uniform con-tinuity.

Proof: Let fn : D → R, D ⊂ R, be continuous functions that converge uni-formly to f : D → R. Take ε > 0. Since fn ⇒ f there exists an N ∈ N suchthat

∀x ∈ D ∀n ≥ N supx∈D|f(x)− fn(x)| < ε/3. (3.1)

45

Let x0 ∈ D and n ≥ N . The continuity of fn implies that there exists a δ > 0such that

|x− x0| < δ =⇒ |fn(x)− fn(x0)| < ε/3. (3.2)

Taking x ∈ B(x0, δ), and using (3.1) and (3.2) we acquire

|f(x)− f(x0)| = |f(x)− fn(x) + fn(x)− fn(x0) + fn(x0)− f(x0)|≤ |f(x)− fn(x)|+ |fn(x0)− fn(x)|+ |fn(x0)− f(x0)|< ε/3 + ε/3 + ε/3 = ε.

Thus f is continuous at x0 and therefore continuous since x0 was arbitrary. Theproof of the second claim is similar. Simply switch the quanti�ers in (3.1) andchoose δ independent of x0.

Remark 3.3. One thing that should be noted is that while uniform convergencepreserves continuity and uniform continuity, it does not necessarily make con-tinuous functions become uniformly continuous. To see this let f be a functionwhich is continuous but not uniformly so, e.g. x 7→ 1/x on (0, 1). Then considerthe functions

fn := f, n ∈ N∗.

Obviously the functions fn are continuous and converge uniformly to f .

x

y

0 1

1

y = f(x) = fn(x)

Figure 3.2: Convergent fn ⇒ f which is not uniformly convergent.


We have established that uniform convergence is su�cient for preserving conti-nuity. But is it also necessary? That is to say, does f being convergent implythat fn converges uniformly to f? The answer is no as we shall see in the nextexample.

Example 3.4. De�ne the functions fn : [0, 2]→ R by

fn(x) =

nx, if 0 ≤ x < 1/n,

2− nx, if 1/n ≤ x < 2/n,

0, if 2/n ≤ x ≤ 2,

n ∈ N∗.

Obviously these functions are continuous and fn(0) = 0 for each n ∈ N∗. Takex ∈ (0, 2], then there exists an N ∈ N∗ such that

N ≥ 2

xi.e. x ≥ 2

N.

This means that fn(x) = 0 for all n ≥ N . Hence fn → 0, which is a continuousfunction. However, this convergence is not uniform since

∀n ∈ N∗ : supx∈[0,2]

|fn(x)| ≥ |fn(1/n)| = 1.

x

y

0 1

1

y = fn(x)

y = f(x)

Figure 3.3: Continuous pointwise limit f with fn 6⇒ f .

47

Thus uniform convergence is not necessary for preserving continuity, not evenwhen the involved functions are de�ned on a compact interval. Can we replaceuniform convergence with something weaker to precisely capture this continuitypreserving property? The answer � at least for functions de�ned on compactintervals � is yes, and this is the motivation for introducing quasi uniform con-vergence in De�nition 2.109. We have the following theorem.

Theorem 3.5. Let (fn) be a sequence of continuous functions fn : [a, b] → Rwhere [a, b] ⊂ R is compact and let fn → f . Then f is continuous if and only if

fnq

⇒ f .

Proof: ⇐= Assume that (fn) is quasi uniformly convergent and take x0 ∈ [a, b].Let ε > 0 and n0 = 0. By de�nition there exist n1, . . . , nr ∈ N and correspond-ing open intervals In1 , . . . , Inr such that

[a, b] ⊂r⋃j=1

Inj, (3.3)

andx ∈ Inj

=⇒ |fnj(x)− f(x)| < ε/3, j = 1, . . . , r. (3.4)

Take x0 ∈ [a, b]. By (3.3) x0 ∈ Inifor some 1 ≤ i ≤ r. Since Ini

is open thereexists a δ1 > 0 such that B(x0, δ1) ⊂ Ini

. By the continuity of fnithere also

exists a δ < δ1 such that

|x− x0| < δ =⇒ |fni(x)− fni

(x0)| < ε/3. (3.5)

Let x ∈ B(x0, δ). By (3.4) and (3.5) we have that

|f(x)− f(x0)| = |f(x)− fni(x) + fni

(x)− fni(x0) + fni

(x0)− f(x0)|≤ |f(x)− fni

(x)|+ |fni(x0)− fni

(x)|+ |fni(x0)− f(x0)|

< ε/3 + ε/3 + ε/3 = ε.

