The Lebesgue Integral (Cambridge Tracts in Mathematics)

THE LEBESGUE INTEGRAL BY

J. C. BURKILL

FellOfD of P�ttrhcnu�,

Cambridge

CAMBR IDGE AT THE UNIVERS ITY PRESS 1963

PUBLISHED BY

THE SYNDICS OF THE CAMBRIDGE UNIVERSITY PRESS Hl•ntlcr House. 200 Euston Roall, London. N'.W. 1

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Rt.pri·ut,./ by ojJNttl·litl,ography by John Dickens� Co. Ltd., Nortluunpton

PREFACE My aim is to give an account of the theory of integration due

to Lebesgue in a form which may appeal to those who have no wish to plumb the depths of the theory of real functions. There

is no novelty of treatment in this tract; the presentation is essentially that of Lebesgue himself.

The groundwork in

analysis and calculus with which the reader is assumed to be

acquainted is, roughly, what is in Hardy's

A Course of Pure

Mathematics.

It has long been clear that anyone who uses the integral calculus in the course of his work, whether it be in pure or applied mathematics, should normally interpret integration in the Lebesgue sense. A few simple principles then govern the manipulation of expressions containing integrals. To appreciate this general remark, the reader is asked to turn to p.

42;

calculations such as are contained in Examples

4-8 might confront anyone having to carry through a mathe­ matical argument. result has the

Consider in more detail Exam pie

4;

the

look of being right rather than wrong, but the

limiting process involved is by no means simple, and the justi­ fication of it without an appeal to Lebesgue's principles would be tiresome. Anyone with a grasp of these principles wiH see that the easily proved fact, that

(1-tfn)n increases to its limit

e-t, ensures the validity of the passage to the limit. The attitude which the working mathematician may take towards the more general concepts of integration has been ex­ pressed by Hardy, Littlewood and Polya in

Inequalities. After

dealing with inequalities between finite sets of numbers and extending them to infinite series, they tum to inequalities between integrals and begin Chapter vr with these preliminary remarks on Lebesgue integrals: The integrals considered in this chapter are Lebesgue integrals, except in §§ 6•15--6·22, where we are concerned with Stieltjes integrals. It may be convenient that we should state here how much knowledge ot the theory we assume. This is tor the most part very little, und all that the reader usually

vi

PREFACE

needs to know is that there is some definition of an integral which possesses the properties specified below. There are naturally many of our theorems

which remain significant and true with the older definitions. but the subjerl becomes «JSier. as well as more comprehensive, if the definitions presupposed have the proper degree of generality.

. Since Lebesgue's original exposition a. number of different approaches to the theory have been discovered, some of then1 having attractions of simplicity or generality. It is possible to anive quickly at the integral without any stress on the idea. of measure. I believe, however, that there is an ultimate gain in knowing the outlines of the theory of measure, and I have "' developed this first in as intuitive a way as possible. During several years of lecturing on this topic I must have adopted ideas from so many of the books and papers on it that detailed acknowledgement would now be difficult. My greatest debts are to the classical books of de ]a Vallee Poussin, Cara­ thoodory and Saks, and the straightforward account (having a similar scope to this) given by Titchmarsh in his Theory of I also wish to record that one of my many debts to G. H. Hardy lay in his encouragement to write this tract.

Functions.

J. C. B.

September, 1949

Reprinting has allowed me to put some details into § 2· 2 which had been left to the reader. The first paragraph of§ 2·7 men­ tioning the role of an axiom of choice in the Lebesgue theory has been recast. I might have helped the reader more by discussing this axiom at its first appearance-on p. 3, in enumerating the qets E,,.. To do this now would disturb the type too much, and I can help him most by urging him to read an account of the foundations of the subject such as is given in the books specified on p. 87. There are other less important alterations. I thank Mr Ingham and Professor Besicovitch for constructive criticism. Jun�, 1958

J. C. B.

CONTENTS

Art.

Author' s Preface

Chapter I. SETS OF POINTS 1·1 1·2 1·3 1·4 1·5

The algebra of sets Infinite sets Sets of points. Descriptive properties Covering theorems Plane sets

Page v

I

3 4 6 7

Chapter II. MEASURE 2·1 2·2 2·3 2·4 2·5 2·6 2·7 2·8 2·9 2·10 2·11 2·12

Measure Measure of open sets Measure of closed sets Open and closed sets Outer and inner measure. Measurable sets The additive property of measure Non-measurable sets Further properties of measure Sequences of sets Plane measure Measurability in the sense of Borel Measurable functions

10 10 11 12 13 14 15 16 18 21 23 23

Chapter ]]]. THE LE BESGUE INTEGRAL 3·1 The Lebesgue integral 3·2 The Riemann integral 3·3 The scope of Lebesgue's definition 3·4 The integral as the limit of approximative sums 3·5 The integra,} of an unbo�nded function 3·6 The integral over an infinite range

26 27 28 30 31 33

.. Art. .

Vlll

3·7 3·8 3·9 3·10

C O NTENTS

Simple properties of the integral Sets of measure zero Sequences of integrals of positive functions Sequences of integrals (integration term by term)

Chapter JV.

4·1 4·2 4·3 4·4 4·5

4·6 4·7 4·8

Page

DIFFERENTIATI O N AND I N T EGRATI O N

Differentiation and integration as inverse processes The derivates of a function Vitali's covering theorem Differentiability of a monotonic function The integral of the derivative of an increasing func· tion Functions of bounded variation Differentiation of the indefinite integral Absolutely continuous functions

Chapter V.

0/w,pter VJ.

6·2 6·3 6·4 6·5

44 44 46 48

49 50 52 54

FURTHER P R O PERTIES O F THE I N TEGRAL

5·1 Integration by parts 5·2 Change of variable 5·3 Multiple integrals 5·4 Fubini's theorem 5·5 Differentiation of multiple integrals 5·6 The class LP 5·7 The metric space LP

6·1

34 37 38 40

58 58 61 63 65 65 61

THE LEBESGUE·ST IELTJES IN TEGRAL

Integration with respect to a. function The variation of an increasing function The Lebesgue·Stieltjes integral Integration by parts Change of variable. Second mean·value theorem

70 71 72 75 77

Solutions of some examples

80

C H A PTER I

SETS OF POINTS The refinements of the differential and integral calculus, which form the topic of this tract, largely depend on the properties of sets of points in one or more dimensions. This chapter con­ tains those properties that will be needed, in so far aa they are deacriptive and not metrical. The mles of algebra applied to setB hold whether the members of the sets are points or are objects or concepts of any kind. All that we require for a set E to be defined is that we can say of any given object x whether it is or is not a member of E. '

1·1. The algebra of sets. Let E be a set,t the members of which may be of any nature. The aum of two sets E1 , E1 is defined to be the set of objects which belong either to E1 or to E1 (or to both); the sum is written E1 + E1• By definition E1 + E1 is the same as E1 + E1, no question of order being involved. The

definition extends to any finite or infinite n umher of sets, E1 + E1 + . being the set of objects belonging to at least one E11• In the definition of an infinite sum there is no appeal to any limiting process. of any number (finite or infinite) of sets The product E1 E1 E1 , E1, is defined to be the set of objects belonging to every one of the sets En. The sets En may have no members common to all of them, and the product is then the null set-the set which has no members. . .

. • •

• • •

t Class and aggregate are synonymous with set; French ensemble, German Menge.

2

S E T S O F P O INTS

If every member of E1 is a member of E0 we say that E1 is contained in E0 and we write E1cE0 (or E0';lE1). The set of members of E0 which do not belong to E1, may be written E0-E1 or a.ltematively as OEb the cumplement of E1• It is easy to see that, the complements being taken with respect to a. fixed E0, O(E1+E2 + ... )= E1• E2 ..., and

O(E1E1

0 0

• • •

)= OE1 +OE2+ ...

.

is an infinite sequence of sets, the Limit seta. If E1, E2, upper limit, lim En, is defined to be the set of objects which belong to infinitely many of the E,.. The lower limit lim En is • • •

defined to be the set of objects, each of which belongs to all but a. finite number of the En. Clearly limE,.';l limE3• H the sets lim E., lim E,. are the same, we say that the sequence E1, E2, has a liinit, lim En. If a sequence of sets is increasing or decreasing, it has a limit: more precisely, • • •

(a) If Enc;En+l' then limEn= E1+E1+ ..., (b) If E3';lE+ n l' then limEn= E1E2 .... To prove (a), write E= E1+E2+ ... and observe that if xis a member of limEn then xis a member of E; hence limEncE. The result will now follow if we prove that EclimE,.. This is true because any member of Eis a member of En for some n and so (since the sets form an increasing sequence) for all greater n and therefore of lim En. A similar proof holds for (b), the product set possibly being null. More generally, the upper and lower limits of any sequence of sets, not necessarily monotonic (increasing or decreasing), ma.y be expressed in terms of sums and products. The formulae are limEn= (E1 +E1+E,+ ..,) (E1+E3+ ...) (E3+ .. ) ••. , limEn = E1E2 E3 ... +E2E3 + E3 + ...• .

• • •

The proof is left to the reader.

• • •

3

I N FI N I T JJ S E T S

1·2. Infinite sets. Two sets are called Bimilar if there is a one-one correspondence between the members of one and the members of the other. Thus two sets with finitely many mem­ bers are similar if and only if each has the same number of members. The idea of similarity is the foundation of any theory of infinite numbers. We shall give here only those outlines of this topic which are essential for later chapters. With infinite sets we have a phenomenon which cannot occur with finite sets, namely, that a set can be similar to a part of itself. For instance, the set of positive integers is similar to the set of even integers or to the set of perfect cubes. Any set which is similar to the set of all positive integers or to a finite sub-set of them is said to be enumerable. The one­ one correspondence may be displayed by using the positive integers as suffixes, so that the members of any enumerable set may be specified as x1,x1,x3, It is clear that any sub-set of an enumerable set is enumer­ able. The metnbers of an enumerable set of enumerable sets E1,Et, . . ••••

.

form an enumerable set. For let the members of Em be enumerated as x,,.1, xm1, xm3, ... The me•nbers of an the sets then form a double array:

Xu

Ztt

Zta

Xst

Xss

X23

Xat

Xat

Xas . . .

••

·

• ••

. ..

This array can be enumerated as a single sequence, for example, by taking terms along the successive diagonals in order X u, Xu, Xzt, X13' Z tt' Xalt •

•

••

As a particular case of this, the set of positive rational numbers is enumerable, for they are all included in the set

1 1 2 I 2 3

I' 2' 1' 3' 2' I'····

4

SETS O F P O INTS

Clearly, the set of all rational numbers (positive, negative, or zero) is enumerable.

A further application of the same argument proves that the

point8 of the plane of whick both co-ordinates are rational form an enumerable set. For if rltr1, ... are the rationals enumerated, the rational points of the plane can be displayed

as

(rt, rt)

(rt, rz)

(rt, ra)

(r1 ,r1)

(r1, r1 )

(r1,r3)

.. .

•

r•

• ••

and form an enumerable set of enumerable sets. The simplest example of a set which is not enumerable is the set of all points of an interval. Take the interval (0, I) and sup· pose, on the contrary, that aU the numbers between 0 and be enumerated as decimal

x1, x1, x3,

. • ••

Let each Xn be expressed

1 can as

a

the u's being numbers from 0 to 9. Write down a new number,

where Vn is determined from Unn by the rule that Vn

unn =I=

1 and vn = 2 if unn

is not the same

as

=

=1

if

1. Then ylies between 0 and 1 and

any Xn, for it differs from Xn in the nth

decimal place. This contradicts the hypothesis that the se· que nee

xux1,

• • •

included all the numbers of (0,

1 ).

1·3. Sets of points. Descripdve properties. that E is a set of points on a line.

