Measure theory and integration-Woodhead _ G De Barra

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MEASURE THEORY AND INTEGRATION

"Talking of education, people have now a-days" (said he) "got a strange opinion that every thing should be taught by lectures. Now, I cannot see that lectures can do so much good as reading the books from which the lectures are taken. I know nothing that can be best taught by lectures, except where experiments are to be shewn. You may teach chymestry by lectures — You might teach making of shoes by lectures!" James Bosweil: Life of Samuel Johnson, 1766

ABOUT OUR AUTHOR Gearoid de Barra was born in the city of Galway, West Ireland and moved as a young boy to Dublin where he spent his schooldays. He then studied mathematics at University College Dublin, National University of Ireland

where he gained his BSc. Moving to England, he graduated from the University of London with a PhD for research on the convergence of random variables, an area of application of some of the material covered in this book. He then transferred to Hull University in Yorkshire for a teaching appointment; and afterwards spent two summers in 1975 and 1988 in Australia, teaching and researching at the University of New South Wales. More recently, he became Senior Lecturer at the Royal Holloway College, University of London, to continue teaching and research related to aspects of operator theory and measure theory involving ideas from the material in this book.

He has enjoyed teaching university mathematics at all undergraduate and postgraduate levels, including many courses on measure theory and its

applications to functional analysis, from which source this book has developed.

The first edition was the standard text in the departments of mathematics at both Cardiff University and Royal Holloway College, and has attracted attention in Canada and Scandinavia. It was also translated into Italian as Teoria del/a Misura è deli integrazione.

Measure Theory and Integration

C. de Barra, PhD Department of Mathematics Royal Holloway University of London

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Contents

Preface

.

Notation

9 11

Chapter 1 Preliminaries Set Theory 1.1 1.2 Topological Ideas 1.3 Sequences and 1Mm 1.4 Functions and Mappings 1.5 Cardinal Numbers and Countability 1.6 Further Properties of Open Sets 1.7 Cantor-like Sets

15

17 18 21

22 23 23

Chapter 2 Measure on the Real Line 2.1

2.2 2.3 2.4 2.5 2.6

Lebesgue Outer Measure Measurable Sets Regularity Measurable Functions Borel and Lebesgue Measurability Hausdorff Measures on the Real Line

Chapter 3 Integration of Functions of a Real Variable Integration of Non-negative Functions 3.1 3.2 The General Integral 3.3 Integration of Series 3.4 Riemann and Lebesgue Integrals Chapter 4 Differentiation The Four Derivates 4.1

27 30

37 42 45

54 60 68 71

77 S

Contents

6

4.2 4.3 4.4 4.5 4.6

Continuous Non-differentiable Functions Functions of Bounded Variation Lebesgue's Differentiation Theorem Differentiation and integration The I.ebesgue Set

Chapter 5 Abstract Measure Spaces 5.1 Measures and Outer Measures 5.2 Extension of a Measure 5.3 Uniqueness of the Extension 5.4 Completion of a Measure 5.5 Measure Spaces 5.6 integration with respect to a Measure

.79 81

84 87

90

93 95

99 100 102 105

Chapter 6 Inequalities and the L" Spaces 6.1

Thel/Spaces

109

6.2 6.3 6.4 6.5

Convex Functions Jensen's Inequality The Inequalities of Holder and Minkowski

111

Completeness of L'ip)

113 115 118

Chapter 7 Convergence 7.1

7.2 7.3 7.4

Convergence in Measure Almost Uniform Convergence Convergence Diagrams Counterexamples

Chapter 8 Signed Measures and their Derivatives 8.1 Signed Measures and the Hahn Decomposition 8.2 The Jordan Decomposition 8.3 The Radon-Nikodym Theorem 8.4 Some Applications of the Radon-Nikodym Theorem 8.5 Bounded Linear Functionals on

121

125 128 131

133 137 139 142 147

Chapter 9 Lebesgue-Stieltjes Integration 9.1

9.2 9.3 9.4 9.5 9.6

Lebesgue-Stieltjes Measure Applications to Hausdorff Measures

Absolutely Continuous Functions Integration by Parts Change of Variable Riesz Representation Theorem for C(J)

153 157 160 163 167 172

Contents

Chapter 10 Measure and Integration in a Product Space 10.1 Measurability in a Product Space 10.2 The Product Measure and Fubini's Theorem 10.3 Lebesgue Measure in Euclidean Space 10.4 Laplace and Fourier Transforms Hints and Answers to Exercises Chapter 1 Chapter 2 Chapter 3 Chapter 4 ChapterS Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10

7

176 179 185 189

197 198

204 209 211 215

220 223 227 230

References

236

Index

237

Preface to First Edition This book has a dual purpose, being designed for a University level course on measure and integration, and also for use as a reference by those more interested in the manipulation of sums and integrals than in the proof of the mathematics involved. Because it is a textbook there are few references to the origins of the subject, which lie in analysis, geometry and probability. The only prerequisite is a first course in analysis and what little topology is required has been developed

within the text. Apart from the central importance of the material in pure mathematics, there are many uses in different branches of applied mathematics and probability. In this book I have chosen to approach integration via measure, rather than the other way round, because in teaching the subject I have found that in this

way the ideas are easier for the student to grasp and appear more concrete. Indeed, the theory is set out in some detail in Chapters 2 and 3 for the case of the real line in a manner which generalizes easily. Then, in Chapter 5, the results for general measure spaces are obtained, often without any new proof. The essential L" results are obtained in Chapter 6; this material can be taken immediately after

Chapters 2 and 3 if the space involved is assumed to be the real line, and the measure Lebesgue measure.

In keeping with the role of the book as a first text on the subject, the proofs are set out in considerable detail. This may make some of the proofs longer than

they might be; but in fact very few of the proofs present any real difficulty. Nevertheless the essentials of the subject are a knowledge of the basic results and an ability to apply them. So at a first reading proofs may, perhaps, be skipped. After reading the statements of the results of the theorems and the numerous worked examples the reader should be able to try the exercises. Over 300 of these are provided and they are an integral part of the book. Fairly detailed solutions are provided at the end of the book, to be looked at as a last resort.