Hence f is continuous at x0 and thus continuous since x0 was arbitrary.

=⇒ Assume that f is continuous. Take x ∈ [a, b], let ε > 0 and let n0 ∈ N.Since f is the pointwise limit of (fn) there exists nx ≥ n0 such that

|fnx(x)− f(x)| < ε/3, x ∈ R. (3.6)

The continuity of the functions fnxand f implies that there exists a δx > 0

such that

y ∈ [a, b] ∩B(x, δx) =⇒

{|fnx

(y)− fnx(x)| < ε/3,

|f(y)− f(x)| < ε/3.(3.7)


Since the open cover {B(x, δx) : x ∈ [a, b]} covers every point x ∈ [a, b], it isan open cover of [a, b]. Thus since [a, b] is compact there is a �nite subcoverB(x1, δ1), . . . , B(xr, δr). Let y ∈ [a, b], then y belongs to the open intervalB(xi, δi) for some 1 ≤ i ≤ r. Thus by (3.6) and (3.7)

|fni(y)− f(y)| ≤ |fni(y)− fni(xi)|+ |fni(xi)− f(xi)|+ |f(xi)− f(y)|< 3 · ε/3 = ε.

(3.8)

Hence fnq

⇒ f and we are done.

Having properly motivated the introduction of quasi uniform convergence, letus try to get a better grip of what it looks like by returning to Example 3.4.

By Theorem 3.5 we know that fnq

⇒ f so let us get to work. Take ε > 0 andn0 ∈ N. Next let n1 = n0 and In1

= [−ε/2n1, ε/2n1]. Since |f1(x)| ≤ |x| wehave that

x ∈ In1∩ [0, 2] =⇒ |fn1

(x)| ≤ ε/2 < ε.

Next choose In2= [ε/3n1,max{ε, 3}] and take n2 ∈ N such that n2 > 6n1/ε i.e.

2/n2 < ε/3n1, then fn2= 0 on In2

and thus

x ∈ In2∩ [0, 2] =⇒ |fn2

(x)| = 0 < ε.

Finally we notice that

[0, 2] ⊂ (−ε/2, 3) ⊂ In1 ∪ In1 .

Hence fnq


49

x

y

0 1

n1

2

n1

2· · · · · ·

ε

2

ε

2n1

ε

2n1

In1

ε

3n1

3

In2

y = fn1(x)y = fn2(x)

Figure 3.4: The sequence (fn) converges quasi uniformly.

Chapter 4

Relation Between

Convergences


52 Chapter 4. Relation Between Convergences

In this chapter we will explore how the di�erent kinds of convergence are related,proving implications and providing illuminating counterexamples. All functionsare assumed to be real value and de�ned on measurable subsets of R. In orderto construct counterexamples, the following property is very useful and will beused without mention.

Proposition 4.1. Letp→ and

q→ be two di�erent kinds of convergence de�ned on

some set X. Suppose that xnp→ x implies xn

q→ x for any sequence (xn) ⊂ X.

If ynq→ y and yn 6

p→ y, then (yn) is not p-convergent.

Proof: Suppose that ynq→ y and yn 6

p→ y. Assume that (yn) is p-convergent,i.e. yn

p→ z for some z ∈ X. By assumption ynq→ y, but since limits are unique

y = z. This is a contradiction since yn 6p→ y. Thus (yn) is not p-convergent.

Properly equipped to construct counterexamples we begin our journey. Let usstart with some easy positive results.

Theorem 4.2. (i) fnq

⇒ f =⇒ fn → f .

(ii) fn → f =⇒ fna.e.→ f .

(iii) fn → f in L∞ =⇒ fna

⇒ f .

Proof:

(i)-(ii) This follows directly from the de�nition.

(iii) Take fn : D → R such that fn → f in L∞. Then

‖f − fn‖ = supD\E|f − fn|

for some E, independent of n, with µ(E) = 0. By Remark 2.108 fn ⇒ fon D \ E which completes the proof.

Theorem 4.3. Suppose that fn : A→ R converges uniformly to f , then

(i) fna

⇒ f .

(ii) If fn ∈ L∞, then fn → f in L∞.

(iii) fn → f .

(iv) If fn : [a, b]→ R, then fnq

⇒ f .

53

Proof:

(i) This follows directly from the de�nition by setting Eε := ∅ for each ε > 0.