E is bounded if all its points intervaL

!_point P,

0

_

c-.,/l;,.h

of abscissa

x,

if there is a neighbourhood which be!�

T to_�--�-�-

-

are

Suppose

included in some finite

is said to be an

interiO!J!_�nt

(x- 8, x + 8) of P_�

of E

every point of

DESCRIPTIVE PROPERTIES

A_ se� E is_.�ai� �_be _{)2,e1! if �v_ e_ty_pgi _ nt Qf!t is�-i�te�or

polnt. The simplest open set is an interval a < x <.!.>... without its en�points .-.. ApointP, of abscissa x (which may or may not be a point of E) is said to be a limit-point of E if any neighbourhoo� (x- 8, x + 8), however small 8, contains a point of E other than P. It follows that every neighbourhood of P contains infinitely many points of E. The set of all limit-points of E is called the derived set of E and -!s denoted by E' . Weierstrass proved that, if E is a bounded set having in� finitely many points, then E' contains at ]east one point. t A set E is said to be closed if E :> E' . (For example, consider the closed interval a� x� b.) It will be convenient to reserve the letters 0 and Q, with suffixes, for open and closed sets respectively. We shall prove first that these two ideas are complementary. If Q is cl08ed, then CQ is open (the complement natural1y mitst be taken with respect to an open interval). For let P be a point of CQ. Since Q contains all its limit­ points, Pis not a limit-point of Q . Therefore there is a neigh­ bourhood of P free of points of Q, and so CQ is open. �·-

-

-

--

----

----

-

If 0 is open, then CO (taken with respect is clo8ed.

----

to

a closed interval)

For no point of 0 is a limit-point of points of CO. It is to be observed that the set of all points of the line ( oo < x
If E is any set, E' is closed. If E' has a limit-point P, let I be a neighbourhood of P. I contains a point of E', say Ji. Let/1 be a neighbourhood of Ji contained in I. Then infinitely many points of E lie in 11 and so in/. Hence Pis a limit-point of E and therefore is a point of E'. The set E + E' may be called the closure of E. t llurtJy, A

Cours� of l'ure lUatltrmatita,

(·h.

1.

6

SETS O F P O I N T S

A linear open set is the sum ofan enumer{lble Bet of open intervals. Every point of 0 is contained in an interval I, co�ting entirely of points of 0, whose end-points belong to 00. No two of the intervals I overl&p. To prove that they form an enumerable set,suppose that P1, Ps, .. are the rationa.l points of the line,enumerated. With each interval I associate the Pn of •

.

smallest suffix contained in it. We thus have a one-one corre­ spondence between the intervals I and a sub-cla.ss of the po�itive integers, and the theorem is proyed.

·

The 8'Um of (a finite number or) an infinity of open sets is open. For if Pis a point of l:On it is a point of some On. It is an interior point of On and a fortiori an interior point of :EOn. The result complementary to the la.st is that the product of

infinitely many closed sets is closed. The product set may be

null. There is one important case in which we can assert that

it is not null,namely, that of a decreasing sequence of bounded sets.

If Qn C (a, b) and Qn:::::>Qn+l' then llQn i8 closed and not null. Let Pn be the left-hand end-point ofQn. The Pn have a limit­ point P. P is contained in every Qn and therefore in llQn. I

1·4. Covering theorems. In proving theorems about real

functions,we often have the situation that every point of a set

E is associated by some property or other with an interval con­ taining it. In general these intervals form a non-enumerable set and we desire

to

select an enumerable (or

finite)

sub-set

which cover every point of E. The reader may well have first met this problem in a. discussion of the properties of a continu­ ous function f(x) in an interval

a� x� b.

lf, for a given

£,

we

associate with a value x the greatest interval within which the function lies betweenf(x) ±�,then it can be shown that a finite number of these intervals will serve to 'cover' (a, b), and this

yields the property of uniform continuity.t We prove two covering theorems.

t Hardy,

Pun M�, pp. 196-201.

C OVBBING THEOREMS

7

LINDELO F's THBOBBM. To each pcnnt x of a set E correaponda an open intervall(x) containing x. Then there ia an enumerable �

aet of tke8e interval8 covering E .

The rational intervals of the line (i.e. intervals with rational end-points) form an enumerable set. An intervali(x) contain­ ing x includes a rational interval containing x. Thus E is covered by an enumerable set of rational intervals, and a fortiori by an enumerable set of the I(x). THE HEINB-BOBEL THBOBEM. Iftke E ofLindeliif'stkeorem is

bou/nded and closed, it can be covered by a finite num ber of the a88ociated 1 (x). Let Ihl,, ... be a Lindelof covering of E . Let En be the part of E outside I1 + I2 + . . . + In . Then E. is closed. The theorem follows if, for a sufficiently large n, E , is null. Suppose that no En is null. Then, since the En form a decreasing sequence of bounded closed sets, there is a point common to all of them. This point is in E and is in no In, and we have a contradiction. l·S.

�lane sets. The reader will satisfy himself that many

of the properties which we have for simplicity established for linear sets are true of sets in higher dimensions. Some care is necessary. The natural plane analogue of a linear interval is a rectangle. It is not true that the general open set in the plane can be decomposed into non-overlapping open rectangles in the same way that a linear open set is the sum of intervals. The standard decomposition of a plane open set is into rectangles, not overlapping but adjoining, i.e. having sides and parts of sides in common. For carrying out this process in a systematic way, and for other purposes, the idea of a network is useful. Take a pair of axes and the lines x = ± n, y = + n for all integral values. Denote by G1 the set of all squares of side l so formed. Each square is to be regarded as closed, i.e. its sides be-long to it. The lines x = + !n, y = ± !n will no\v form a set of squares of side !; call this set 01• Continue this process of bisect.ion; for every integral m we have a set of squares Gm each

8

S E T S OF POINTS ,

1 of side 1/2*- . Call this construction a. network and any square appearing in it a 1ne8h. The set of meshes is enumerable. Any point of the plane is then determined by & sequence of meshes g1=:Jg1=:J ,where Um is a mesh of G,.. We apply this construction to prove the theorem about plane open sets. A plane open 8et i8 the sum of an enumerable set. of closed rectangles. Let 0 be the set, and P any point of it. Since 0 is open, a mesh Um containing P will, if m is sufficiently large, be wholly contained in 0. Then 0 consists of the meshes of 01 contained in it, the meshes of 01 contained in 0 but not in Gb the meshes of 03 contained in 0 but not in 01 or 01, and so on. This proves the theorem. These remarks about sets and networks in a plane can be extended to higher dimensions. • • •

EXAMPLES ON CHAP T E R I

( 1) If E3 and Fn

are

two sequences of sets, prove that

lim E.,+ lim Fn clim(E.,+ Fn) C lim En+ lim Fn clim (En+ F.) = limE3 +lim Fn.

Establish a similar chain of inequalities for products. Deduce that, if limEn and lim Fn exist, then so do lim(En+ F.) and lim(E,.,Fn). . (2) If,for any choice of a finite number of values x1,x1, ,x3, • • .

ft.

the sum :E j(x,.) is bounded, prove that the set of values of x r=l

for which j(x) =t= 0 is enumerable. Deduce that the values of x for which a given increasing function is discontinuous form an enumerable set. (3 ) Let E be the set of points 1 1 1 X = 1+ 3m+ 5n' 2

•

EXAI\IPLES

9

where l, m, n. have all positive integral values. What are the first, second and third derived sets E',E",E'"? (4) If f(x) is a continuous function, and A is any constant, prove that the set E (f�A) of values of x for which J(x) �A is closed. Prove that the same result is true under the more general hypothesis that f(x) is upper 8emi-continUOU8, i.e. for each e, lim f(x) �/(f).

�'

(5) A point of E which is not a limit-point is called isolated. A set all of whose points are isolated is called an isolated set. Prove that an isolated set is enumerable. (6) Prove that the set of maxima of a given function f(x) is enumerable. (7) A set E for which E E' is called perfect . (The simplest perfect set is a closed interval.) Prove that the following con­ struction gives a perfect set (Cantor's ternary set). From the closed interval (0, l) remove the middle third, the interval (l, j), taken as open. Remove the (open) middle thirds of each of the two remaining intervals (0, !) and (f, 1). This will leave four intervals. Remove the middle third of each of them. Continue the process indefinitely. The set of points which remain is perfect. (8) Prove that the perfect set of Ex. 7 is not enumerable. =

CHAPTER II

M EASt:TRE 2·1. Measure. Following Borel and Lebesgue we shall aim

·

at assigning to a set of points on a line a number called its mt4c9t.We which shall generalize the idea of length. The measure of an interval will be its length. For the theory to be satis­ factory we shall want, for example, the measure of the sum of two sets without common points to be equal to the sum of the measures of the sets, that is to say, measure is to be an addititJe ftJ:ndion of sets of points. If 0 is an open set, the additive property requires us to define its measure mO to be the sum of the lengths of its constituent intervals; it is assumed that this sum is convergent (it will certainly be convergent if 0 is contained in a finite jnterval). 2·2. Measure of open sets.

Let 01 and 01 be open sets with no common points. Since the measure of an open set is the sum of the lengths of its intervals we have m( 01+ 01)

= m 01+m 01•

More generally, we shall prove that, if 01 and 02 are any two open sets, then m( 01+ 02)+m(01 01) m 01+m 01• =

Taking first a special case, if 01 and 01 consist of a finite number (n) of intervals, we can prove the result by induction onn. Generally, let 01 and 02 be any open sets; we can suppose their intervals enumerated (p. 6). Let e.,-+0 as V-+00. Taken such that In and J., the sums of the firstn intervals of 01 and 01 respectively, satisfy m 01-m�. <e.,

and m 01- mJ. <e.,.

II

OPBN SET S - C L O SBD SETS

Then I,. Jfl C 0102, and any prescribed interval of 0102 is in I,. J, if v is large enough. Hence, as v-+ oo. The inequalities

'tn(01 01) = lim m(1,. J,t ).

I"' +Jn C01 +01 c (In +J,.)+ (01-1,.)+ (02-J,) show that, as v � oo, m(01 +01) = lim m(l., +J,). By the special case of the theorem, m(I11 +J,t) +tn(l,�J,J 1nl,t +ntJ,. . Let v�oo, and we ha.ve the theorem for 01, 01• If Take now an enumerable infinity of open sets 01.. 01, they are disjoint (i.e. no two have coinmo:a points), then -just as for two setsm( 0 1+01+ ... ) m01 +n�01+ ... , if the right-hand side converges. If the sets overlap. the sign < replaces This is an occasion to mention infinite measw-e. If the lengths of the interv als of an open set form a divergent series we could say that i ts measure is oo and could attach extended tneanings to some theorems of this chapter. It is generally, however, more convenient to restrict measure to be finite. •

=

• . ••

=

=.

2·3. Measure of closed sets. If Q is a closed set, and 0 is any open set of finite measure containing Q, then 0-Q will be an open set and the additive property requires that

mij

=

mO-m(O-Q).

We take this as the definition of the measure of a closed set. It must be shown that the value of mij is independent of the particular 0 chosen. Let 01, 01 be two different O's. From §2·2

m01+ m(01-Q) and

=

m(01 +01)+m( 0102-Q),

m01+m( 0 1-Q) = m(01 +01)+m( 0 1 02'-Q).

By subtraction, m01-m(01-Q) m01-m(02-Q), and the definition of mij is j ustified =

•

•

12

MEASURE

We now prove the analogue for closed sets of the property of§ 2·2, namely, mQ1+mQs = m(Q1+Q,)+m(Q1Qt )· Let 0 be an open set of finite measure containing Q1 +Q1• Let 01 = OQh and 02 = OQ,, the complements being taken with respect to 0. Then 0101 = C(Q1 +Q1) and 01 +01 = C(Q1Q1). The result of§ 2·2 then gives {m0-m01}+{m0-m02} ={m0-m(01 +01)}+{m0-m(0101)}, ... and this is what we want. We shall in future refer to a proof on these lines as a proof by complementB. As a corollary, measure is additive for closed sets. (Take Q1 and Q1 to have no common points.) 2·4. Open and closed sets.