Different combinations of the chapters can be read depending on the student's interests and needs. Chapter 1 is introductory and parts of it can be read in detail according as the definitions, etc., are used later. Then Chapters 2 and 3 provide a basic course in Lebesgue measure and integration. Then Chapter 4 gives essential results on differentiation and functions of bounded variation, all for functions on the real line. Chapters 1, 2, 3, 5, 6 take the reader as far as general measure spaces and the L" results. Altematively, Chapters 1, 2, 3, 5, 7 9

10

introduce the

Preface

reader to convergence in measure and almost uniform

convergence. To get to the Radon-Nikodym results and related material the reader needs Chapters 1, 2, 3, 5, 6, 8. For a course with the emphasis on differentiation and Lebesgue-Stieltjes integrals one reads Chapters 1, 2, 3, 4, 5, 8, 9. Finally, to get to measure and integration on product spaces the appropriate

route is Chapters 1, 2, 3, 5, 6, 10. Some sections can be omitted at a first reading, for example: Section 2.6 on Hausdorff measures; Section 4.6 on the Lebesgue set; Sections 8.5 and 9.6 on Riesz Representation Theorems and Section 9.2 on Hausdorff measures. Much of the material in the book has been used in courses on measure theory

at Royal Holloway College (University of London). This book has developed

out of its predecessor introduction to Measure Theory by the same author (1974), and has now been rewritten in a considerably extended, revised and updated form. There are numerous proofs and a reorganization of structure. The important new material now added includes Hausdorif measures in Chapters 2 and 9 and the Riesz Representation Theorems in Chapters 8 and 9.

Ode Barra Preface to Second Edition The material in this book covers several aspects of classical analysis including measure, integration with respect to a measure and differentiation. These topics lead on to many branches of modem mathematics. So some notes have been added which indicate some of the directions in which the material leads. These

are less formal than the main text and contain some references for further reading.

Ode Barra Royal Holloway, University of London January 2003

Notation

Notation is listed in the order in which it appears in the text.

0: end of proof. iff: if, and only if. 3: there exists. v: given any. x CA: x is a member of the set A. A c B, (A 2 B): set A is included in (includes) the set B. A C B: set A is a proper subset of the set B. Ex: P(x)I. the set of those x with property P. CA: the complement of A. 0: the empty set. U, fl: union, intersection (of sets). A —B. the set of elements of A not in B. A A B = (A — B) U (B — A): the symmetric difference of the sets A, B. Z: integers (positive or negative). N: positive integers. Q: rationals. R: real numbers. ?(A): the power set of A, i.e. the set of subsets of A. A X B: the Cartesian product of the sets, A, B. [xl: the equivalence class containing x (Chapter 1), or, in Chapter 2, etc., the closed interval consisting of the real number x. Ix, the metric space consisting of the space X with metric p. A: the closure of the set A. G5 set: one which is a countable intersection of open sets. set: one which is a countable union of closed sets. 11

Notation

12

inf A, sup A: infirnum and supremum of the set A. Urn sup x,,, liin upper and lower limits of the sequence (x,,). x(a—),

x(a+): left-hand, right-hand limits of x at a. So x(a+) is the function

whose value at a is lim x,, = o(n"): ,f1' x,, 0. x,, = O(n"): x,, is bounded.

-* a,

> a. Similarly flx+),ftx—), etc.

characteristic function of the set A (= I on A, = 0 on CA). Card A: the cardinality of A. the cardinality of N. (2(a): (in Theorem 9, Chapter 1) the equivalence class containing a. Cantor-like sets. etc.: the 'removed intervals'; '1,1 the 'residual intervals', for the Cantor-like sets.

N(x,e): the set [z': It—xi 0 such that xa F(x). So suppose x and y lie in is nowhere dense, there than as is an interval 'k c (x,y) for some n,k. But then F(y) — F(x)> 1/3", so F is strictly increasing. Exerci9es

7. Find the length of the intervals of Example 5. 8. Using the notation of the construction of the Lebesgue function, p. 25, show that the estimate of ILm — L I may be improved to ILm — LI 1/3.2m.

CHAPTER 2

Measure on the Real Line

We consider a class of sets (measurable sets) on the real line and the functions (measurable functions) arising from them. It is for this large class of functions that we will construct a theory of integration in the next chapter. On this class of sets, which includes the intervals, we show how to define Lebesgue measure which is a generalization of the idea of length, is suitably additive and is invariant under translations of the set. Apart from integration theory the methods are of independent interest as tools for studying sets on the real line. Indeed for sets which are 'scanty' we further refine the idea of measure in the last section §2.6 and construct Hausdorff measures particularly appropriate for the Cantor-like sets constructed in the last chapter. Sections 2.5 and 2.6 will not be used in the integration theory of the next chapter. 2.1 LEBESGIJE OUTER MEASURE

All the sets considered in this chapter are contained in R, the real line, unless stated otherwise. We wifi be concerned particularly with intervals I of the form

I=

[a,b), where a and b are finite, and unless otherwise specified, intervals may

be supposed to be of this type. When a = b, I is the empty set 0. We will write 1(f) for the length of I, namely b —a.

Definition 1: The Lebesgue outer measure, or more briefly the outer measure, of a set is given by m(A) = inf where the infimum is taken over all finite or countable collections of intervals [I,,J such that A c U For notational convenience we need only deal with countable coverings of A; the finite case is included since we may take = 0 except for a finite number of integers n.

Theoremi: (ill) m*(A)Cm*(B) if A c B,

(iv) m*([x])=OforanyxeR. 27

[th.2

Measure onthe Real Line

28

Proof: (i), (ii) and (iii) are obvious. Since x = 1/n, (iv) follows. 0 and

[x, x + (1/n)) for each n,

I,,

Example 1: Show that for any set A, m*(A) = m*(A + x) where A + x = , that is: outer measure is translation invariant. [y + x: y and Solution: For each e > 0 there exists a collection [Ia] such that A ç m*(A) + x). So, for each e, m(A +x) 1(1,,) — e. But clearly A + x c

m*(A + x).

so we have

Theorem 2: The outer measure of an interval equals its length.

Proof: Case 1. Suppose that I is a closed interval, [a,bJ, say. Then, for each e >0, we have from Theorem 1 and Definition 1 that

+

—a + e,

m*(T)0,! may be covered by a collec-

tion of mtervals [Ia] such that m*(f)> For each n, let i;, = Theorem,

—

€12",

— e,

ba), then U i;,

n1

where I,, =

say.

I, so by the

p. 18, a finite subcollection of the 1,, say

(Ck, dk), covers I. Then, as we may suppose that no 1k

is

. ., where 1k = contained in any other,

wehave,supposingthatc1 l(f)—2e.

(2.2)

Then (2.1) and (2.2) give the result.

If a = b, Theorem 1(u) gives the result.Ifa 0, there is an open set 0 containingA and such that m*(O) 0 there is nothing to prove. If m(G — F) = 0, then since G — F is open we must, by (1), have G F. But G contains Q whose closure

is R, soF= Rand m(F)=oo. But m(G)0, 0 open, 0 2 E such (2.11)

—

E) can be made arbitrarily small, then E is measur-

able by the last theorem. Write J =

U

and U =

0

ii J. Then by Example 1,

p. 16, and subadditivity

(2.12)

Since U ç J we have U — E ç J — E and since E c Owe have E — U = E — J.

ButE ç

So

+ e. So by (2.11),m(O U)=m(O—U)=m(O)—m(U) al = A, A or 0, and the result

Solution: Depending on a, the set [x: follows.

Example 12: Continuous functions are measurable.