(ii) By de�nition

||fn − f ||∞ = ess supA|fn − f | ≤ sup

A|fn − f |.

We now notice that supA |fn − f | → 0 by Remark 2.108, which completesthe proof.

(iii) Let ε > 0, there exists an N ∈ N such that

∀n ≥ N ∀x ∈ A : |fn(x)− f(x)| < ε.

Speci�cally we have

∀x ∈ A ∃Nx := N ∀n ≥ Nx : |fn(x)− f(x)| < ε.

Hence fn(x)→ f(x) for each x ∈ A, and we are done.

(iv) Assume that fn : [a, b] → R. Let ε > 0, then there exists an n0 ∈ N suchthat

∀x ∈ [a, b] : |fn0(x)− f(x)| < ε.

Letting n1 ≥ n0 and letting I1 be an open interval containing [a, b], weacquire

∀x ∈ I1 ∩ [a, b] : |fn1(x)− f(x)| < ε.

Thus fnq


Having established these relatively easy results, we wonder about the converse.It turns out none of the converse statements of either of the theorems hold. Letus start easy.

Example 4.4. Recalling Theorem 3.5 we see that Example 3.4 shows that

fnq

⇒ f does not imply fn ⇒ f .

Thus we already constructed the counterexample in Chapter 3, how foresightedof us. Next we actually construct a new counterexample.


Example 4.5. Let f ∈ L∞([a, b]), take α ∈ D and set

fn :=

{f(x), if x 6= α,

f(x) + 1, if x = α,n ∈ N.

Then fn ∼ f , so fna.e.→ f . Notice also that

∀n ∈ N : ||fn − f ||∞ = 0,

therefore fn → f in L∞([a, b]). However,

∀n ∈ N : |fn(α)− f(α)| = 1,

so fn 6→ f pointwise. Thus neither convergence in L∞, nor almost everywhereconvergence implies pointwise convergence.

This example essentially only uses that f ∼ g is not the same as f = g. Let usconsider a more interesting counterexample, with nonconstant sequences.

Example 4.6. De�ne fn : [0, 1] → R by fn(x) := xn. Let ε ∈ (0, 1) and setEε := [1− ε/2, 1]. Notice that µ(Eε) = ε/2 < ε, and that

sup[0,1]\Eε

|fn| = (1− ε/2)n → 0, as n→∞.

Thus fna

⇒ f . However,

∀n ∈ N∗ : ‖fn‖∞ = ess sup[0,1]

|fn| = sup(0,1)

|xn| = 1,

so fn 6→ f in L∞. This also shows that fn 6⇒ f . Hence we have that almostuniform convergence implies neither uniform convergence, nor convergence inL∞.

Next we consider a sequence of functions (fn) in Lp, where 1 ≤ p < ∞, suchthat fn

a.e.→ f . A reasonable claim might be that fn → f in Lp. This howeverturns out to be false as we shall see in the next example.

55

x

y

0 11− ε

Eε

y = fn(x)

y = f(x)

Figure 4.1: fna

⇒ f but fn 6→ f in L∞.

Example 4.7. Let fn : R→ R be de�ned by

fn(x) :=

{n, if 0 < x < 1/n,

0, otherwise,n ∈ N∗.

Notice that

∀n ∈ N∗ :∫R|fn(x)|pdx =

∫ 1/n

0

npdx = np−1 <∞, (4.1)

so the sequence (fn) lies in Lp(R). We will now show that fna.e.→ 0. If x ≤ 0 then

fn(x) = 0 for each n ∈ N∗. Therefore assume that x > 0 and set N := d1/xe,we then have that 1/N ≤ x so

∀n ≥ N :1

n≤ x =⇒ ∀n ≥ N : fn(x) = 0.

Thus fn(x)→ 0 pointwise, implying fna.e.→ 0. However,

‖fn − 0‖p = ‖fn‖p(4.1)== n

p−1p = n1−

1p 6→ 0, n→∞,

and we conclude that fn 6→ 0 in Lp(R). Hence fna.e.→ f 6=⇒ fn → f in Lp(R)

in general.

Notice that

∀ε ∈ (0, 1) : µ({x : |f(x)| > ε}) = µ([0, 1/n])→ 0, as n→∞.