(i) (a) lfQcO, then mQ<mO.

(0 is assumed not nuU.)

(b) IfOcQ, then mO�mQ. To prove (a), observe that, taking complements with respect to a closed set containing 0, we have C(O-Q)

=

OO+Q,

and that 00 and Q are closed sets with no common points. From§ 2·3, m(CO)+mQ = m{O(O-Q)}. Hence

mO-mQ = m(O-Q),

and the right·hand side is greater than zero. The proof of (b) is sftnilar. (ii) (a) For a given 0 and given E, a Q can be ccm8tructed, con­ tained in 0, such that mO-mQ< E. (b) For a given Q and given E, an 0 can be constructed, ccm· taining Q, BUCh that mO-mf:J<E.

OUTER AND INNER MEASURE

(a) Let /1, /2,

• ••

l

=

13

be the intervals of 0. Choose n so large that m/1+ml1 + ... +mln>m0-!E.

From each end of each /,. (r = I, 2, . . , n) cut off an open interval of length Eml,f4l. This leaves a finite number of closed intervals forming a closed set Q such that .

mQ

=

l- }E>m0-E.

This proves (a ), and (b ) follows by complements. COROLLARY. The measure of an open set is the upper bound of the measures of closed set� contained in it. The measure of a closed set is the lower bound of the measures of open sets containing it.

2·5. Outer and inner measure. Measurable sets. Given a set E, we define its ouUr measure m*E to be the lower bound of the measures of open sets containing E. The inner measure m.E is defined to be the upper bound of the measures of closed sets contained in E. If m*E m.E, then E is said to be meaBUrable and its measure mE is the common value of m*E and m.E. Some results follow at once from the definitions: (i) If Ec(a, b), then m*E b-a-m.(CE). For, if 0;;)E, then ao, taken with respect to the closed interval (a, b), is a Q contained in CE. (ii) m*E�m.E. For, if 0;;)E;;)Q, then, by § 2•4 (i) (a), mO>mQ. (iii) If E1;;)E2, m*E1�m*E1 and m E1�m.E2• * For any 0 containing E1 contains E1 and so the lower bound of the measures of O's containing E1 is greater than or equal to the lower bound of the measures of O's containing E1 • We must assure ourselves that measure, as defined in this section, when applied to open and closed sets is consistent with the measure of such sets, as originally defined. The measure of an 0, defined as the sum of its intervals, is equal to m*0 since 0 itself is an open set containing 0. It is equal to m.O by =

=

14

M E ABUBE

§ 2•4(ii )(oorollary). Similarly the two definitions of the of a closed set are consistent.

meas

ure

2·6. The additive property of measure. (i) (a) m*(E1+E1)+m*(E1E1) 'm*E1+m*E1,

( b) m.(E1+E1)+m.(E1 E1)� m.E1+m.E,. Let 01, 01 be open sets containing E1, E1, such that m01-m*E1 <E, Then

m01-m*E1<E.

m*(E1+E1)+m*(E1E1)�m(01+01)+m(0101) = m01+m01�m*E1+m*E1+2E.

Since E is arbitrary, (a) follows; (b) is complementary. CoROLLARY. IfE1,E1 are mea8Urable and disjoint, then E1+E1 is metJBt4.table a'TUl m(E1+E1) = mE1+mE1• This may be extended to a finite number of sets. An essential feature of the Borel-Lebesgue concept of meas­ ure is thatjt is additive, not merely for a finite number of sets, but for an enumerable infinity. This is a consequence of the following pair of inequalities. (ii ) (a) m*(E1+E1+... )�m*E1 +m*E1+ .... (b) Ifno two E's /w,ve wmmon pointa, m.(E1 +E1+ ... )�m.E1 +m.E1+

...•

Let 011 be an open set containing En such that mo. -m*E. < E/2". Then

m*(E1+E1+

•. •

)�m(01 +01+ ... ) �m01+m01+

•••

from §.2·2

�E+m*E1+m*E1+

...•

Since E is arbitrary, (a) is proved. For (b), m.(E1+E1+ ... )�m.(E1+E1+...+E.)�m..E1+••.+m.E., from the finite case.

N O N -M E A S URABLE SBTS

15

Since this is true for all n, the result follows. We have thus proved the fundamental theorem. If E1, E1, are 1Ma�Urable and disjoint, tken E1+ E1+ .. is measurable and m(E1+E1 + ... ) mE1+mE1+ .... Note as a special case that any enumerable set of points has measure zero. • • •

.

==

The reader will ask whether sets exist for which the outer and inner measures differ, and, if so, that examples of such sets shall be produced. The answer depends on fundamental questions in the theory of aggregates which we cannot discuss. (See pp. vi, 87.) An 'axiom of choice' admits as being adequately defined an aggregate containing infinitely many members for which no rule of selection can be laid down in advance. The Lebesgue theory repeatedly appeals to the admis· sibility of an enumerable infinity of arbitrary choices (e.g. the choice of the sets On in § 2·6 (ii)). To defn i e a non·meaaurable set we have to assume a stronger axiom of choice admitting as being adequately defined an aggregate depending on a non·enumerable infinity of arbitrary choices. Cqutru.ction of a ntm·mecuurable set. For simplicity of state­ ment we shall define the set on the circumference of a circle; the circle may then be supposed 'unrolled' into an interval and the set becomes a linear set. Take a circle of diameter 1. Divide the points of the circum­ ference into sets in such a way that, if P is a point of any particular set, then the points distant 1,2, . from P in either . direction along the arc belong to the same set. Take one repre­ sentative of each of these sets, forming an aggregate E; in this step we appeal to the axiom of choice since it is not possible to specify how the representative is to be chosen. Let E," E_1l be the sets obtained from E by rotations of n in the two direc­ tions round the arc. Then it is clear that the sets E, E1, E_1, are disjoint and together they fill the circumfer­ E1, E_1, ence. Moreover, they are congruent with one another, and if measurable must all have the same measure. This measure

2·7. Non-measurable sets.

. •

• • •

16

MEASURE

cannot be zero, since then the measure of the sum would be zero, whereas it is in fact'"· Nor can the measure be positive, for then the measure of the sum would be infinite. Hence E cannot. be measurable. In higher dimensions there are still more curious phenomena. Hausdorff showed that 'a half of a sphere can be congruent with a third of a sphere'. More precisely, the surface 8 of a sphere can be divided into four sets A , B, C, D, where D is enumerable, and

A=B=C, A=B + C,

where - denotes congruence by rotation about suitable axes through the centre of the sphere. The proof of this is outside the scope of this tract;t we have said enough to show the interest and difficulty of the problem of measure.

2·8. Further properties of measure. In this section some miscellaneous re�Jult$ are collected. It is possible to extend to general sets the ideas ·about infinite meaBUre which were put forward for open sets in§ 2· 2 . A set E may be said to be measurable and to have infinite measure if, for every r, the part of it, Er, contained in the interval (- r, r) is measurable and mE,.-.oo as r-.oo. This convention would give a meaning, for example, to the equation expressing the additive property (at the end of§ 2·6) when I:mE,. diverges. We now pass to an inequality which supplements those of § 2·6 (i). (i) If E1, E2 are di8joint,

m*(E1 + E1)�m*E1 + m.E,�m* (E1 + E1). It will suffice if we write out the proof of the former of these inequalities. If F-:J G, we have from§ 2 ·6 (i) (b), ..

m.F�m.G + m.(F - 0). t Hausdorff, Mengenlehre (1914), p. 469.

Let

F = OE1

FURTHER PROPERTIES (with respect to some containing

Then, since

17 0) and G = E1•

OE1 - E2 = C(E1 + E1),

m0 - m*E1�m.E2 + m0 - m*(E1 + E1),

we have

giving the result. (ii)

If M i8 M,ea81trable, then for any E with m*E ji1tite, m*E = m*(EM) + m*(E- EM).

The inequali�y § 2·6 (i)

(a)

gives at once the relation

�

be·

tween the left-hand and right-hand sides. Again, from § 2·6 (i)

(a),

m*E + m*M�m*(EM) + m*(E - EM + M), and, from (i) above,

m*(E- EM + M)�m*(E - EM) + m* M. Since this gives

m*M=m.M, m*E�m*(EM) + m*(E- EM),

which is an inequality in the reverse direction

to the one we

have already. This property of measurable

sets has been taken as funda·

mental by Caratheodoryt in his elegant and general exposition of measure.

Lebesgue's condition for meaaumbility. A neceasary and &UJficiem condition for E to be mea&urable is that, given £, (iii)

E = I + e1 - e1, where 8 i8 the 8'Um of a finite number of interval8, and m*e1 < £, m*e1 < £. To prove necessity, choose 0, containing E, such that mO < mE + £. Take 0 - E = e1, and take as 8 enough intervals of 0 for e1 = 0 - C to have measure less than £. f Vorlesungen iifJer TUlle Funktitn-.en.

18

MEASURE

To prove sufficiency, we have on the one hand

m*E�m*(l+e1)�ml+E, and on the other

m*(CE)�m*(CI+e1) � m(CI)+E, m.E�mi-E.

and so

Since E is arbitrary, E is measurable. The gist of this criterion for measurability is that a

mtB&U

r­ able set is 'nearly' (in a m-etrical seue) a finite set of intmJall.

(For two other simple guiding principles in the theory of real functions see Examples 3 and 4: at the end of this chapter.) Other forms of the necessary and sufficient condition are (1) E = O-e1 where m*e1<E, or (2) E or

=

Q+e1 where m*e1<E,

(3) there exist O, Q with O=:JE=:JQ and mO-mQ<E.

(iv) If E1, E1,

E1 E1

• • •

are measurable, so are E1+E1+. . . and

(So far only the special case of the sum of diajoint sets has • • • •

been dealt with, in § 2·6 (ii).) We show that the product E1 E1 is measurable. By the condition of (iii) in the form (2), given E, we can find closed sets Q1,Q1, such that E,.. = Q. +e., where m.e.<E/2•. • • •

•• •

Then

• • •

= Qt Qa

• · ·

+e,

ece1+e1+... ,

where and

E1 Ea

so

m*e<E.

is closed (from § 1 ·3), and so, by the sufficiency of the condition (iii) (2), E1 E2 • is measurable. The result for the sum E1+E2+ ... is complementary.

Q1 Q1

• • •

• •

2·9. Sequences of sets. (i) Let E1cE1c ... , and let E

lim E.. Then, if each E. is mea&ttrable, E is measurable and mE lim mE•. =

-

ft.-+CO

SEQUENCES

For

E

=

OF

19

SETS

E1+(E1-E1)+(E3-E2)+ . . . ,

and so mE == mE1 +m(E2-E1)+ ... =mE1 +(mE2-mE1)+ . . . ,

giving the result. (ii) Let E1�E,� ... be a decrea8ing sequence of meaB'Urable sets. Tken m(limEn) =lim mE•. The proof is from (i) by complements. (It must be under­ stood in (ii) that the sets E. are not aU of infinite me&aure.) We extend (i) and (ii) from limits to upper and lower limits. (iii) If tJt.e En are mea8'Urable and are all contained in a set of finite meaaure, then lim mEn E; m(limEn). Write Fn = E,..+E,.. +1 +. . . . Then F. is measurable, F.:::> FA+l -

and Therefore

-

lim E.

= lim F•.

m(lim En) = lim mF•.