Solution: 1ff is continuous, [x: flx) > a] is open and therefore measurable. Theorem 13: Let c be any real number and let f and g be real.valued measurable functions defined on the same measurable set E. Then f + c, cf, f + g, f— g and fg are also measurable.

Proof: For each a, [x: f(x) + c > aJ = [x: flx) >a —c] , a measurable set. So f + c is measurable. If c 0, cf is measurable as in Example 10 above; otherwise, if c> 0, [x: cf(x) > al = [x: flx) > c1a1, a measurable set, and similarly for c a] only if f(x) > a —g(x), that is, only if there exists a rational r1 such thatf(x) > > a —g(x), where fri, 1 1, 2,. . . } is an enumera-

tion of Q. But theng(x) >a —rj and soxEE [x:f(x)>rj] fl [x:g(x)>a —rj]. B=

Hence A

U ([x: f(x) > rjj

[x: g(x) > a — r,]), a measurable set.

1=1

Since A clearly contains B we have A = B and so f + g is measurable. Then f—g =f+ (—g)is also measurable. + g)2 — (f — g)2), so it is sufficient to show that Finally: fg =

[x:f2(x)>aJ =

f is. if a < 0, [x: f2(x) > a] U

R is measurable. If a

0,

set. U

Corollary: The results hold for extended real-valued measurable functions except or vice versa, and similarly that f + g is not defined wheneverf= andg

forf—g. For [x:f(x) +g(x)>aJ = U ([x: 11x) > i=

ñ [x: g(x)>a —r,1)U

1

measurable set. The case of f—g is similar. 0

Theorem 14: Let {f,, } be a sequence of measurable functions defined on the same measurable set. Then

sup fjismeasurableforeachn,

(i) I

(ii)

inf

(iii) sup

is measurable for each ii, is

is measurable, (v) lim sup is measurable, is measurable. (vi) km

(iv)

[Ci'. 2

Measure on the Real Line

40 Proof: (1)

Since [x:

sup f1(x)> al =

U [x: f,(x) > al , we have 1=1

sup f1 1 al ,so is measurable. n= I = —sup (iv) and so is measurable. (v) lim sup = inf(sup f1), a measurable function by (iii) and (Iv). i>n = — lim sup (vi) urn and so is measurable. 0

Definition 8: In line with Definition 5, we say that the function f is Borel measurable or a Borel function if V a, [x: f(x) > a] is a Bore! set.

Theorems 12, 13, 14 and their proofs, apply also to Bore! functions when 'measurable function' and 'measurable set' are replaced throughout by 'Borel measurable function' and 'Borel set' respectively. The next theorem cannot be adapted in this way: see Exercise 43, p. 45.

Definition 9: If a property holds except on a set of measure zero, we say that it holds almost everwhere, usually abbreviated to a.e.

Theorem 15: Let f be a measurable function and let f= g a.e. Theng is measurable.

Proof: [x: flx)> al [x: g(x) > a] ç [x: flx) *g(x)] and the result follows immediately from Example 4, p. 30, and Example 5, p. 31. 0

Example 13: Let be a sequence of measurable functions converging a.e. to f; then f is measurable, since f = lim sup f1, a.e.

Example 14: If f is a measurable function, then so are If = max (f, 0) and

f=-minQO).

Solution: Example !0 and Theorem 14, (i) and (ii), give the result.

Example 15: The set of points on which a sequence of measurable functions { f,, } converges, is measurable.

Solution: The set in question is [x: lim sup — lim measurable by Theorem 14, (v) and (vi), and Example 9, p. 38.

Definition 10: Let f be a measurable function; then inf(a: f the essential supremum of f, denoted by ess supf. Example 16: Show that

ess sup f, a.e.

Solution: If ess sup f =

the

= 01 which is a a.e.] is catted

result is obvious. Suppose ess sup f =

Then

Measurable Functions

Sec. 2.41

41

a.e., as required. So suppose n a.e. from Definition 10. Sof = = ess sup and E = [x: flx)> [x: fIx)> 1/n + that ess sup f is finite. Write

V n E Z, f

ess

supfj ,so E =

U

,t=

=0,so m(E) = 0.

But from its definition

1

Example 17: Show that for any measurable functions f and g

supf+ ess supg,

ess sup

and give an example of strict inequality.

Solution: From the last example

f+g 0. Find the measure of the set 29. Show that monotone functions are measurable.

30. Let f= g a.e. where f is a continuous function. Show that ess supf= ess supg = supf. supf. 31. Show that for any measurable functionf, ess sup functions and g 0. Show that 32. Let f and g be a.e.

Measure on the Real Line

42

[Qi. 2

f

be a measurable function not almost everywhere infinite. Show that there exists a set of positive measure on which f is bounded. 34. Let f be a measurable function on [a,bj and let fbe differentiable a.e. Show that there is a function, measurable on [a,b] , which equalsf a.e. 35. Let f be a continuous function and g a measurable function. Show that the composite function f o g is measurable. 36. Let x E [0,1] have the expansion to the base 1, x = 0 x1x2.. . x,, . . . for some integer 1, the non-terminating expansion being used in cases of ambix for each n. guity. Show that

33. Let

2.5 BOREL AND LEBESGUE MEASURABILITY

of Bore! sets, the class This section considers the relation between the class of Lebesgue-measurable sets and the class 'P(R) of all subsets of R. The section may be omitted without loss of logical continuity, but it provides tions between these classes without which the theory, though stifi valid, would be rather artificial. ¶P(R). Using Theorem 16 From Theorem 8, p. 32, we have that and in Theorem 18 that * i)(. we show in Theorem 17 that is measurable or Borel measurable if, and Since the characteristic function only if, A is measurable or is a Bore! set, respectively (Example 11, p. 39), we have corresponding relations between the two classes of measurable functions and the class of all real-valued functions.

Theorem 16: Let E be a measurable set. Then for eachy the setE +y x EE] is measurable and the measures are the same. Proof: By Theorem 10, p. 36, ye >0,

e. Then the set 0 + y is open and 0 +y

[x +y:

an open set 0,0

E+y. But (0+y)—(E+y)=

if) + y. By Example 1, p. 28, m((0 — E) + y)

e and the measurability of E + y follows, using Theorem 10 again. That the measures are equal follows on using Example 1 again. D (0 —

Theorem 17: There exists a non-measurable set.