Hence fnµ−→ 0, so fn

µ−→ f 6=⇒ fn → f in Lp, in general. Thus we have twocounterexamples in one, how e�cient!


x

0 1

y

x

0 1

y

x

0 1

y

x

0 1

y

y = 1/x

y = fn(x) · · ·

Figure 4.2: fna

⇒ 0 but fn 6→ 0 in Lp.

One thing to note in the above example is that ‖fn‖ 6→ ‖f‖. Indeed this seemsto be where it all goes wrong. What then happens if we require this to be true?Will the assertion then hold? The answer is yes and is presented in the followinglemma due to Riesz.

Lemma 4.8. (Kusolitsch [8]) Let (fn) be a sequence of measurable functions

such that fna.e.→ f and let 1 ≤ p < ∞. Then fn → f in Lp if and only if

‖fn‖p → ‖f‖p.

To prove this we will use the following lemma, the proof of which is the topicof Appendix A.

Lemma 4.9. Let b ≥ a ≥ 0 and let 1 ≤ p <∞, then (b− a)p ≤ bp − ap.

Proof of Lemma 4.8: To begin with assume that fn → f in Lp. Then by thereverse triangle inequality∣∣‖fn‖p − ‖f‖p∣∣ ≤ ‖fn − f‖p → 0,

implying ‖fn‖p → ‖f‖p. Conversely assume that ‖fn‖p → ‖f‖, and de�ne thefunctions f∗n by

f∗n :=

{fn, if |fn| ≤ |f |,|f | sgn fn, if |fn| > |f |,

n ∈ N.

Clearly since fna.e.→ f we have that f∗n

a.e.→ f , i.e.

|f∗n − f |a.e.→ 0. (4.2)

57

Observe that |f − f∗n| ≤ 2|f | so |f − f∗n|p ≤ 2p|f |p. Thus the functions |f − f∗n|pare integrable and dominated by 2p|f |p. The dominated convergence theoremthen implies that

limn→∞

‖f − f∗n‖p = limn→∞

(∫R|f − f∗n|pdx

)1/p

=

(∫R

limn→∞

|f − f∗n|pdx)1/p (4.2)

== 0.

(4.3)

Using this, we have that∣∣‖f∗n‖p − ‖f‖p∣∣ ≤ ‖f∗n − f‖p → 0, implying

‖f∗n‖p → ‖f‖p. (4.4)

Moving on, we know that |fn − f∗n| = |fn| − |f∗n|. Thus Lemma 4.9 implies that|fn − f∗n|p ≤ |fn|p − |f∗n|p, so

‖fn − f∗n‖pp =∫R|fn − f∗n|pdx ≤

∫R|fn|pdx−

∫R|f∗n|pdx = ‖fn‖pp − ‖f∗n‖pp.

By assumption ‖fn‖pp → ‖f‖pp as n → ∞, and (4.4) states that ‖f∗n‖pp → ‖f‖ppas n→∞. Hence limn→∞ ‖f∗n − fn‖p = 0. Using this and (4.3) we acquire

‖f − fn‖p ≤ ‖f − f∗n‖p + ‖fn − f∗n‖p → 0 as n→∞,

which �nishes the proof.

We now know that under certain conditions almost everywhere convergenceimplies convergence in Lp, but what about the converse? Does convergence inLp imply almost everywhere convergence? The answer is no as we shall see inthe next example.

Example 4.10. De�ne the functions fn,m : [0, 1]→ R by

fn,m(x) :=

{1, if m2−n ≤ x ≤ (m+ 1)2−n,

0, otherwise,m = 0, . . . , 2n − 1, n ∈ N.

Next de�ne the functions n,m : N∗ → N by

n(j) = blog2(j)c, m(j) = j − 2n(j),

and set f̃j = fn(j),m(j) for j ∈ N∗. We then have that

(f̃j) = f0,0, f1,0, f1,1, f2,0, . . . , f2,3, f3,0, . . . .


Let us show that f̃n → 0 in Lp. First observe that limj→∞

n(j) =∞ and

‖f̃j‖pp =∫ 1

0

|f̃j |pdx =

∫ 1

0

|fn(j),m(j)|pdx =

∫ 1

0

|fn(j),0|pdx

=

∫ 2−n(j)

0

dx = 2−n(j) → 0 as j →∞.

Hence f̃n → 0 in Lp. Now let us examine the pointwise limit. Take x ∈ [0, 1],this number has a binary expansion

x = 0.b1b2 . . . , bj ∈ {0, 1},

where we set 1 := 0.11 . . .. Next set

S(j) :=

j∑k=1

bk2−k, j ∈ N∗,

then x ∈ [S(j), S(j) + 2−j ] for j ∈ N∗, thus

∀j ∈ N∗ : f2j ,2jS(j)(x) = 1.