But, since Fn contains each of E", E•+h . . .

,

mFn �bound (mE., mE•+l' .. .), lim mF" � lim mE", and so giving the result. The necessity of the hypothesis that the E" are contained in a set of finite measure is illustrated by the example E. equal to the interval (n, n+ 1), for which the conclusion is false. (iv) Lim mE,.�m(liJ!! EnJ· The proof is oomplemental'J to that of (ill). Combining (iii) and (iv) we have (v) If the En are mea8urable and are all contained in a set of flniu meaB'Ure, and if they have a limit set E, then lim mEn eNt8

arul eq'IUll8 mE.

We extend these results to sequences of sets which are not assumed to be measura.ble. The following lemma will be useful. (vi) If E i6 any set, tMre are meaaurable sets F and G IJ'UCh

that

and

Fc:Ec:G mF =m.E,

m*E =mG.

MEASURE

20

F and G may be called respectively a mea8'Urable kernel and a measurable envelope of E. Let En be a decreasing sequence tending to zero. We can choose a sequence On of open sets all containing E The sets

such that

G be the product contains E. We have Let

set 0102•••• Then

G is

measurable and

mG�m*E mG�mOn

and for every

n.

Hence

mG =m*E. A complementary construction yields a set F with the

properties stated. (vii)

Let E1 cE2c ... , and let E = lim En. Then

EncE for every n, m*En�m*E and so lim m*En�m* E. It remains to prove the reverse inequality. Let Gn be a measurable envelope of En, so that mG, = m*En• Since

Hn = GnGn+J•••

Write then

E,�.cHncGn m*En = mHn =mGn.

and so Now the

Hn form

an increasing sequence of measurable sets,

and if His their limit,

EcH.

Therefore

m*E�mH = limmHn

(from (i))

=limm*En.

This proves (vii). The complementary result is

(viii) Let E1�E2=:J •••• Then m (lim En) =lim m* En, if the

right-hand Bide i8 finite.

.

PLANE MEASURE

21

2·10. Plane measure. The theory which has been given in detail for linear sets in §§ 2·1-2·9 can be extended to higher dimensions. It is sufficient to consider the plane. The founda­ tion stone is the open set, as it is for linear sets. The measure of a rectangle is its area, and as the (plane) measure of any linear set must be zero .(since it can be enclosed in rectangles of arbitrarily small area), it is indifferent whether the sides of a rectangle are regarded as belonging to it or not. The measure of an open set is then defined to be the sum of the measures of the meshes of a network of which it is com­ posed (§ 1•5). To justify this definition we must prove that the measure so defined is independent of the particular network taken. Take two networks, which may have different origins and axes in different directions. Let the open set 0 be composed 00

�

of the meshes � r" of the first network or of the meshes � &�c of 1

the second. Using the symbol 00

1

p,

for two-dimensional measure, let us

00

suppose that � p,rt > � fLBk · Choose n large enough to make 1

1

Inside each r�, we can place a closed rectangle R, and about each 8k we can circumscribe an open rectangle Sk, maintaining the inequality

But the Heine-Borel theorem (§ 1·4) shows that the last inn

equality is impossible. For every point of the closed set � R, n

1

is a point of 0 and is interior to some Sk. The set � R1 can 1

therefore be covered by a finite number of the Sk, and the sum n

of the areas of these Sk will be greater than � p.Ri. This 1

22

MEASUBE

contradicts the preceding inequality, and the definition of the measure of an open set is justified. The theory of measure of plane sets (or sets in n dimensions) can be developed as in §§ 2•1-2·9.

Ordi'fUJU set8. Let E be a linear set, which we shall suppose to lie on the �-axis. Let 0 be the set of ordinates of height k erected on E, i.e. the set of points (x, y) such that x is in E and 0�y�'"· We shall prove that 1'*0 = km*E

and

IL•O = km.E.

In particular, 0 is meaBUrable if and only if E ia, arul then fLO. = kmE. Enclose E in an open set 0 with mO <m*E + £. Let I be a typical interval of0, say a < x
a <x
- "1


Then l: R is an open set enclosing 0 and having measure less than (k + 2TJ) (m*E + £). Since £ and 1J are arbitrarily small, it follows that JL*O E;; hm*E. We must now prove that IL *O � hm*E. Take an open set containing 0; express it as the sum of meshes of a network formed by parallels to the axes. Enlarge each mesh slightly into an open rectangle. This construction can be performed so as to give a set of open rectangles l: 8,.., containing 0, with

l: IL8. < IL *0 + £. Since each ordinate ( x = x0, 0 � y�k) is a closed line�r set, the Reine-Borel theorem shows that a finite number of the 8,.. can be selected which cover it. Choose a single open rectangle T(�0) containing the ordinate and contained in this finite set of S's which cover it. Then l: T(x0), summed for x0 in E, is an open set, and its section by the axis y 0 is a linear open set, say l: J,.. On each 1,. erect an open rectangle R8 of height h. =

B OREL MEASURE

Then

km*E � km(�J.) p.(� R.) � p.(l: T ) �p.(l;8n) � �JL8n < p.*O + £.

23

==

Since £ is arbitrary, this gives the inequality p.*O. � hm*E, and so we have ,.t*O = km*E. We now give a 'proof by complements' that p. O. = km.E. * Enclose E in 0, and let 00 be the ordinate set of height k on0. It is easy to see that 0.0 is measurable and p.0.0 kmO. The first part of the theorem, applied to 0.0 - 0 shows that ==

p.*(!l0 - 0.) = km*(O- E).

Hence, by subtraction,

p..n = km.E.

2•11. Measurability in the sense of Borel. The approach to the measure of a set by way of outer and inner measure (§ 2·5) was made by Lebesgue in 1902. In 1898 Borel, having defined the measure of open sets and closed sets, had extended the definition to sets obtainable from them by a finite number or an enumerable infinity of operations of addition, subtraction, multiplication and taking a limit. Sets thus definable are said to be measurable (B); they form a sub·class of the class of sets measurable according to Lebesgue's procedure. The advan�ge of knowing that a set is measurable (B) is that a definite construction for obtaining it from intervals is implied. Given any meJUurable 8et E, there are Beta F and G, meaBUrable (B), &uck that FcEcG and mF = mE = mG. For the sets constructed in § 2·9 (vi) are meaaurable (B) and have the properties stated.

2·12. Measurable funcdons. Let j(x) be defined in a

measurable set E0• Denote by E(/> A ) the set of points of E0 at which f(x) > A , with a similar meaning for E(j� A), etc. The function f(x) is said to be mea&urable if, for all values of the constant A , the sets of one of the four families E(f> A ), E(j < A ), E(j� A ) , E(j � A )

are measurable.

24

1\ol E A S U B E

We prove that any one of these four conditions implies the other three. The sets of the first condition are complementary to those of the fourth, and similarly for the second and third conditions. To prove that the first condition implies the third, we observe that the set E(/ ?J:. A ) is the product of the sets E(/ > .A - *) for n I, 2, . . . , and so, being the product of measurable sets, is measurable. Similar arguments will complete the proof of the equivalence of the four conditions. The following sequence of results establishes the general principle that elementary operations performed on measurable functions yield measurable functions . (i) Iff i8 measurable and c is a constant, then J + c and cf are =

meaBU

rable.

This follows easily from the definition. (ii) If I and g are finite and mWJurable, the set E(f> g) is

mea&U

rable.

. If, for a particular x, f(x) > g(x), there is a rational r lying between them. Hence

E(l>g)

=

:E E(I > r) E(g < r) f'

summed over all rationals r, and is the sum of measurable sets. (iii) If I and g are finite and measurable, 80 are l+g and f-g. For E(l+ g > A) = E(f> A - g). The function A - g is measur­ able from (i) and the set E(l> A - g) is measurable from (ii). (iv) With the hypotkesia of (iii), fg is measu rable. If A > O, E(r > A) E(J> .jA) + E(J< - ,JA), which shows that the square of a measurable function is measurable. A pro' duct is reduced to squares by the identity =

4lg

=

(/+g)2 - (/-g) 2.

be a sequence of 'meaBU rable junctiona. PM:n M(x), the upper bound of the valu& at x of f1 , /2, i8 mea8'U rable. So iB the lower bound. (v) Let f1 ,j1,

• • •

• • •

M E A S U RABLE F ll N C T I O N S

25

E(M > A ) = ");E(fn > A),

For

n

which is the sum of measurable sets.

(vi) The limit of a

tio'IUI is mea8U rable.

mono

tonic sequence of measurable func­

For suppose the sequence is increasing (/n �/n+1). Then the limit is the same as the upper bound, and (v) gives the result. (vii) If f1,j2, are measurable, 80 are the upper and lower • • •

limit functio'M of the Bequence.

Define Mn(x) to be the upper bound of fn(x), fn+l (x), . . . . Then, by (v), .lJfn (x) is measurable. Also Mn�lJtn+l· By (vi), limMn is measurable. But this is the same function as limfn · (viii) A continu01t8 Junction i8 measurable. If f is continuous, it is easy to see that the set E(f�A) con­ tains its limit-points, i.e. it is closed and therefore measurable. E X A M P L E S O N C H APTER II

( 1) Prove that Cantor's ternary set (Ch. I, Ex. 7) has

measure zero. (2) Let j(x, y) be a measurable function of x for each y, and continuous in y for each x. Prove that lim/(x, y) and limf(x, y) 11�a 11�a are measurable functions of x. (3) Let f(x) be a measurable function in (a, b). Prove that, given E, there is a continuous function �(x) such that J/(x) - �(x) I < E except in a set of measure less than E. (In general terms, 'any measurable function is nearly a continuous function'.) (4) Egoro!f'8 theorem. Let the sequence of measurable func­ . tions fn (x) tend to the finite limit f(x) in E. Prove that, given �' we can find a sub-set of E of measure greater than mE - 8 in which the convergence is uniform. A rough expression of this important theorem is that ' every convergent sequence of measurable functions is nearly uniformly convergent '. 3

CHAPTER III T H E L E B E S G U E INT E G R A L

3•1.

The Lebesgue integral.

The idea of the definite

integral which has come down through the centuries associates

J: (x ck ! )

with the

area

bounded by the curve

x-axis and the ordinates x

=

a, x = b.

y

=

/(z), the

Having developed in the

last chapter the concept of measure of a plane set of points we can, following Lebesgue, present the idea in a refined form. Let

E be a

set of points

x (which

may in a special case be an

inter .ral), and j(x) a function, supposed in the first instance to be positive.

Let fi be the plane set of points

(X, y)

such that

X takes

all

E and 0 � y � j(x). n can be described as the ordinate­ set of the function j(x) on E. If n is plane-measurable we shall say that j(x) IUUJ a Lebesgue integralt in E, written

values in

JJ
or

and its value is ,.,.n. If /(x) is not restricted to

be

(L)

JJ
positive, it can be expressed as

the difference of two positive funotions

j(x)

=

f+(x) -j_(x),

where /+ (x) is zero for values of

x for which j(x) is negative and is elsewhere equal to j(x), and f_ (x) is zero if j(x) is positive and is otherwise equal to - j(x). If n+, n_ are the ordinate sets of f+,f_, then the integral of f is defined to be p.!l+- ,.,.n_. f 14'or some years summable

was

commonly used to mean inlegrabk

(L).

THE R I E M A N N IN TEG RAL

27

In many cases we shall give proofs of theorems only for positive f, on the understanding that the extension to a general f can be made by the decomposition given in the last para· graph. The ordinate set n of j(x) has been defined by the 'closed' inequality 0 � y �f(x). We shall show that, so far as measure is concerned, it is indifferent whether the points on the curve y f(x) are included or not. (This fact will be needed in § 3·9.) Let no be the set defined by X in E and 0 � y < f(x). The transformation x' x, y' ( 1 + 8) y alters the areas of rectangles in the ratio 1 : 1 + 8 and hence the outer and inner measures of any plane sets in the same ratio. This transforma­ tion turns no into an ordinate set of ( 1 + 8)/(x). n is contained in this and so =

=

=

,..,• n � p.*n0 {( 1 + 8)/}

=

( I + 8)p. * � (j).