Proof: Ifx,yE [0,lJ,letx '\'yify—xEQ1 =Qfl[—l,1].Then"iseasily seen to be an equivalence relation on [0,11 and so by Chapter 1, p. 17, [0,11 = UEa, Ea disjoint sets such that x andy are in the same Ea if, and only if, X Since is countable, each Ea is a countable set. Since [0,11 is uncountable

there are uncountable many sets Ea. Using the Axiom of Choice, p. 17, we consider a set V in [0,11 containing just one element xa from each Ea. Let {rj} be an enumeration of Q1, and for each n write V,, V + r,,. If y E fl Vm there exist xa, x0 E V such thaty = + andy = x0 + Tm. But thenx0 — xc, E 01, so = xc, by definition of V and we have n = m. So V,, ñ Vm 0

Sec. 2.51

for n # m. Also [0,11

43

Borel and Lebesgue Measurability

U

n=

ç [—1,21 , since

V,,

V

x E [0,1] , x E

for some

1

a and then x =

xE

is each V,, and m(V) the V Then using the measurability of the sets V,, we have

But this sum can only be 0 or

So

V is not measurable. 0

Theorem 18: Not every measurable set is a Borel set.

Proof: Write each x E [0,11 in binary form

= 0 or 1, choosing a non4erminating expansion for each x> 0. Define the function fby with

the values of f, which is known as Cantor's function, lie entirely in the is a measurable function of x (Exercise 36), f is Cantor set P, p. 24. Since measurable. Also f is a one4o-one mapping from [0,1] onto its range, since the Then

value fl:x) defines the sequence

in the expansion

uniquely, so x is

determined uniquely. were the same, then by Example 19, p. 41, r' (B) would be If and measurable for any measurable set B and any measurable function f. Let f be the Cantor function and V a non-measurable set in [0,11. Then B fiV) lies in P and so has imasure zero. So B is measurable. But since f is one-to-one, f' (B) 0 We conclude that is strictly contained in V which is We now give two examples showing unexpected implications of measurability.

Example 20: Let Tbe a measurable set of positive measure and let T* = [x —y: x E T,y E TI. Show that T* contains an interval (—a,a) for some a >0.

Solution: By Theorem 10, p. 36, T contains a closed set Cof positive measure. Since m(C) = lim m(C fl [—n,n]) we may assume that C is a bounded set. By Theorem 10, again, there exists an open set U, U D C, such that m(U — C) < m(C). Define the distance between two sets A and B to be d(A ,B) = inf[lx —y 1: Since lx — y I is a continuous function of x andy, the distance x E A, y E between A and B is positive if A and B are disjoint closed sets one of which is

[Qi. 2

Measure on the Real Line

44

bounded. Let a be the distance between the closed sets C and CU, so that a> 0. Let x be any point of (—a,a). We wish to show that Cfl(C—x) #0. For then, z E C such that since C — x = [y: y + x E CJ, we have that v x E

z'=z+xECandsox=z'—zEr. Since

lxi 0 and Ift 0. So if lxi 0) is contained in an open interval /' of length e(1 + Then

= inf = (1 +

1(4)

, J,

closed, A c U 4

inf

open,A culk,

(1

= (1 Let

0 to get

as required.

Theorem 20: Let H(A)p. Then H(A)= 0. Proof: Let

> 0 and let Elk] be a covering of A by intervals with l(ik)

each k. Then l(IkY'

—

11 (k)

for

Hausdorff Measures on the Real Line

Sec. 2.61

Taking the infirnum of the last term over

(A) So all such coverings we get

(A) < eq-p

(A)

Letting 6

47

H(A).

0, the result follows. D

then H(A)= oo for q 0 there exists a family of intervals /=1

(E,)

1=1

with

U Ijj for each i, and

such that

Then U

H(E1).

E1

—

/=1

2

c U U Ijj,so E1) 1=1

+ e and the result follows. 0

So

We now show that Hausdorff measure includes Lebesgue measure as a special case.

= m*(A).

Theorem 22: For any set A,

Proof: From their definitions we have m*(A)

for all

> O,so m*(A)

Hr(A). So we wish to show that m*(A) and clearly we may assume m*(A) < Then for any given e >0 we have for some family of intervals [Ia]

that A U n1

and m*(A)

n1

— e.

Now if I is any finite interval and

[Qi.2

Measure onthe RealLine

48

>0 then (2.14) For let b >0 and choose open intervals J1, and

with 1(J1)

0, and, letting -÷0

c' = c 1

?11 by the last theorem, giving the result. []

Corollary 1: Let I be an interval of positive or infinite length. Then H(I) = Proof: Suppose that I is a finite interval. Then the result follows from Theorem 20 and its Corollary, and Theorem 21. So suppose that I is an infinite interval and p> I. Since I C Ulk, 'k fImte intervals, we have H(I) =0. So the result is true for p 1 and Theorem 20 ensures that H(J) = for 0 0), either H(A) 0 for allp >0, or for some Po (0 0 then H(A U B) = Proof: It is sufficient to show that the same identity holds for

& But the identity is immediate for H,6 provided

for all small

0. So

n—i

k1

11(D2k)-*oo

contradicting the finiteness of H(A). Similarly the second series in (2.16) must converge and the result follows. U Corollary: All Borel sets are H-measurable. For a Bore! setA we will use the notations

and Hpb(A).

[Ch. 2

Measure on the Real Line

50

Definition 15: The Hausdorffdiinension of E is inf[p: H;(E) = 01. By Corollary 4 to Theorem 22 Hausdorff dimension is a well defIned number in the interval [0,11 for any set A in the real line. Also, except for the trivial case of sets E with = 0 for all p, we have that Hausdorff dimension equals

sup[p:H(E)=oo]. Theorem 27: The Hausdorff dimension of the Cantor-like set

is — log

2/log

Proof: We refer to the construction of

given in p. 24. The first residual intervals J1,1 and J1,2 are translates of multiples of [0,1] and contain subsets, say p('I and p(2), of Since, to get we dissect J1,1 in the same way that [0,1] was dissected, we have that (and $2)) isa translate of a multiple (by of Then, by Theorem using Examples 22 and 23. So either = 0 or or 1 giving p 2/ log = say. We will show that 0 (Ps) 0. The distance between the sets p(') and p(2) is at — Then as in Theorem 25, any cover [Ii] of by intervals of length at most may be decomposed into covers , / = 1, 2, of least 1 —

Let

1

the sets $fl,fi2) and

+

= I

I

(2.18) i

Suppose the first sum of the right of (2.18) is the lesser. Since p(2) is a translate of $1) the same translation applied to the intervals [11,1] gives a cover say of Then as for (2.17), but in reverse, we may map the intervals onto intervals ['], say, covering and with l(1) = for each i. So i(I;y'O = + as for (2.17) I

I

i

,

by construction.

So if any one of the intervals I is of length

(2.19) 1—

we have

>

Hausdorff Measures on the Real Line

Sec.

51

Now since is compact we may suppose, by the Heine Borel Theorem, p. 18, that all the coverings considered are finite, so nun l(Ik) > 0. Since the intervals of a subset of the intervals uk1 we have [11 are multiples (by I —I

1

(2.20)

minl(Ik).

If each interval 1 is of length less than I — we apply the same process to the which was applied to the cover [I,] , and we must obtain after a fInite )Po l(J1)Po number of steps, a cover 1'?] with max 1(1?) 1 — and cover

(I

as in (2.19). So in any case we have required. D

Corollary: For each diimnsion a.