Hence the subsequence f2j ,2jS(j)(x) converges to 1 but this means that f̃j(x) 6→ 0.

Thus since µ([0, 1]) = 1 6= 0 we conclude that f̃ 6a.e.→ 0.

Note that∀ε ∈ (0, 1) ∀j ∈ N∗ : µ(x : |f̃j | > 1/2) = ‖fj‖1,

so we have that f̃jµ−→ 0, as well. This shows that fn

µ−→ f 6=⇒ fna.e.→ f in

general.

Having ruled out these implications, we do have a somewhat weaker result.

Theorem 4.11. Let fn → f in Lp, then there is a subsequence fσ(n) whichconverges to f almost everywhere.

This follows from two results which are important in their own right.

Theorem 4.12. Let fn → f in Lp, then fnµ−→ f .

Theorem 4.13. If fnµ−→ f then there is a subsequence fσ(n) which converges

to f almost everywhere.

59

x

0 1

y

x

0 1

y

x

0 1

y

x

0 1

y

x

0 1

y

x

0 1

y

x

0 1

y

y = fj(x)

· · ·

Figure 4.3: f̃j → 0 in Lp but f̃j 6a.e.→ 0.

Proof of Theorem 4.12: Take ε > 0 and let

An := {x : |fn(x)− f(x)| > ε}.

We then have that

‖fn − f‖pp =∫R|fn − f |pdx ≥

∫An

|fn − f |pdx >∫An

εpdx = µ(An)εp.

The assumption ‖fn−f‖p → 0 now implies that µ(An)→ 0 and we are done.

Proof of Theorem 4.13: Assume that fnµ−→ f , i.e.

∀ε > 0 : µ({x : |fn(x)− f(x)| > ε})→ 0.

Then there exists a subsequence σ(n) such that

µ({x : |fσ(n)(x)− f(x)| > 1/n}) < 2−n, n ∈ N.

Let An = {x : |fσ(n)(x)− f(x)| > 1/n} and set A := limj→∞⋃∞n=j An. Then

µ(A) = limj→∞

µ

( ∞⋃n=j

An

)≤ limj→∞

∞∑n=j

µ(An) < limj→∞

∞∑n=j

2−n = limj→∞

21−j = 0.


Hence A is of measure zero. Now take x /∈ A, then x /∈⋃∞n=lAn for some l ∈ N.

But this means that

∀n ≥ l : |fσ(n)(x)− f(x)| ≤ 1/n.

Thus fσ(n)(x)→ f(x) as n→∞, and we conclude that fσ(n)a.e.→ f .

In Theorem 4.13 we saw how convergence in measure almost implies conver-gence almost everywhere. What about the converse? Does almost everywhereconvergence imply convergence in measure? The answer is no as we shall see inthe next example.

Example 4.14. Let fn = χ[n,n+1], n ∈ N, and take x ∈ R. Then fn(x) = 0 forall n ≥ x. Hence fn → 0 pointwise. Let ε = 1/2 and set An := {x : |fn(x)| > ε},then

∀n ∈ N : µ(An) = µ([n, n+ 1]) = 1.

Therefore µ(An) 6→ 0, implying fn 6µ−→ 0. Thus fn → f pointwise 6=⇒ fn

µ−→ f

in general. It follows that fna.e.→ f 6=⇒ fn

µ−→ f either.

x

y

0

1f0 f2 f4 f6 f8

· · ·

Figure 4.4: fna.e.→ 0 but fn 6

µ−→ 0.

In the above counterexample pointwise convergence is achieved by pushing thenonzero part of our functions towards in�nity. If the measure is �nite this is nolonger possible and then the implication does hold.

61

Theorem 4.15. (Bass [2], Proposition 10.2) Suppose that µ is a �nite measure.

If fna.e.→ f then fn

µ−→ f .

Proof: Let ε > 0 and assume that fna.e.→ f . De�ne

An := {x : |fn(x)− f(x)| > ε}, n ∈ N,

then χAn

a.e.→ 0. This function is also dominated by the integrable functiong = 1, so the dominated convergence theorem now implies that

limn→∞

µ(An) = limn→∞

∫χAn

dx =

∫0dx = 0.

Hence fnµ−→ f .