Since 8 is arbitrary, p.*n � p.*n0• But 00c 0 and so p.*n �-t*n0• Similarly �' n �-'• no, and hence if either 0 or no * is measurable so is the other and ,.., n ,.., n0• =

=

=

3·2.

The Riemann integral. It is assumed that the reader

is acquainted with Riemann's idea of the definite integral of a bounded function. t We summarize his definition, in the form given to it by Darboux. Divide the interval (a, b) into a finite number of parts 81, • . • , 8n; let M,, m,. be the upper and lower bounds of j(x) for values of x in 8,.. Form upper and lower approximative sums 8

=

I: M"8,., ft

1

s = � m,.8,.. '"

1

Let J be the lower bound of sums 8 and j the upper bound of sums 8 for all modes of subdividing (a, b). If J = j, this number is the Riemann integral (R)r/(x)tk.

t See, e.g., \Vhittaker and \\•atson, Atodern Analysis� ch. IV; or Hardy, \"11 ((or a eontinuou� integrand ).

Pure Mathematics, ch.

28

THE L E B E S G U E I N T EG R AL

It can be proved that J and j are not merely bounds of sums 8 and 8 but are the unique limits of these sums as the length of the greatest sub·interval 8,. tends to zero. It is easy to think of a function which is integrable (L) in an interval, but not integrable (R). Let E be the set of rational numbers in (a, b). Let /(x) be the ckaraeteri8tic Junction of E, that is to say, /(x) I for x in E and f(x) = 0 for other values of x. Then, whatever the mode of subdivision of (a, b), every M,. is 1 and every m,. is 0. Hence J b - a and j = 0, and there is no Riemann integral. But since mE = 0, it follows from § 2·10 ==

=

and § 3·1 that (L)

J:f(x)ck

=

0.

We shall pro�e the following consistency theorem.

J:

If (R) f(x)ck exi8t&,

80

doe8 (L)

J:f(x)ck and tJtey are equal.

By the Riemann definition, the inter�al (a, b) can be sub. divided into parts 8r such that

The rectang]es of base 8r and height M,. form a set containing the ordinate set n off(x). Similarly the rectangles of base 8r and height mr form a set contained in !l. Each of these sets of rectangles•is plane·measurable, and we have A

A

� mr 8r � ,.,.n � p.*!l � � M,.8,.. 1

1

Since • is arbitrary, n is measurable and p.O which is what we set out to prove.

=

,

(R)

fJ(x)ck, a

3·3. The scope of Lebesgue's definition. We inquire

what restrictions are to be laid on the field of integration E and on the integrandj(x) for the Lebesgue integral a meaning.

J j(x)ck to have E

S C O P E OF L E B E S G U E ' S D E F I N I T I O N

29

It is clear from § 2·10 that E must be supposed measurable, for if it is not, the integral of a constant would not have a definite value. If the ordinate-set n of a functicm j(x) is meaaurable , then f(x) is a measurable functicm. Consider the part B!l of n contained in the strip 8 of the plane for which l � y < l+h. Let E1 denote the set of values of x in E for which f(x) > l. Then, as h takes a sequence of values decreasing to zero, the sets El+h form an increasing sequence with E1 as limit·set. If 8E1 and 8E1+h are the points of the strip 8 whose abscissae are respectively in E1 and E1+h' then BE,+n, cSncsE,.

Remembering that sn is plane-measurable and using the theorem of § 2· 10, we have

hm*El+h � p.(S!l) � hm.E1 and so

m*Ez+h � nt.E1•

Let /�; tend to zero and we have, from § 2·9 (vii), m*E1 = lim m*E,+h � m.E1• h-+0 Therefore E1 is measurable, and it wi1l follow that f(x) is measurable provided that the set in which f(x) = 0 is measur· able. It is to be observed that the set E(f = 0), whether measur· able or not, would give an ordinate·set of measure zero. The set E(f = 0) is in fact measurable being the complement of E{/> 0) with regard to E. E was originally assumed measurable and E(f> 0) has been proved measurable. Hence f(x) is measurable. The next section will show that the converse of the last theorem is true, though it is clear that, without some restric­ tion on f(x), p.!l may be infinite. This possibility will be ruled out by assuming for the present that j(x) is bounded, and that E has finite measure.

THE L E B ESG U E I N T E G RAL

30

3·4. The integral as the limit of approximative Suppose that f(x) is measurable and that

sums.

A �f(x) � B. Take a scale of numbers subdividing the range of variation (A, B) of f(x), Let E., let

=

E(l., �f(x) < lr+1), e,. = mE,. for En = E(f(x)

1111

B),

r =

1, 2,

. . •

, n - 1, and

e" = mE".

ft.

8 = l: lr+1 e,.,

Write

1

(where ln+l is taken to be B).

t,..,.t-----:r---�---,1-­

l,t----+-�---S.-�---+-r--

0

£,

£,

FJg. 1.

We shall prove that: As the greatut sub-interval (lr+1 - l,.) tend&

aftd 8 Aatle tile common limit

Jj(x)tk.

..

to

zero, the sums 8

For any scale of subdivision, the numbers S, s satisfy

AmE � s � S � BmE.

31

APPROXIMAT I V E S U M S

Let J be the lower bound of sums 8 and j the upper bound of sums 8 for aU modes of subdivision of (A, B). Let 81, 81 be the sums corresponding to any set of numbers dividing (A,B), and 82, 82 the sums for another set. Consider now the scale obtained by taking aU the numbers of both these sets, and let 8', 8' be the corresponding sums. The introduction of a fresh dividing number cannot increase a sum 8 or decrease a sum 8. For suppose a number � inserted between l,. and l,.+1; the increase in 8 is

which is negative or zero. Therefore 81�8' �8'�81, that is to· say, any upper sum is greater than or equal to any lower sum. Hence J �j. Moreover, 8 - 8 = l: (l,+1 - l,) e, ' EmE if max (1,.+1 - l,.) � £. Therefore J j and 8, 8 tend to this limit as max (l,+1 - l,.) tends to zero. We stiU have to prove that this limit is equal to the Lebesgue integral of j(x) as defined in § 3·1. Suppose first that f(x) �0, and let n be its ordinate set. Let t/l(x) be the function equal to l,. in E,(r 1 , 2, . . . , n) and 1/J(x) the function equal to lr+1, in each E,; let 0;, flt be their respective ordinate sets. Then fl; c D. c D..v, and (from § 2·10) 8 p,D.Y, s = p,D.; . Hence the common limit of 8, 8 is p,D.. There is no difficulty in extending the proof to cover a bounded function taking both positive and negative values. We shall now show how the extension to unbounded functions may conveniently be made. =

=

=

3·5. The integral of an unbounded function. Suppose that j(x) is measurable and positive. Define an auxiliary function {f(x)}n {/(x) },�

=

j(x) for x such that f(x) � n,

{/(x) }11

=

n for x such that f(x) > n.

32

T H E L E B E S G U E I N T E G RAL

Define

J f(x) dx E

J {f(x)} dx fj<x)lk

as lim

n-+co

with the definition of

E

n

aa

(assu med finite). This agree s

p.O., where n is its ordinate-

set, for if (0)" is the ordinate·set of

{f(x)},"

then (0),. has

aa

being equal to

Q f(x) /+(a:) -J_(a:) in § 3·1, J/(x)lk agree the Jj+(x)lk-Jj-<x)lk. This If f(x) is integrable in E, 80 is 1/(x) I· The converse ill true, provided t1uU f(x) i8 mea8Urable. Also J/(x)lk tO J,p<x) I da:. /(x) = f+(x)-f_(x), I f(x) I = f+ (x) +f_ (x).

and p,(!l)A tendS tO p,!l. is not restricte d to be positive, decompose it into

limit If

a.s

and define

again is in

ment with

definition by ordinate-sets.

If, as above, we write

It is easy that

then

to see by considering the appropriate ordinate- sets

f l f(x) ldx = J f+(x) dx +J f_(x)dx. f(x) are f+(x) E

E

For the converse, note that, if

andf_(x).

E

is measurable, so

Finally, the inequality comes from

I a - b I � a +b,

where

a

and b are the integrals in E of

f+(x) f_(x). and

It may be said, following the usage for infinite series, that the Lebesgue integral is

that is to say, it integrates only those functions whose modulus is� also

ab80lutely convergent;

integrable.

Examples { 1 ) Verify from the definition that

1

J. :r«lk

exists if oc < 1.

INFINITE BAN G E

33

J(x) be positive and bounded in (£, I) for each posit . dx tends to a finite limit l as £--+0, then fl/(x) dx tive If f�J(x) J0 J etists in the sense of this section and is equal to l. (2) Let £.

(3) In Example 2, it is not possible to omit the condition that

j(a:) is positive.

Prove that the function

j(a:) ! (a:1 sin�) =

is not integrable in Lebesgue's sense over (0, 1).

3·6. The integral over

an

infinite range. We must

x dx. -«J «J i dx i0 x J:x. J:x. --dx .

re­

view in the light of the Lebesgue theory the interpretation of sin integrals such as 2 and The simplest way of X 1 X assigning a meaning to them is as limits, as X tends to infinity, dx sin of the integrals 1 and It is to be observed that X X o 1 the second of the integrals is not 'absolutely convergent', for Sin dx tends to a it is not difficult to prove that, although o . XI I SID fi nite limit, dx tends to infinity. 0 X The argument of the first paragraph of § 3·5 will convince the reader that it is indifferent whether an integral of a positive

i

x

function such as

Ifj(x)

x x i .x

x lim f � or directly as the J..,dX� is regarded asx�..,Jl X 1

measure of the appropriate (unbounded) ordinate-set. is not restricted to be positive, then the decomposition

J(x) /+(x)-f_ (x) =

of·§ 3·1 defines its integral over an infinite range (say - co, oo) as, p.fl+ - ,_,.n_ provided that these are finite, in which case

Jj

,.n+ +,.�L I I is also finite, and the integral is absolutely canvergent. The method of the last paragraph would not hold for Sln dx, for which the interpretation lim � dx must be =

i

. .., x

0

X retained.

. x x�..,J.0 X

34

T H E L E B E S G U E I N T EG R A L

In future it will be assumed (in the absence of any state­ ment to the contrary) that any range of integration is allowed to be infinite so long as the integral in question is absolutely · convergent. 3·7. Simple properties of the integral. This section con­ tains some properties which would be expected of the Lebesgue integral if it is to be a useful concept. The theorems apply to functions bounded or unbounded, and to ranges of integration finite or infinite. We shall suppose without explicit mention that the functions and sets are measurable. Any proof can be expressed either in terms of ordinateMsets or of approximative sums; we shall choose which­ ever presentation seems the more direct. The variable of integration will usually for brevity be omitted, and written simply as

J/

IJ<:r:)

dx

(i) Iff�g, then

I/" fi· For, iff is positive, O.(f, E), the ordinate-set of f based on E, is contained in Q(g, E) and so p,O(f) � pfl(g). To extend the proof to functions taking both signs, write f = f+-f_ and g g+ -g_, the suffixes having the meanings 888igned to them in § Then, in E, f+�g+ and f_ � g_. =

3·1.

Hence

Therefore

p.O.(f+) - p.O.(f_) � p.O.(g+) - p.fl(g_),

which is what we have to prove. COBOI.LARY. 11

set E

offinite

If A ��� B in then AmE �I/� BmE.