—

1, there exists a set Q

0

(Ps) >0, as

So

R with Hausdorff

Po. The follow(A)< equal oofor 0 ff dx.

(I) and (ii) are immediate from DefInitions 2 and 3 (p. 55) respectively. (ill) is obvious if a = 0. If a> 0, is a measurable simple function with ip af where is simple, ti f, and then f p dx a f dx if, and only if, p = (Theorem 1(iii), above). So f af dx = sup f p dx = a sup f dx a If dx. and apply (i). 0 (iv): Note that The following result will be basic in proving convergence theorems.

Jntegration of Non-negative Functions

Sec. 3.fl

Theorem 3 (Fatou's Lemma): Let {f,,, n =

1, 2,

.

.

57

.} be a sequence of non-

negative measurable functions. Then

Iiminfff,,dx>fliminff,,dx.

(3.2)

Fivof: Let f =

urn inf f,,. Then f is a non-negative measurable function. From Definition 2, p. 55, the result follows if, for each measurable simple function we have with

(33) Then from Definition 1, p. 54, for some measurable set Case I. f dx a >0 on A. Write = inf fj(x), and A,, = A, we have m(A) =00 and k

a measurable set. Then A,, ç a, all k is monotone increasing and lim gJ((x) = J1x)

Ix: each x,

each n. But, for

p(x). So A c

U ,,1 ff,, dx>fg,, dx>am(A,,). So liin miff,, dx =

and (3.3) holds.

f p dx 0]. Then m(B) f,, by hypothesis, so by Theorem 2(i), p. 56, ff dx ) f f,, dx, and hence (3.5) dx. Relations (3.4) and (3.5) give the result. 0

Theorem 5: Let f be a non-negative measurable function. Then there exists a sequence {p,, } of measurable simple functions such that, for each x, p,,(x) t Proof: By construction. Write, for each n, E,,k = [x: (k — 1)12" nJ. Put n2"

k—I

+

2" Then the functions so,, are measurable simple

=

k= 1

Also, since the dissection it is easily seen is a refinement of that giving p,,(x) for each x. If fix) is finite, x E CF,, for all large n, and then

of the range off giving that

(x)

T". So v,,,(x) tjl:x).Iffix)= oo,thenx E

F,,, so p,,(x) = n

for all n, and again p,,(x) t fix). 0 Corollary: lim ftp,, dx = ff dx, where tp,, and fare as in Theorem 5. This application of Theorem 5, with Theorem 4, gives us a method of evaluating f f dx alternative to that of Definition 2, p. 55. Theorem 6: Let f and g be non-negative measurable functions. Then

ffdx+fgdx—f(f+g)dx.

(3.6)

Proof: Consider (3.6) for measurable simple functions s.p and

Let the values

oftpbea1,...,a,,takenonsetsA1,...,A,,,andletthevaluesofi/ibeb1,..., bm taken on sets B,, . . . , B,,,. Then the simple function tp + i,i' has the value a1 + b1 on the measurable set A1 ñ B1, so from Theorem 1(i), p. 56, we obtain

fA1flB1

A1flB,

tpdx+f

i,&dx.

(3.7)

But the union of the nm disjoint sets A1 fl Bj is R, so Theorem 1(u) applied to both sides of (3.7) gives (3.8)

Let f and g be any non-negative measurable functions. Let ftp,,), {ç",, } be t sequences of measurable simple functions, tp,, f, t g. f (tp,, + ip,,) dx = f tp,, dx + f dx. So, letting n tend to infinity, Theorem 4 gives the result. 0

Sec. 3.IJ

Integration of Non-negative Functions

59

Theorem 7: Let { f,,) be a sequence of non-negative measurable functions. Then

By induction, (3 .6) applies to a sum of n functions. So if Sn

f=

fS,,dx

fj, then

=

fj, so the result follows from Theorem

Example 4: Give an example where strict inequality occurs in (3.2) (Fatou's Lemma). n = 1,2,... . Then lim inffn(x) = 0, Xfo,ij ,f2n = for all x, but ffn(X) dx = 1, for all n. Such an example is helpful in remembering the 'direction of the inequality' in (3.2).

Solution: Let f2n-1

Examples:Showthatf

clxfx=oo.

Solution: x' is a continuous function for x > 0, and so is measurable. It is

positive, so the integral is defined. Also f X' dx >

f

dx. But x' > k'

1n on [k — 1,k),

dx

6: f(x), 0

/(x) =

x

k=211 k' X[k—i,k) dx

k2

k1

as

n

1, is defined by:f1x) =0 for x rational, if x is irrational,

n, where n is the number of zeros immediately after the decimal point,

in the representation of x on the decimal scale. Show that f is

and

find I fdx. Solution: For x E (0,!] let g(x) = n if x < 1(T", n = 0, 0. Then g a.e., so f is measurable and by Example 3,

f

g(1)

ffdx= fgdx. f'gdx= — —

r

9n

n1 '"

1

1,... ,and

Integration of Functions of a Real Variable

60

[(1.3

Exercises

1. Prove (3.8) directly from Definition 1 without reference to Theorem 1. (Note that the numbers a1 + b1 are not necessarily distinct.)

2. Let f, g

0 be measurable, with I

fgdx=f(f—g)dx. 0 be measurable, Urn

3. Let

ffdx=limfffl

=

g, f g dx 0 is measurable. Show thatff,, dx t ffdx. 0, measurable. Construct a sequence of measurable simple func. tions, such that t I and m [x: 01 0,

=0. So Theorem 10 gives the result.

u

n—p

if a =0, the same substitution gives f,,(x)dx =

f

u

+ u2/n2)'

u

du

du = 1/2,

using Theorem 10.

Example 20: Let f be a non-negative integrable function on [0,11. Then there function p(x) such that is integrable on [0,1] and exists a

= Solution: It follows easily from Example 15 that tim f a—'O

dx

0. So Y n, 3 x,,

0

xn

(0

o, then

af

I 20. Show that if

,

>

then

(x) for x >0 and n =

1,

2, . . . Find whether the limit of the integral equals the integral of the limit in the following cases, and evaluate the limits involved. ex/2 dx,

(i) I

(ii)

I

f

dx.

22. Show that if a>0, then Urn

f

(1 —xfnr

dx =

f

dx.

23. Show that

Jim f

'41n £?*X2 dx

=

f

urn

se/n

e?*X2

fora>Obutnot fora=O. 24. Find the range of values of a for which

'0

tim

—x)x" dx = tim f

—x)x" dx.

Find also the range of values of a for which the conditions of Theorem 10

are satisfied.

3.3 INTEGRATION OF SERIES

In the following examples we wish to write f j1x) dx as a series. We expand f ftx) dx = f dx, provided we can justify the interchange of and f. If the functions are of constant sign for each n and x we can appeal to Theorem 7, p. 59. If is an alternating series, Theorem 10, p. 63, may apply. If I dx can be shown to converge, Theorem 11,

flx) =

p. 64, applies. In many cases more than one method is available.