We now move on to almost uniform convergence. To begin with, we have thetheorem that motivates the concept.

Theorem 4.16. Egoro�'s theorem (Bass [2], Theorem 10.8) Suppose that µ

is a �nite measure and fna.e.→ f . Then fn converges almost uniformly to f .

Proof: Let µ be �nite and assume that fna.e.→ f . Take ε > 0 and de�ne

An,k :=

∞⋃m=n

{x : |fm(x)− f(x)| > 1/k}, n, k ∈ N∗.

Fix k. Since fna.e.→ f , we know that for almost every x there exists an M such

that∀m ≥M : |fm(x)− f(x)| ≤ 1/k.

Hence µ(An,k) → 0 as n → ∞, which means that we can choose nk such thatµ(Ank,k) < ε2−k. Set A :=

⋃∞k=1Ank,k, then

µ(A) ≤∞∑k=1

µ(Ank,k) <

∞∑k=1

2−kε = ε.

Thus µ(A) < ε. Now take x /∈ A. Then x /∈ Ank,k, which means that

∀n ≥ nk : |fn(x)− f(x)| ≤ 1/k.

Letting k → ∞ we get that fn(x) → f(x) with nk independent of x. Hencefn ⇒ f on Ac and we are done.

What about the converse then? For once it actually holds and we don not evenneed the measure to be �nite!


Theorem 4.17. If fna

⇒ f then fna.e.→ f .

Proof: Assume that fna⇒ f , then for each ε > 0 there exists Eε with µ(Eε) < ε

such that fn ⇒ f on Ecε . Speci�cally, we have the sets E 1nfor n ∈ N∗. De�ne

E :=⋂∞n=0E 1

n, then

µ(E) ≤ µ(E 1n) <

1

n, ∀n ∈ N∗.

Hence µ(E) = 0. Now take x ∈ Ec, then x /∈ E 1kfor some k ∈ N∗ and therefore

fn(x)→ f(x) since fn ⇒ f on E 1k

c. Thus fna.e.→ f and we are done.

The converse statement was been shown to be true for �nite measures µ in The-orem 4.16. However, if we drop the requirement that µ be �nite then Example

4.10 together with Theorem 4.17 shows that fna.e.→ f 6=⇒ fn

a

⇒ f in general.

Theorem 4.18. If fna

⇒ f then fnµ−→ f .

To prove this theorem we �rst need the following lemma.

Lemma 4.19. If fn ⇒ f then fnµ−→ f .

Proof: Take ε > 0 and de�ne

An := {x : |fn(x)− f(x)| > ε}.

Assume that fn ⇒ f , then there exists an N ∈ N such that

∀x ∀n ≥ N : |fn(x)− f(x)| < ε.

Hence µ(An) = 0 for all n ≥ N so limn→∞

µ(An) = 0 and we are done.

Proof of Theorem 4.18: Assume that fna⇒ f , then there exists a set E with

µ(E) < 1 such that fn ⇒ f on Ec. Notice that the restriction of µ to E is �nite,and by Theorem 4.17 fn

a.e.→ f . Thus Theorem 4.15 implies that fnµ−→ f on E.

Furthermore fn ⇒ f on Ec, so by Lemma 4.19 fnµ−→ f on Ec and we conclude

that fnµ−→ f .

Recall that in Example 4.14 we showed that fna.e.→ f 6=⇒ fn

µ−→ f in general.

In light of Theorem 4.18 this means that fnµ−→ f 6=⇒ fn

a⇒ f in general either.

Thus converse of Theorem 4.18 is false.

Finally we consider weak and weak* convergence. To begin with we have thefollowing results.

63

Theorem 4.20. Let (xn) be a sequence in a normed space X. If xn → x thenxn ⇁ x.

Proof: Assume that xn → x and let f ∈ X∗, then

| 〈xn − x, f〉 | ≤ ‖f‖ ‖xn − x‖ → 0 as n→∞.

Thus xn ⇁ x.

Theorem 4.21. Suppose that X is a normed space whose dual X∗ is also anormed space. Let fn be a sequence in X∗. If fn → f , then fn

∗⇁ f .

Proof: Assume that fn → f and let x ∈ X, then

| 〈x, fn − f〉 | ≤ ‖x‖ ‖fn − f‖ → 0 as n→∞.

Thus fn∗⇁ f .

Recall that if X is a re�exive space, then the weak and weak* convergence arethe same. Thus we shall produce a simultaneous counterexample to the converseof the above theorems.