The first mean-value theoretn. meiJ8Uf'e,

35

SIMPLE PROPERT I E S

(ii) (The additive property for seta.) If E is the BUm of a finite numher or an enumerable infinity of non-overlapping meaBurable sets E1, E1, , then 10 J f = J f+ j f+ ... if the left-hand Bide exists. If f is positive, let 0 = 0.(/, E), Qn = D.(f,En). Q = :E On, and so p,D. = :E p,Q,.. Then (iii) If c U a contJtant, J cf = cJ/ We give the proof for positive f. Let c be positive. The ordinate-set D.(cf) is derived from 0(/) by the transformation y' = cy (homogeneous strain). This alters the areas of = rectangles in the ratio 1 c, and hence the measures of any sets • • •

E

E,

E.

,

E

'

x

x,

:

in the same ratio. This proves the theorem. If c is negative, say c = - then =

k, p,ll(k/) kp,O(/) and Jc! = p.D.(kf) = - kp.D.(J) = cJ!· (iv) (The additive property for function.s.) If J and g are integrable in E, so is f+g and t(/+g) = fj+ fl· We shall suppose first that (i) f and g are bounded, and (ii) E has finite measure, and remove these restrictions in tum. We start by proving the result when g is a constant k. Take for f a scale of subdivision, of which l, is a typical member, giving an approximative sum, :El,e, = say, for J f. The numbers l,.+k then form a scale of subdivision for the function J+ k, giving an approximative sum, say, for (/+ k). t Then a' = }"; (l,. + k)e,. = s-+:lcmE. -

·

8

8

1

E

T H E L E B E S G U E I N TEGRAL

36

Taking the limit as max

(l,+1 - l,.) tends to

zero,

we have

g to be any bounded function. Let E,. be the set E(l, ��< l,+1). Then J (f+ g) = :f1 J (/+ g) from (ii) � t J (l,.+ g) from (i) 1 Take now

E

B, E,

l1'+1 in place of l,., we have J,p+ g) <S; 8 + J,,uTaking limits as max (l,+1 -l,.) tends to Similarly, putting

zero,

we deduce that

The result must now be extended to unbounded functions. Suppose first that I and are both positive. Then, with the meaning assigned to the suffix n in § 3·5,

g (/+ g),. � (f),.+ (g),. � �n·

Integrating and letting n tend to infinity, we have the result. and are not always positive, decompose E into measurable are of constant sign. sub-sets (six at most) in which Consider for example the set in which I and J+ are positive is the sum of two and is negative. Then / = positive functions and we have only to apply the last case and

Iff g

J, g, J+ g (f+g)+(-g)

g

transpose the term

Jg.

g

37

SETS OF MEASURE ZERO

If, finally, contained in

.. E has infinite measure, let Ex be the part of E

( - X, X). Then, by what we have already proved, f (/+ g) = J !+ J g. J Ez

Bz

Letting

Ez

X tend to infinity, we have the result.

Example If J, g are integrable, establish the integrability of Jg under suitable conditions.

3·8. Sets of measure zero. A property which holds ex­ cept in a set of measure zero is said to hold The accepted abbreviation for 'almost everywhere' is p.p.

almost everywhere.

(presque partout). ( i) Iff= o p.p. in E, I/ = 0. Let E1 be the set of measure zero in which J =1= 0. Then J != J !+ J J. The first term on the right-hand side is zero because mE1 = 0 and the second because J = 0. CoROt.t.ARY. Iff= g p.p. in E, and one of them i8 integrable, so is the other and IEf = Il· It is thus possible-and often convenient-to speak of zero I/ although f may be undefined in a subset of of E. E

E,

E-E1

.IJlea&ure

The next two theorems are in the nature of converses of the last.

(ii)

Iff;.. 0 and I/ = 0, then f = 0 p.p. in E.

T H E L E B E S G U E I NTEGRAL

38

E(f> 0) = l: E(f> 1/n) = l:E3, say. For each n, mE. = 0,for, if not,J fdx?: J fdx?: n!mEn > 0, which contradiets the hypothesis. Hence E(f> 0) has measure zero,and the theorem is proved. (iii) If J i8 integrable, and J:f 0 for every x in a x.;; b, then I= 0 p.p. in (a, b). If the conclusion is false,then either f > 0 in a set of positive measure or f 0 in a set of positive meas uppose the former. There is then a closed set Q of positive measure con­ tained in the set in which I > 0. Let 0 be the complement of Q ; and l:In its constituent intervals. If I,, is (an, bn), then J I = f.,.I- J.a..l = o, Cl)

The set

a-1 E

Ea

=

<

by hypothesis. Hence

E;

ure-s

I,.

J/ = l: frf = (ii), I=

0'

a

and

a

J/= J:!- J/= O.

From 0 p.p. in Q, which is a contradiction, and the theorem is proved. 3·9. Sequences of integrals of positive functions. The theorems of this section and the next are concerned with the question:

If, a& n�, 1.�1, under what conditions can we assert tlust (a, b) the standard uniform convergence of the sequence

For Riemann integrals over a finite interval sufficient condition is the to its limit function

f.

f.

S B Q U E NOBS

OF

INT EGRALS

39

The conditions for Lebesgue integrals are more general and simpler. The most useful and most easily applied test is that and x in a of § 3·1 0 (D), namely • that /n is bounded for all set E of finite measure. We begin by observing that the results of§ 2·9, (i)-(v), applied to a sequence of ordinate-sets n. and their limit n (or more generally their upper and lower limits) yield theorems about the integration of a sequence of positive. functions. Some lemmas are required. On nn. f j,., then If � o f M(x) m(x), x of/1 (x),f2(x), . . . . lim � off. If 0 'ffl, ,/,.+1, oM f = limf3, of f., lim n. lim Qfl, of lim /11r lim ft�r· The assertion about M(x) is easily seen to be true if the ordinate-set n. is taken to be 'open at the top' ' i.e. the set (x in E, o , y < /n(x)) and the ordinate set of M(x) defined < M (x)). similarly by the relations (x in E, 0 For m(x) take 'closed' ordinate-sets with defining inequalities 0 � �f.(x) and 0 � ' m(x). The second and third paragraphs follow from the first in .the same way that (vi) and (vii) of § 2·12 follow from (v ) . Using these lemmas, we deduce the following results from § 2·9, (i)-(v ). that f. � 0 f f lim fn, (i) If fn �/n+l'

n

l:fl. and J. and is an ordinate-set o are rupectively ordinate-set& o and the upper and lower bounds at then is an ordinate-Bet and For any sequence poaitive functiotuJ are ordinate-setB and 'y

y

y

in the range o integration. and then J! = limJf . (ii) If /11 � /n+l ' and f = limj,., then J! = limfJ,.. (iii) If, for all n, fn (x) � t/J(x), where .P is int�rabk, then limJ!,..;; J(lim f ).

Suppose

=

..

.

40

THE L E B E S G U E INTE GRAL

(iv)

For any &eq..unu ofpositive function8 f,._, J
(v) It follows from, (iii) and (iv) that, if 0 �f. � .p, where .P i8 imegrable, and ftr. tends to a limit f, then

From the principle of § 3·8, thi8 stiU hold8 if it i8 only

true

p.p.

that fn tends to f. These statements (i)-(v) are for the most part steps towards the more general results for functions of arbitrary sign given in the next section. The one which has the greatest inde­ pendent interest is (iv), and we restate it in its most useful form. FATOU 'S LEMMA. If fn � 0, and fn�f p.p., then

J!� limJf...

3·10 Sequences of integrals (integradon term by term). We now suppose that the sign of fn is unrestricted and give the most useful results.

(A) BEPPO LEVI's THEOREM. If/11r � /,+1, and f = limf., then The theorem (i) of § 3·9 applied to the sequence f, -/1 gives \

f!- f!l = limf!..- f!l.

and hence the result. Notice that if

Jf,.-+CO, the theorem admits of the inter­

pretation that then p.O. is infinite, n being the ordinate-set off.

INTE GRATION T E RM BY TERM

41

This theorem may be restated in the language of series instead of sequences.

(B) If I:un (x) is a serie& of positive

terms,

then

provided that either side is finite. (0) THEOREM OF DOMINATED CONVERGENCE. If, for aU n, I /n (x) I � ep(x), where .P is integrable, and In�f p. p . , then J! = umJf . Take the sequence In + .p, which is positive and is 'dominated' by the integrable function 2.p. ..

Then § 3·9 (ill) gives

f!.. + f., � f!+ f.,. And § 3·9 (iv) gives J!+ J1/1 � limJ! . + J1/J. lim

This pair of inequalities proves the theorem. The following special case is constantly used. (D) THEoREM OF BOUNDED CONVERGENCE.

finite re, 1 /n (x) f � M for all n, x. and fn�f p.p., then J/ = lim J/•· mea.m

If, in a set E of

A variant of the theorem of dominated convergence which is repeatedly appealed to in practice is

Let the 8erie& �un(x) be boundedly ronvergent to sum s(x) n (i.e. ifsn (x) � u,. (x), then lsn (x) / ' M). Let cp(x) be integrabk. Then (E)

=

1

42

THE L E B E S G U E INTEGRAL

Another test for the inversion of summation and integration which is convenient of application is the following extension

(B). (F) If either Il: I u,. l l:II u,.l i8.finite, then Il: u,. = l:Iu,..

of

or

That the alternative hypotheses are equivalent is shown by

(B) itself.

Il: I u,.l

Suppose tha.t

l: I u,. (x) l:un(x) l: I u8 (x)

Then So

is finite.

1 converges for almost all

converges for almost all

the function

f.

x,

x.

and it is dominated by

The theorem of dominated convergence

gives the result.

( In (x)

EXAMPLES O N CHAPTER I l l

1) Show that the theorem of bounded convergence applies

to

In (x) n311x/( 1 + n1x2) (R) I/,. -+(R) If =

nx/( 1 +n1x1), and the theorem of dominated con­

vergence to

=

, for

0� x � 1.

[As these

sequences do not converge uniformly, the standard criterion for

does not apply.]

(2) Give an example in which th� sign < is to be taken in Fatou's lemma, § 3·9 (v). (3) Use Egoroff's theorem (Ch. II, Ex. 4) to prove the theorem of bounded convergence.

( 4) Prove that, if "' > 0,

n x«-1 dx

x lim f• (1- ) n ... «>Jo

=

f «>e-z�.a:-ltiz. ' Jo

(This i s the main step in the proof that the integral and product

definitions of the r function are equivalent. See, e.g., Whittaker and Watson,

Modem

A

naly8is, p.

235.)

( 6)

EXAMPLES If '"" (x)

=

43

ae-naz - be-nk (0 < a < b),

:E

prove that

L"'u,. (x) tk L"'{l:u,. (x) :E J: u,. (x) tk +

Verify directly tha.t

diverges .

I

I

)tk.

By the appropriate expansions and term-by-term integra­ tions, establish the following results, of the theorems

(6)

(7)

(.A )-(F) of § 3·10 you appeal.

fl xP log ( 1) 1 1 + . . (p > - 1). + dx J 1- x x (p + 1)' (p + 2)t . } fr.o sinh bx { 1 1 1 + + + J sinhax dx = 2b a1- b' (3a)1- bt (5a)1- bt ··· o

=

0

1t

(8 )

(6)-(8), stating to which

f r.osin axdx J ez - 1 = o

= 2a

tan

( ka1

.,

nb 2a

(0 < b < a).

)

1 1 - m� + 2 -1 2

e

as

(9) For a given f(x), define M(x)

(a

real ).

lim {upper bound of f(x) in (x- 8,x + 8)}, ��

and

m(x) as

0

the limit of the lower bound.