Sec. 3.3J

Integration of Series 00

1

fo

Exampk2l:Showthat Solution:

x1'3

l—x

log — = x1'3 log x

1—x

69

log

dx =

x

1

(j

+ 1)2

x,10 x" (0 —1,j'

o

o

E

(—1)"

p+nq

b

34.

Show that urn f

0 n1

n1

35.

1+xet

o

36.

n1

x"

n2+2 n(n2+4)'

n! dx = E (—1)" (2n)!

Show that f

n0

o

37. if m is an integer,m )0,letJm(x)= Bessel function of order rn.) (i)

Show that if a is a constant, 2 f Jm(2ax)xm+t 0

(ii)

Showthatifa>1,f 0

38.

Showthatf

&=2 E

39.Showthatifa>1,thenf

(2n +

a" 0 n1 40. 1ff is integrable over (a, b) and ri < 1,

(a

= g/fl

fb

fb

nx

= n=O

3.4

71

Riemann and Lebesgue Integrals

Sec. 3.4J

dx.

RIEMANN AND LEBESGUE INTEGRALS

We consider the Rieinann integral of a bounded function foyer a finite interval

=bbeapartition,D,of[a,b].Write SD

where M, equal

sup fin

s,], i =

1,...

,

n. Similarly on replacing M, by m1

to inff over the corresponding interval, we obtain SD

— SI-i).

= Then f is said to be Riemann integrable over [a, b] if given e> 0, there exists D such that — SD 0. By the corollary to Theorem 5, p. 58, there exists a sequence of measurable simple functions such that and f ip,, dx t ff dx. But then (çOfl)h tfh, and so by monotone convergence

f

dx=lim

1b1b

dx.

ExercIses 41. Let S be a

set, rn(S) 0, g continuous, such thatg =0 outside a finite interval and such that f If—gI dx O

—

x

0

x=0

,

.21—+bxcos2— axsin x

x 0, 3 5 >0 such that fix + t) —

ftx)l S

fl

sofa is measurable

The converse follows from:

[x:f(x)>a] =IJ Theorem 11: The measurability of f is equivalent to (i) (ii) (ui)

Va,[x:f(x) h E L(X, ii). Then fg dp exists in the sense of Definition 16, p. 106. kill a.e., where 18. 1ff E L(X, p) and g is a measurable function such that k is a constant, then g E L(X, 11).

19. Let E and F be measurable sets, f E L(E) and p(E and

F) =

0

thenfE L(F)

SF

20. (Tchebychev's inequality). Let f be a measurable function and let A =

forc>0,p[x:ftx)>cJ 0, we define L"(X, p), or more briefly to be the class of measurable functions [f: f)fl dp 0 and 0 0 write q

p

(lfI,,)"

'

in (6.8), to get

If N,, 'g1q

q p The right.hand side is integrable, so fg EL' (p). Integrate both sides to get

IfgI, 1 and let f, g E

di)'4'

(1

Proof: The case p =

1

fl"

then

+ (J lgI"

is trivial. So suppose that p> I and that p and q are con-

jugate indices. Then —

flf+gI"dp fIfI tf+gI"' IItlf+g)"' Na +

dp 11(1

(6.12)

[Ui. 6

Inequalities and the L" Spaces

by Holder's inequality. But (p — (IIfll,, + So

= p, so the right-hand side of (6A2) equals Afil,, +

as

required. 0

Example 11: Show that equality occurs in Minkowski's inequality for p = I if, and only if, we have almost everywhere either f(x) g(x) = 0 or sgn f(x) = sgn g(x); for p> 1 if, and only if, sf = tg a.e., where s and t are non-negative constants, not both zero. — 0 with equality if, and Solution: (i) p = 1: we have f (If I + only if, fl + = if+ a.e., so the condition is necessary and sufficient. (ii) p> 1: The condition is seen to be sufficient on substitution. Conversely, for equality to occur in (6.11) we must have equality in (6.12). Then, outside

a set of measure zero we have, for some a, b, c, d

alfi = blf+

and cigi = dlf+

So we always have af = ± bg a.e. and substituting in (6.11) shows that the signs are the same, giving the result. If p and q are conjugate indices and q 1, then p oo• This suggests analogues of Theorems 7 and 8 for the case p = oo. We recall Definition 3, p. 110.

Theorem 9: 1ff EL' (p) and g

thenfg EL' (p) and

ess sup a.e. we have (fg( Proof: Since and on integrating we get the result. D

Example 12: Of+ gL

1f1

Jig IL a.e. Sofg is integrable

NfL + IIgL

Solution: This follows immediately from Example 17, p. 41.

Example 13: If we write p(f,g) = Hf—gil,,, then forp> 1, p is a metric on L'(ji) that is, (i)

p(a,b)>O,

(III) p(a, b)

(ii)

p(b, a),

(iv)

p(a,b)=Oif,andonlyif,a=b, p(a, b) + p(b, c) > p(a, c).

Solution: (ii) holds by virtue of the convention regarding elements of L"(p); (iv) follows immediately from Minkowski's inequality; the remainder are obvious. Exercises

Give discrete analogues of HOlder's and Minkowski's inequality, as provided in Example 7 for Jensen's inequality. 12. Show that the following inequalities are inconsistent for functionsf EL2 (0, ir) 11

10

(f(x)—sinx)2 dx

13. Show that f x14

sin x

dx

f (flx)—cosx)2

1/9.

The Inequalities of Holder and Minkowski

Sec. 6.4] 14.

e L'

Show that 1ff, g E L' (p), then (i)

p,qE(0,1),p+q =

(ii) jf1P

E L' (ii)

if

1.

E L2(a, b), n =

15

117

1,

2, .

. .

,letfEL2(a, b)and let urn

= 0.

Show that

f2 dx = urn (ii)

I

if a and b are finite, then f f dx = urn

f f dx, a

t

b.

(ill)

flx)=x/2.

If

16. Let p

1 and let

17. Let f

0,fEL(x, 1) for each x E (0,11. Suppose that ti''

-÷0. Show that

where p> 1. Show that F(x) =

f

f dt satisfies F(x) = o(log

18. Let p> If> 0,fEL" (0, oo) and F(x) = conjugate indices, then F(x) = 19.

Show that jfk1 ,

-÷

as

f

EL(0,1),

1/x)'" as

fdt. Show that if p and q are

x -÷0 and as x

oo.

I/k, = I, then if f, E

1 and

for

each i,

20.