Example 4.22. Let 1 < p <∞, then Lp is re�exive with (Lp)∗ = Lq. Considerthe sequence of functions (χ[n,n+1]). These functions lie in Lp(R), and we haveseen that the sequence does not converge in Lp. We will show however, that itis weakly (and therefore weak*) convergent with xn ⇁ 0. Let g ∈ Lq(R). Since∫R |g|

qdx = ‖g‖qq <∞, we have that∫∞n|g|qdx→ 0 as n→∞. Thus

|〈χ[n,n+1], g〉| =∣∣∣∣∫

Rχ[n,n+1]gdx

∣∣∣∣ = ∣∣∣∣∫ n+1

n

gdx

∣∣∣∣ ≤ ∫ n+1

n

1 · |g|dx

≤(∫ n+1

n

1pdx

)1/p(∫ n+1

n

|g|qdx)1/q

=

(∫ n+1

n

|g|qdx)1/q

≤(∫ ∞

n

|g|qdx)1/q

→ 0 as n→∞.

Wherefore χ[n,n+1] ⇁ 0.

Though this double counterexample is e�cient, it may be somewhat unsatisfyingto only have a counterexample for when the weak and weak* convergence coin-cides. We will therefore give two more counterexamples in the two non-re�exivespaces L1(R) and L∞(R). These will use the following result.


x

y

0

1f2 f4 f6 f8

y = g(x)

· · ·

Figure 4.5: 〈fn, g〉 → 0 but fn 6→ 0 in Lp.

Theorem 4.23. Riemann-Lebesgue's lemma (Gasquet-Witomski [6], The-orem 17.1.3) Let f ∈ L1(R), then∫

Rf(t)e−intdt→ 0 as n→∞.

Example 4.24. Consider the sequence (xn) = (e−int)n. Note that

∀n ∈ N ‖xn‖∞ = ess supR|e−int| = ess sup

R1 = 1, (4.5)

thus (xn) ⊂ L∞(R). Now by Riemann-Lebesgue's lemma 4.23, e−int∗⇁ 0.

However, (4.5) implies that xn 6→ x in norm. Thus in L∞, xn∗⇁ x 6=⇒

xn → x.

Example 4.25. Next consider the sequence (xn) = (χ[0,π](t)e−int), notice that

∀n ∈ N : ‖xn‖1 =

∫R|χ[0,π](t)e

−int|dt =∫ π

0

dt = π, (4.6)

thus (xn) ⊂ L1(R). Let f ∈ L∞(R), then

‖χ[0,π]f‖1 =

∫R|χ[0,π]f |dt ≤ ‖f‖∞

∫ π

0

dx = π‖f‖∞ <∞.

Hence χ[0,π]f ∈ L1(R), so by Riemann-Lebesgue's lemma 4.23

〈xn, f〉 =∫R(χ[0,π]f)(t)e

−intdt→ 0, as n→∞.

Thus xn ⇁ 0. However, by (4.6) xn 6→ 0 in norm. Hence in L1, xn∗⇁ x 6=⇒

xn → x.

65

We �nish this section by providing an overview of the results in �gures 4.6 and4.7.

Uniform Quasi uniform Pointwise

L∞ Almost everywhere Weak(*) L∞

Almost uniform Measure Lp

Implies

Implies if ‖fn‖p → ‖f‖p <∞

Implies for �nite measures

Implies convergence of subsequence

4.3

4.3

4.3

4.3

4.2

4.2

4.2

4.16

4.84.15

4.20 4.21

4.18

4.174.13

4.12

4.20 4.21

Figure 4.6: Positive results.


Uniform Quasi uniform Pointwise

L∞ Almost everywhere Weak(*) L∞

Almost uniform Measure Lp

Counterexample provided

Follows logically

/

4.4

/

/

/

4.14

/4.5

/

/

/4.5

/4.7

/ /

4.25 4.22

/4.24

/4.6

/

4.6

/4.10

/

/4.7

/

4.10

/

/

Figure 4.7: Negative results.

Chapter 5

Further Research

This thesis was mostly concerned with measurable functions f : A→ R, A ⊂ R,with respect to the Lebesgue measure. A majority of the results however, stillhold if we consider an arbitrary measure µ. We could still de�ne an integralusing µ and thus acquire Banach spaces Lp(µ). These spaces are still normedso sequences su�ce to describe the topology. In more general topological spaceshowever a generalization of sequences called nets is needed. One natural topicfor further research is therefore exploring how di�erent types of nets are re-lated (by convergence) on more general spaces. Quasi uniform convergence inparticular can be generalized quite extensively as seen in [4].