With the notation of §

J (10)

=

3· 2,

prove that, if f(x) is bounded,

(L) J:M(x)tk,

Deduce from Example 9 that

condition that

f(x)

J:

j = (L) m(x) tk. a,

necessary and sufficient

should be R-integra.ble is that it

is con­

tinuous p.p.

(11)

Let f(x) be L-integrable in

there is a continuous function

Prove tha.t

Prove that, given

such that

(x)

J: f(x) - t/>(x) tk < •· J:if(x + k) -f(x) l dx I

( 1 2)

(a, b).

E,

I

tends to 0 aa k tends to 0.

OHAPTER I V

D I F F E RE N T I A T I O N AND I NT E G RA T I O N

4·1. DJ«erendadon and integradon processes. If f(x) is continuous and F(x) =

cp(x),

then

inverse

dt,

F'(x) = f(x). if it is given that (x) has a continuous derivative

then it is well known that Moreover,

J:!(t)

as

4>(6)- 4>(a) = J:(x) tk.

We shall investigate how far these results hold for Lebesgue integrals.

If no assumption is made about f(x) beyond its integrability

F(x) is its indefinite integral, then F'(x) may fail to be equal to j(x) in a set of measure zero. (For if E is any set of measure zero, and f(x) is its characteristic function (see § 3·2), then F' (x) = 0 everywhere and F' (x) -4: f(x) in E.) We shall prove in § 4·7 that F'(x) f(x) p.p. (L), and

=

4·2. The derlvates of a funcdon. As

general a unique limit, as we consider

k

there is not in

tends to zero, of the ratio

{ cp(x + k) - cp(x) } fk, -

--

lim , lim , lim , lim

�+o ,.-:;:f:o IH-o ,. ,. o of this ratio, these upper and lower limits always existing as finite numbers or + oo or

- oo.

They are called the upper and

45

D E RIVATES

deri

vate8 (or derived numbers) on the right and on the

lower

left respectively, and written as

D- tfo(x),

D+ t/J(x) ,

D+t/J(x),

the sign of the suffix showing whether

D_t/J(x), h

tends

to 0 through

positive or negative values and its position indicating an upper or a lower limit. It is necessary and sufficient for the existence of the deriva· tive

t/J'(x)

that the four derivates should be equal.

It can be proved that the derivates of any measurable func­ tion

are

which

measurable. We confine ourselves to two special cases,

are

sufficient for our purposes and which admit of

simpler proofs.

is either (i) monotonic (ii) conti derivate8 are mea&urable. If

or

t/J(x)

, then its

1t'UO'U8

to give the detail of the proof for D+t/J(x). Write M (x, S) and M,. (x, S) for the upper bounds of

It is sufficient

for

0 < k < S,

{t/J(x + h) - t/J(x)}/k

k

when

takes respectively all values and only

rational values.

D+ t/J(x)

is then lim M (x, S).

1-+0

We shall prove that

M,. � M.

M(x, S)

==

M,. (x, S).

It is clear that

To prove the reverse inequality, suppose that p. is any

number less than

M (x, S).

Then, for some positive k less than S,

t/J(x+ h) - t/J(x) '> P.· h

r

If tfo(x) is an increasing function, take a sequence of rationals approaching

k from above.

Then

tfo(x + r ) - t/J(x) +<x + k) - + (x)

r

�

r

-----

and the latter quotient is greater than p. if

r

is near enough to h.

If tfo(x) is a decreasing function, the same inequality is ob­

tained by letting

J4(x, S) � p..

r

tend to

k

from below.

In either case

46

D I FFERENTIATION A N D I N T E G RA T I O N

If (x+r) -H/>(x + h) as r-+h, and so, again, � (x, 8) � I'· Since I' is any number less than M(x, 8) it follows that M,. (x, 8)�M(x, 8). Hence M,. = M. For each fixed r, {tf>(x + r) -
E:mmplu (1) Evaluate the derivates at x = 0 of tf>(x) = x sin (1/x) (x =1= 0). nstruct a. function having given numbers as derivates for a. given value of x =

•

. 4·3. Vitali's covering theorem. Let E be a set of points x and :E a. set of intervals /. Suppose that, given any x of E and any €, we can find a.n interval I of :E, with length less than €, having x as a.n interior or end-point. We shall then say that :E covers E in the seMe of Vitali. It is clear that, if 0 is any open set containing E, the subset of intervals of :E contained in 0 also covers E in the sense of Vitali. VITALI's THEOREM. Let E be a set of point8 of finite outer metJBUre, and :E a set of inter1Xll8 I covering E in the Ben&e of Vitali. Pken we can find an enumerable set of inter1Xll8 11, Is, of :E, non-overlapping, 8'UCh that the set of points of E not in any I,., has mea8ure zero. We shall define the intervals 11, Is, . . . inductively. Let 0 be a.n open set of finite measure containing E. Then

closed

. . •

we need retain only intervals of :E contained in 0. Choose 11 to be any interval of :E. Suppose that 11, , 1,. have been defined. Let k,., be the upper bound of the lengths • • •

,.

of intervals of l: which have no point in common with � �1 ' Choose for J•+I any such interval, of length greater than lk•. This choice can always be made unless, for some n, in which

case

11 + ... + l,.';;) E, the theorem is proved.

VITALI ' S T H E O R E M

47 00

It will be shown that the set R of points of E not in � I., has 1 measure zero. Suppose on the contrary that m*R 0. Let ln be the length of I.. Denote by J. the interval of length 5l,. concentric with ln.

>

00

00

Since � I.cO, �ln converges and we can choose 1 1

N such that

� mJn = 5 � l" < m* R, N+l N+l 00

00

>N.

and so there are points of R which belong to no J" for n Let x be such a point. Since x belongs to no In, it belongs to an /, of length l say, of � such that II, = 0 for n 1, But I has points in common with an I, for some n > for if not, l � kn < 2l.+1 for every n. Since lim l. = 0, this is impos­ sible. Let n0 be the smallest value of n for which I and I" have common points. l � k,._1• Then =

..., N.

N,

>N

But n0 and so, from the definition of x, x does not belong to Jn,• Since I contains both a point of In, and a point not belonging to Jn,, l � 2l._ > k,._l , and this contradicts the preceding inequality. Hence mR = 0 and Vitali's theorem is proved. We add two coronaries, the first of which embodies the form of the theorem most useful in applications. £,

CoRoLLARY I. Under the hypotke8es of Vitali's theorem, given we can find a, finite number of disjoint interval8 11 , In of � • • •

,

such that the ooter measure of the set of points of E not COtJered by them iB le&s titan E. CoROLLARY 2. The theorem (and the proof) hold in

more) dimen8iO'Tls if intertJal8 a,re interpreted

m.).

atJ

(or sqootvA (cubes, two

48

DIFFERENTIATION A N D INTEGRATION

4·4. Dift"erentiabllity of a monotonic: func:tion. The object of the next two theorems is to prove that a monotonic function has almost everywhere a finite derivative. We shall assume the function (x) to be increasing. The set of values of x for which one of the upper tlerivates 'Of (x) is +oo ha8 measure zero. Let E be the set in (a, b) at which D+4> + oo or D-q, + oo, and suppose mE k. Let K be any (large) number. With each point x of E can be associated a sequence of intervals for which 114> > Kax (where 11 is written for (x+h) - 4>(x)). By Vitali's covering theorem a finite number of these intervals can be selected, non·overlapping, and of measure greater than !k. Summing over these intervals we have "l:A > JKk, or, since q, is increasing, (b) - (a) > lKk. This is false for sufficiently large K unless k 0. The set of pcnnts at which an upper derivate of an increa8ing function i8 greater than a lower derivate ha8 mea&Ure zero. Consider for definiteness D+4> and D_ 4> and suppose that the set of values of x for which D+ > D- 4> has measure greater than zero. This set is the sum of sets E(u, v) in which ' D+q, > u > v > D_ q, =

=

=

=

and u, v are rational numbers. There is then a pair (u, v) for which E(u, v), E say, has measure greater than 0. Enclose E in an open set 0 of measure less than k + E. Any point x of E is the right·hand end·point of arbitrarily small intervals (x- h, x) for which fi 4>(x) -
k,

=

=

• • •

4:9

MO N O T O N I C F U N C T I O N S

� \

Each point of E1 is the left-hand end-point of in rvals for which uh. By Vitali's theorem we can pick out from them disjoint intervals � ' . . . , J", contained in 11 + . . + Im and of total measure greater than lc - 2E. 2E). Summing over these intervals, we have �11 But, since
u(k -

v(k + E) >

u(k- 2e).

If e is small enough, this contradicts the hypothesis > v. This argument is applicable to any pair of upper and lower derivates on the same or opposite sidffi. The two theorems of this section together show that, if
u

4·5. The integraJ of the derivadve of an increasing funcdon. If
is increasing in (a, b), then is integrable and J:4>'(:r:)lk� (b)- (a).


but, from § 3·8 (i), this will not affect the integral. Let k take a decreasing sequence of values tending to 0. For values of x greater than it will be convenient to sup­ pose, as we may, that
b,

.p,.

(x)

=


.

From § 4·4, as h tends to zero, .p,.(x) tends to
J:.,, (:r:) lk � limJ:"'"(:r;) lk.

50

(k" J

DIFFERENTIATION A N D I N T E G R A T I O N b+A b b

f

4 t/J�a (x)dx =

But

=

1

f !{J�\£(x)fk- J: a+A

�(x)dx -

4

}

�(x)dx u

)

�(x)fk .

b+ 1f A k As tends w 0, lim kj , �(x)dx = .f>(b) and, since

f

J

a+h

a

1 a+A

(x)dx � luf>(a), we have lim k

4

(x)dx � �(a).

This proves the theorem. It is interesting to construct a continuous function (x) for which the sign < is w be taken in the theorem. Let E be Cantor's ternary set in (0, 1) (Ch. I, Ex. 7). In the intervals of OE (and at their end-points) define (x) as follows. For l � x � !, (x) l. For l � x � :, �(x) l, and for i � x � :, (x) = :. For /r � x � /.,, (x) = }; for t.77 � x � /.r, (x) f; and so on, repeatedly trisecting along Ox and bi­ secting along Oy. (A diagram will help the reader.) At a point x of E which is not an end-point of an interval of OE, �(x) can be uniquely defined as the limit of values of � taken in a sequence of intervals approaching x, and (x) is continuous at all points of ( 0, 1). Since '(x) = 0 at all points within an interval of OE and the sum of the lengths of these intervals is 1 , c/J'(x) = 0 p.p., =

=

=

and so

J:�'(x)fk

=

0. But �(1 ) - �(0) = 1.

4·6. Functions of bounded variation. Let j(x) be de­ fined for a � x �b. Take a set of points of division a

=

x0 < x1 < . . . < xk = b.

Let p be the sum ofthose differencesj(x,+1) -f(x,) for r = 0, 1, , k - 1, which are positive, and - n the sum of those which are negative. Then j(b) -/(a) = p -n 1 kand l: I /(x,+1) -j(x,) I = p + n = t, say. .•.

r-o

B O U N D E D VARIATION

51

Suppose that P, N, '1' are the upper bounds of p, n, t for all modes of subdivision of (a, b). It is plain that

P(or N) � '1' � P + N. Call the numbers P, N, '1' respectively the positive, negative and total variations ofj(x) in (a, b). Either these three numbers are all finite or '1' and at least one of P and N are infinite; if '1' is finite we say that f(x) ka8 variation in (a, b), and that '1' is its variation. variation, '1' = P + N. Whatever the If j(x) ka8 mode of subdivision, we have

total

bounded

bounded

p

=

n + j(b) -j(a), � N + j(b)-j(a).