5 1f1f2 Equality occurs in the inequality of the last exercise iff one of the or for each pair i,j there exist non-zero constants C,, c, such that

=

0 a.e.,

(6.13)

21. If a,

"v> 0 and p < 1/(a + j3 + rny), then

j2

N,

I Iffml"

f If,,1

fmI" 1. oo). 35. Let p and q be conjugate indices and let fE LP(—oo, oo), g E Show that F(t) =5 f(x + t)g(x) dx is a continuous function oft. 36. Let p and q be conjugate indices and let -*gin -*fgin L1(jz). 11,, = 0, and Show 37. Let f, E L2(p) for each n; then we say that f,, weakly if urn f — 0 and a sequence In1) such that, for each i, ff,,1 dp cJ.Thenforany€,0 0, there exists a sequence If,, such that!,, in L"(p) butf,, a.u. In the list of references which follows and gives the location of the proofs we refer to the case (4) (6); the special case (2) (6) is then implied. Also if we refer to the cases A B and B C we omit reference to the case A C. Finally,

if f,,

if a result has been shown for case (a) we need not refer to it in cases (fi) and ('y).

Using these devices the following list of results is sufficient to construct the diagrams.

Case (a). (3)

(5): Theorem 6, p. 125; Theorem S,p. 123;

Case (j3).

Case ('y).

(5)*(1):

Theorem 8, p.125;

(5) (1) (1)

(6): Theorem 7, p. (5):

Theorem 9,

p.

125. 126.

(5):

Theorem 9, p.

126;

(6)

(2):

Theorem 4, p. 123.

Exercises

12. Let [X, S

be a measurable space and a sequence of measures on 8 (F) p,,(E) for each n. Write p(E) = tim p,1(E).

such that given F E 8, Show that (i) (ii)

pisameasureon S, iffE L(X, p), then for each n,fE L(X, p,,) and ffdp = lirn

13. In the previous exercise, let

...

L(X, P2) 14.

p,,

p,, = p. Show that L(X,p1)

L(X, p), but that in general L(X, p) #

If p and v are finite measures on the measurable space Ix, then p — v is a measure on

L(X, pt,).

J and p >v,

.5 J.

15. Let

be measures on the measurable space IIX, J, let for Write urn p,,(E) = p(E) for each F ES . Show that p is a measure on ,5. 16. Show that the finiteness condition in the last exercise is necessary if p is to be a measure. 17. Show that under the conditions of Exercise 15, L(X, p) ... 2 L(X, 2 each

n, and let p1(X) <

L(X,p1)and show that iffEL(X,p,,) for eachn, then =ffdp. 18. Let IX, S be a measurable space and fp,,J a sequence of finite measures J)

on it such that urn p,, = p, uniformly on S . Show that the set function g.i is a measure.

19. Show that under the conditions of Exercise 18, if f E L(X, IL,)) for each ii,

thenfEL(X,p) and

In Exercise 18 'uniformly' may be omitted. This is implied by the VitaliHahn.Saks Theorem, [17] , p. 70.] [Note:

Counterexamples

Sec. 7.41

131

20. Show that for each of the six kinds of convergence considered in this section, if a sequence is convergent, it is fundamental. 7.4 COUNTEREXAMPLES

In the section we give a set of counterexamples to show that the set of implications given by the diagrams of Section 73 is the most possible. Extra examples can be found in exercises which follow. For example, 'convergence a.e. does not in general imply convergence in the mean of order p (p > 0)' means that there and a sequence of functions If,,) with exists p > 0, a measure space S, does not L"(p) or, if they do, Hf,, limit f a.e. such that f,, L"(p) or tend to zero. As in Section 73, it is not necessary to give a counterexample whenever an C, to show A B it is sufficient to show implication is missing. Thus, if B then clearly A B in case (a). A * C. Also if A B in either case (I) or Also, if a mode of convergence does not imply (2), that is: convergence in the mean, then it cannot imply (4), that is: convergence in the mean of order p, any p > 0; and conversely to show, for example, that (4) (5) it is sufficient to observe that (2) * (5). The counterexamples numbered (i) to (viii) which follow, demonstrate the following 'non-implications' between the kinds of convergence (1) to (6), from which the remaining non-implications may be deduced.

Case (a). (1)*(6) (iv). (v), (2) (1): (vi), (2) * (4): Case (fl). '(1) * (4) * (2): (viii), (5) * (3): (iii). Case (y). (2)

(1): (vi), (3)

(4): (ii), (5)

(vii),

(3)

(2): (i),

(3): (iii).

(i) Let X = [0,1] , f(0) 0, f(x) = 1/x if 0 OJ,B=

then

Ex:flx)=O1.Then

AUB=X,AflBoandp(B)=jfdA=O. Define measures v0, V1 by v0(E) = v(E Cl B), v1(E) = v(E hA) for each so that v = v0 + v1. Since v0(A) = 0 we have v0 I p. Also v1 for

E E cS,

if 1u(E) = 0

we have f f dA = 0 and so, on E,f= 0 a.e. (A). Butf is positive on

E ñ A so A(E fl A) = v(EflA)=O.

0.

From the definition of A we have v

To show that the decomposition is unique we suppose that v

i4+v'1,wherev0lp,z41p,v1

A so v1 (E) =

v0 + p1 =

such thatX=AUB=A'UB',AClB=A'ClB'=O,andvo(B)=p(A) 0. LetEE S,then

E=(EflBflB')U(EflA'flB)u(EnA flA')U(EflA riB'). Clearly p is zero on the last three sets in this union and hence v1 and p'1 are zero by absolute continuity. Since v — = — we have (z4

—v1XEflBflB')=(vo —i4XEnBnB')=O,

as v0(B) = 0. So v1 (E) = (E), which implies v0(E) = i4(E) and the uniqueness of the decomposition. 0

Example 12: Let Hi, S

be a measurable space such that the points (xl of I

Bounded LinearFunctionals onLy'

Sec. 8.51

147

are measurable sets, and let v and p be a-finite measures on IX, 8 Then p, v2 + v3 I p, v1 1. p1 for I and v3([xj)= 0 p = p1 + p2 + p3 where p1 for eaclix EEX. Solution: The last theorem gives v = p0 + where v1 p and J. p. Let D fl D), p3(E) = v0(E fl CD) for each [x: 01 and write p2(E) = E E 8. Then I. p3 and p2 1 p, p3 1 p so p2 1 p1 , p3 1 p1 since p1 p. Clearly Ofor eachx.

Exercises

22. In the special case where p is c,-finite use Theorem 5, p. 139, to prove Theorem 6, p. 142.

23. Let

be sequences of positive numbers such that inf 0, >0. Let p, v be the measures on such that p([nJ)= v([nJ) = Show that p p but that the result of Theorem 6 does not hold.