Appendix A

Convex Functions


70 Appendix A. Convex Functions

This appendix introduces convex functions in order to prove Lemma 4.9. Letus start with the de�nition.

De�nition A.1. (Thompson [10], De�nition 7.33) Let f : I → R where I is aninterval. The function f is convex on I if

∀x, y ∈ I, λ ∈ [0, 1] : f(λx+ (1− λ)y) ≤ λf(x) + (1− λ)f(y).

In order to prove Lemma 4.9 we will �rst need the following lemma.

Lemma A.2. Let f ∈ C2(I) where I is an interval and suppose that f ′′(x) ≥ 0for all x ∈ I. Then f is convex.

Proof: Let a, b ∈ I, we may without loss of generality assume that a < b. Takex0 ∈ [a, b], then

x0 = λa+ (1− λ)b (A.1)

for some λ ∈ [0, 1]. Since f ∈ C2(I), Lagrange's remainder theorem (Abbott [1],Theorem 6.6.3) implies that

f(x) = f(x0) + f ′(x0)(x− x0) + f ′′(c)(x− x0)2, x ∈ I,

for some c ∈ I. We know that f ′′(c) ≥ 0, therefore f ′′(c)(x − x0)2 ≥ 0 andremoving this term yields the inequality

f(x) ≥ f(x0) + f ′(x0)(x− x0). (A.2)

Using (A.1) together with the above equation, and setting x to a and b respec-tively we arrive at the two inequalities

f(a) ≥ f(x0) + f ′(x0)(1− λ)(a− b), (A.3)

andf(b) ≥ f(x0)− f ′(x0)λ(a− b). (A.4)

Taking λ · (A.3)+ (1− λ) · (A.4) we get that

λf(a) + (1− λ)f(b) ≥ λf(x0) + f ′(x0)λ(1− λ)(a− b)+ (1− λ)f(x0)− f ′(x0)λ(1− λ)(a− b)

= f(x0) = f(λa+ (1− λ)b).

Thus f is convex and we are done.

Finally we are ready for the main proof.

71

Proof of Lemma 4.9: Let b ≥ a ≥ 0 and let 1 ≤ p <∞. Set f(x) := xp, x ≥ 0,we want to show that f(b− a) ≤ f(b)− f(a). To begin with notice that

∀x : f ′′(x) = p(p− 1)xp−2 ≥ 0,

so by Lemma A.2 the function f is convex. De�ne F (x) := f(b)b x. Since f is

convex we have that

f(x) ≤ F (x), x ∈ [0, b], as seen in �gure A.1.

Speci�cally f(b− a) ≤ F (b− a) and −F (a) ≤ −f(a), so by the linearity of F

f(b− a) ≤ F (b− a) = F (b)− F (a) ≤ f(b)− f(a).

x

y

0 a b− a b

y = f(x)

y = F (x)(b− a)p

bp − ap

Figure A.1: f(x) = xp.

Bibliography

[1] Stephen Abbott. Understanding analysis. Springer, 2001.

[2] Richard F. Bass. Real analysis for graduate students. Western University,2011.

[3] Richard Beals. Analysis: an introduction. Cambridge University Press,2004.

[4] Agata Caserta, Giuseppe Di Maio, and Lubica Holá. Arzelà's theoremand strong uniform convergence on bornologies. Journal of MathematicalAnalysis and Applications, 371(1):384�392, 2010.

[5] Gerald B Folland. Real analysis: modern techniques and their applications.John Wiley & Sons, 1999.

[6] Claude Gasquet and Patrick Witomski. Fourier analysis and applications:�ltering, numerical computation, wavelets. Springer Science & BusinessMedia, 2013.

[7] Erwin Kreyszig. Introductory functional analysis with applications. JohnWiley & Sons, 1989.

[8] Norbert Kusolitsch. Why the theorem of Sche�é should be rather called atheorem of Riesz. Periodica Mathematica Hungarica, 61(1):225�229, 2010.

[9] Gert K Pedersen. Analysis now. Springer Science & Business Media, 2012.

[10] Brian S Thomson, Judith B Bruckner, and Andrew M Bruckner. Elemen-tary real analysis. ClassicalRealAnalysis.com, 2008.


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