24. Let

be

a mean fundamental sequence in L(X, p); show that V e> 0,

]&>Osuchthatforallk,flfkldp b > 0. This shows that p is

and

a positive measure. From (9.7) we

the modulus of continuity of the at most &j belong either to the same

can calculate at distance

(9.7)

h(61)

w,(61) 2h(61). So if 0 2. By definition, is the a-algebra generated by sets of the form FE E X F where E E So by hypothesis contains all the sets A1 X .. . X A k+1 , A1 E Now take F, and in this restricted class of sets consider countable unions and complements in X F. We generate, by the inductive hypothesis, all sets of the form E X F, E E FEE so the a-algebra generated by the sets A1 X X is just and by induction this holds for all positive integers k. (II) Let be the a-algebra obtained by this completion, so Also X A2 C c A1 belong to S . So S containsj)(X X X 6B and hence by induction So S as required. (ni) As on the line the Borel sets are the a-algebra generated by the open 3ets, by definition. Now 6B X . . X 6/3 is generated by the open sets G = G1 X Gft, G1 open, and so consists of Borel sets. But every open set may be written as a countable union of sets G so ¶13 X . X 6B is just the Borel sets. Now (ill)

follows from (ii). 0 Corollary 1: The measure mft is given by its values on the half-open 'intervals' [x: a1 3) on t> 1, we obtain a dominating function, so limit =0 by Theorem 10. 1) to get an integrand and use 1 — x 0, E2 has a

subset Fof positive measure on whichf> I + 6; so ,i(E2) = 0. If i(E3)> 0 and while

0 we have that f f" f" dp = (—1)" sum to exist,

=

(—

oscillates finitely,

oscillates infInitely. So for the limit

=

=

0,

since by Theorem 21, tim f f"

Chapter 6

215

33. Write E1 = [x: 1(x)> 1J and E2 = [x: fl:x)< 1J. Applying Theorem 21 to ñ E1), and Theorem 15 the integral over E ñ E1 we get the limit = applied to that over E fl E2 gives iz(E ñ E2).

HINTS AND ANSWERS TO EXERCISES: CHAPTER 6 1.

+ g2

0, that the integral g is infinite (log piT' with a finite limit as n -* logx -*0 asx -*0. follows from the fact that if a>

2. (i) is obvious. (ii) For a =

0,

for all smallx. -* as t -*0, we have > x' for all small x. 4. Since, for a> 0, ta to get a fmite integral. integrate explicitly over (0,1) and (1,00) 5. (i) For p = 2 )17

3.

(ii) If p > 2, f

(1 — log

x)" dx is seen to be infinite as in Exercise

2(11).

(iii) If p

f

Ia1IP)1/r'

a1 + b1V)'11'

(iii) If a1 >0, b1> 0, i = 1,.

.

.

, n,

1

so that

a1 =

a) max b,. The

a1b1

11 of(i), for example, is obtained by takingX = [1,

.

proof

1 0, 3 y such that

fP

(x

—

But f fP dt

for all large x. So Fix) < 19. For n = 2 we get HOlder's inequality. So suppose that the result holds for n = m — 1, m

a

> 2. Then if

fm-i Ia dp

1=1

=

(1 —

we have

(J If 1k1 dpy*i.

But from HOlder's inequality

..

.

.. Ifm...iI"

(JIfmV'm di)l*m.

So the result holds for m, and by induction the result follows.

20. If any f1 = 0 a.e. or if (6.13) holds, equality is trivial. Conversely, the case

ii =

2

is given in Example 10, p. 115. We suppose that the result is true for

n — 1 functions and consider the case of n functions, supposing that nof, =

a.e. Then the fact that (6.13) holds for n — 1 functions, together with the chain of inequalities which gives the result for n functions, gives us (6.13) 0

in this case.

21. In Exercise 19, take n = 3, (x) = f2 (x) = 21" and if = p(a + 0 + y) < 1 take k1 = to get the result.

22. Let p >1 and letf, E L"(p), 1

1,. . . ,n. Then

—1

k2

= Ix — = 5/(pfl), k3 =

E

follows

immediately, by induction, from the case n = 2. 23. As we may consider f/M, we may suppose that ess sup f = 1. Then f" a.e., so the required limit exists and is not greater than 1. But If" dp =

ff". I djz 0, a continuous function k vanishing outside a bounded set, such that Nf — ku €/2]. 2. If ci = 0,

—

(f +

> €1

[x: If,, —11 > €121 U

the result is obvious; otherwise

= [x:

[x:

the measure of which tends to zero as n -+ 3.

From the definition of convergence in measure it can be seen that the limit

function f is finite valued a.e. So V e> 0, set G, and K> 0 such that p(G) 0,

jointsetsof S ;then lim

= lim

p,,(B,) =

Jim p,,(Bj)

=

p(Bj.

The interchange of summation and lim is allowable by Theorem 15, p. 105,

so jz is a measure.

(ii) It is sufficient to prove the result for f> 0. As p

p,,, f f dp,, p,, easily gives f f ifi take X to be a single point and p,,(X) = n. Then every finite function belongs to each L(,p,,), but 0 L(p). 14. Routine verification. 15. By the last exercise — is a measure, for each n. Applying Exercise 12 to this sequence of measures we get that — p is a measure. So the last exercise, again, gives that p = is a measure. — —

16. Take I =

[1, oo) and define p,,(E) =

f

dx. Let

i= 1,2,.. .Thenp,,(B,)t 0 asn B,) =

each

= [i, i + 1), = 0. But

n, so z(CJBi) =

and

so p is not countably additive.

17. The first part is obvious. For the second, considering simple functions and taking limits, we get f f dp1 — f f dp,, = — p1,, is a — p,,), since measure by Exercise 14. Applying Exercise 12(11) to — we get the

ff

result.

18. Finite additivity and the other properties being obvious, we wish to check that p is countably additive. Let be disjoint measurable sets. Then B,) — 1=1

i.i(L) B,) —

p(B1) 1=1

1=1

B,)l i—I

+

U B,)

I=N+1

+ We have

0 as k

suchthatizn( L)

1=N+ 1

so

B,)

since

converges uniformly, Ye>

< e/3 for all n and for

Now

choose N2 such that for n N2 the first and third terms are < e/3, so that the right-hand side I — 1. But this gives ffdiln > I — and a contradiction, so f EL' (p). Hence f øm < for all m > m, (€). Let m2 = m0 + m,. Then f 0m2 dun — I 0m2 duil n0, as is finitevalued a.e. So 1ff d1i — I 0. Then f Ifn J'm V' d1i + f If,1 — fV' gives the f fm — I" dii, and letting n, m 1

ff

f

result. 21, 22. Obvious.

23. V €> 0,

in [€,

1] is

But

n

f

dx =

1

—

—f

0

as n -+ 24. Similar to Exercise 23.

25. By HOlder's inequality fn, and the limit function f, are in L' (p), and Ifn —f11 where p. q are conjugate indices.

EL'(l, 00) but, if n >m, Nfn

26. n,

27.

29.

=

(f" x' dx)2 -f 0

as

oo.

If t> 0, et < on

28.

m

fm

x' -÷0

l/t; so n3'2

computation, f

-÷0°, forx

>0. But,

dx —f 0.

As for Exercise 27, using e_t