This being true for all values of p, we deduce that

P � N +j(b) -j(a), P - N �f(b) -j(a).

or

A similar argument gives and so Then

'l' � p + n

=

N - P �j(a) - j(b), P - N = f(b) -f(a). p +p - {f(b ) - j(a)}

=

2p + N - P,

and since this is true for any choice of points of division, we may replace p by its upper bound P, giving

T � P + N. This combined with the obvious inequality

'l' � P + N

gives the result.

A junction of bounded variation is the difference between bounded i-ru;rea8ing junctions.

two

Let j(x) have bounded variation in (a, b) and a � x � b. Let P(x) and N(x) be the positive and negative variations of j(x) in the interval (a, x). The proof of the last theorem shows that

j(:x:)

=

{j(a) +P(x)} - N(x)

52

D I FFERENTIATION A N D IN TEGRATION

and the right-hand side is the difference between two bounded increasing functions of x. Conversely, if f(x) = cfo(x) - t/J(x), where cfo(x) and t/J (x) are bounded increasing functions, it follows from summing the inequalities

!/(x,.+I ) - f(x,.) I � {cfo (x,+l ) - tfo(x, )} + {ifJ(x,.+I ) - t/J(xr)} that f(x) has bounded variation. Examples (1) Prove that the sum and product of two functions of bounded variation have-bounded variation. (2) Prove that, if f(x) is of bounded variation and continu­ ous, the functions P(x), N(x) and T(x) are continuous. (3) Prove that the functions xsin (1/x), xlsin (1/zl) (defined to be 0 for x 0) are not of bounded variation in any interval containing x 0. Prove that xi sin (l/x8'1) has bounded varia­ tion. (4) Prove that a necessary and sufficient condition for the curve x == x(t), y y(t), a � t � b, to have finite length is that x(t) and y(t) have bounded variation. =

=

=

4·7. Differentiation of the indeftnite integral. If F(x)

=

F(a) +

J:!(t)dt, then F'(x)

=

f(x) almost everywhere.

F(x), being the integral of f+ - f_ , where /+ and f_ are the positive functions defined in § 3·1, is the difference between two increasing functions. Therefore F'(x) exists p.p., and it remains to show that F'(x) f(x) p.p. Suppose first that f(x) is bounded, say rJ(x) r � K. Let h take a sequence of values tending to 0. , =

Then and

! F(x+ htF(x) = u:+ltf(t)dt F(x+ h�- F(x) ' �F (x) p.p.

$!, K

DIFFERENTIATION O F THE INTEGRAL

53

By the theorem of bounded convergence, if c is any point of

(a, b), I.e }I.e{F(x+k)-F(x)}dx lim F'(x)dx = h--+0 � {![".li'(x)lk- !J:+".li'(x)lk} = F(c)-F(a), on account of the continuity of F. Hence J:{F'(x)-/(x)}lk 0 for all values of in (a, b). From § 3·8 (vii), F'(x) /(x) p.p. Now suppose that f(x) is unbounded. By the usual decom­ position of f into its positive and negative parts (/ f+ -f_ ), it is sufficient to give the proof for positive /(x). Let U<x)}*' be defined as in § 3·5. Then J:[J(t) -{/(t)},.] dt, being the integral of a positive function, increases with x. 1 , ,

(I

(I

=

=

c

=

=

Therefore its derivative exists p.p. and is never negative. By the result for bounded functions

d J.z{/(t)}.dt = fJ(x)}n p.p. dx F'(x) � {J(x)}*' p.p. 4

Hence

or, since n is arbitrarily large,

p.p. F'(x) f(x) � From this, f.li''(x)lk � fJ(x)lk = J'(b)-J'(a). •

54

DlFFEB E N TlATlON A N D lNTEGBA.TION

But the theorem of § 4·5 gives the reverse of this inequality, and so

J:{F'(x)-/(x)}dx

0.

=

Since the integrand is p.p. greater than or equal to zero, it is, by § 3·8 (ii), zero p.p., and the theorem is proved. AB the example given in § 4·1 shows, there is a sense in which this theorem is the best possible. The theorem contains as a special case a fundamental metric property of sets of points. If E is measurable we define the average density of E in an interval I to be m(EI)fml. The upper and lower right-hand densities of E at a point are the upper and lower limits of the average density in as k-+ 0. Similarly the left-hand densities are the limits of the average density in If all these four num­ bers are equal we speak simply of the den&ity of E at By applying the theorem of this section to the integral of the characteristic function of E, we have: The den&ity o a measurable set E i8 1 at almost all points o E, and is 0 at olmtJ8t aJ,l points o OE.

x

(x, x +k)

+

(x-k, x).

f

x.

f

f

f(x) (xn, x,. + kn)

4·8. Absolutely continuous functions. Let be de­ fined for a � are , Suppose that 1) , non-overlapping intervals in (a, H, given E, we can find 8 such that

x 'b.

(x1, x1 + k b). l: I f(x,. + k,.) - f(x,.) I for all choices of intervals with l:k,. 8, f(x) is said to be abso­ lutely contin1W'U8. interval (x, x +k), we see that an absolutely By taking • • .

<E

"

r-1

<

one

continuous function is continuous. It is clear that the sum of two absolutely continuous func­ tions is absolutely continuous. An absolutely contin1W'U8 function has variation. Given any subdivision of (a, we can, by inserting fresh points

b),

•

bO'Unded

ABSOLUTE CONTINUITY of division, split up

(a, b) into N

55

sets of intervals, each set being

of total length less than 8, where

N � b-a + I. 8 a fortiori N

Then :E I

summed over all these sub·intervals, and original sub-intervals, is less than

L\f f,

over the

E.

The importance of the class of absolutely continuous func­ tions is that it is the same

aa

the class of indefinite Lebesgue

integrals. We proceed to prove this.

If 'U8 in an interoal, and
The lemma will follow if, for any i

hypotheses hold,

=

Let E be the set, of measure With each point

al

in which

in which the =

of E is associated an interval

(arbitrarily small) such that

where "1 is any positive number. Given

E,

we can by Vitali's theorem (§ 4·3) pick out

a

finite

set, J1 say, of these intervals, non-overlapping, such that the measure of the complementary intervals (�) is less than Then

E.

where �1 and �� denote summations over the intervals of J1 And, by absolute and .fa respectively. Now 0 �1 continuity, �� tends to 0 as E tends to 0. Hence

� � 'fl(b-a). I (b) tf>(a). A necessary and BU.fficient condition that a function shmdd be an indefinite integral is that it shuuld be absolutely continUOU8. Jj(x)lk Since "' is arbitrarily small,

=

THEOREJ.\1.

We shall show that, for any set E of sufficiently small

measure contained in E.,

can be made arbitrarily

56

D I FFEREN TIATION A N D I N T E G R A T I O N

small. The special case in which E is the sum of intervals gives the necessity of the condition in the theorem. If l(x) is bounded, the first mean-value theorem (§ 3· 7 (i))

J! .;; J1 1 1. we ma.y suppose I positive. Then, given E, we can find n such that f I � J (/) n + JE. J Take 8 E/2n. Then, for any E with measure less than 8, fl
gives the result. For unbounded f, since

E.

Ee

=

We have now to prove the sufficiency half of the theorem, i.e. that, if t/J(x) is absolutely continuous, .P'(x) is integrable and

1/J(x) - 1/J(a)

=

J:1/J'(x)th:.

t/J(x), having bounded variation, is the difference

between

two increasing functions

ep(x) = e/11(x) - t/12(x).

From § 4·5, .Pi(x) and .p;(x), existing p.p. are integrable and so .P' (x) is integrable. Then �(x)

=

.f.(x) -

J:1/J'(x)th: is the difference between two

absolutely continuous functions and so is abso1utely con­ tinuous. Also, from § 4·7, �� (x) 0 p.p. Hence, by the lemma of this section, q,(x) is constant and so =

1/J(x) - 1/J(a) '

J:1/J'(x) th:,

,. and the theorem is proved. We now see that the continuous increasing function (x) defined at the end of § 4·5 is not absolutely continuous in (0, 1) since

EXAMPLES

57

EXAMPLES O N CHAPTER I V ( I ) Prove that the product of two absolutely continuous

functions is absolutely continuous.

(2)

By calculating the sum of its increments over suitable

intervals, give a direct proof that the function defined in §

4·5

is not absolutely continuous. (3) Construct a direct proof of the density theorem (end of §

4·7) without using properties of integrals. (4) From the density theorem deduce the

for bounded f.

(5)

F'

=

f theorem

Prove that the Lebesgue integral will integrate any

bounded derivative. (Ex.

3

of §

3·5 shows that the

hypothesis

of boundedness cannot be omitted.)

(6) Prove that any increasing function j(x) can be decom­ posed into the sum of three functions

tP(X) + ep(X) + X(X), where

.P'(x)

=

+(x) is absolutely continuous, ep(x) is continuous with 0 p. p. , x(x) = l: {J(f,. + 0) - J(f,. - 0)} summed over dis-

continuities

e,..

fr
CHAPTER V FURT H E R P R O PE R T I E S O F T H E I N T E G R A L This chapter contains theorems of a familiar type used in

working with integrals. It will be noted that the conditions under which they can be established for Lebesgue integrals are very wide.

5·1. Integradon by parts. If F(x), G(x) are respectively

indefinite integraL! of f(x), g(x), then

ibFgdx [FGt -ibfGdx. =

o:

G

o:

a

From Example 1 of Chapter IV the product of the absolutely

continuous functions F,

is absolutely continuous. From § 4·8,

G

to Fg+ JG almost everywhere. (Since F, are continuous, it is easy to see that Fg and fG are integrable. See example of § 3·7 . )

FG

is the integral of its derivative, which is equal

5·2. Change of variable. Let x

=

x(t) be a non-lkcrea&ing absolutely contin'UOU8 juMtion oft, with x( ex) = a, x(p) = b. Then, if f(x) is integrable,

fJ(x)

Let F(x)

=

J:f(x) rk

rk

=

f:J{x(t)}x'(t)dt.

. We start by proving t'w o lemmas.

If F(x), x(t) are absolutely contin'UO'U8 fumtion8 of x arul t re�Jpectively and x(t) is -monotonic, then F{x(t)} is an ab8olutely contin'UOU8 Junction of t. LEMMA 1.

59

C H A N G E O F VARIABLE

t,., t� is a typical interval of a set contained in ( {J) and x,. x(t,.), x; x(t;), then the absolute continuity of x(t) implies that, as �(t;-t,.) tends to zero, so does �(x;-x,.) . But � I F(x;)-F(x,.) f tends to zero �(x;-x,.) tends to zero, and this proves Lemma I . LEMMA 2. Let x( t) be a non-decreaBing absolutely con.tin'UO'U8 function. Let X be a set of valueB of x, and '1' the set of valuu of t for whick x x(t) i8 in X. 'l'ken mX 0 implies that x'(t) 0 p.p. in '1'. With the notation of Lemma I, since x(t) is absolutely con­ For, if =

ex,

=

as

=

=

=

tinuous,

Hence, if Oz, 01 are corresponding open sets of which typical intervals are respectively and

(x,., x;) (t,., t�), mOx J. x'(t)dt. =

o,

Let Oz,n be a decreasing sequence of open sets containing X such that lim mOz,n 0.

11-+CIO

=

Then the product of the corresponding sets 01•11, say '1'0, contains '1' and

f x'(t)dt � f. x'(t)dt mox.n for n no(€). Therefore J x'(t) dt Since x'(t) we have, from § 3·8 (ii), x'(t) p.p. in Po 'I'.

=

o,...

To

< €

>

= 0.

� 0, 0 and so p. p. in '1', and the lemma is proved. We now prove the main theorem. At almost all points of (cx, {J) a finite derivative Let 2i be the set of t for which =1= 0. We have, if =

x'(t) exists.

x'(t) x(t+h) =t=x(t), F{x(t +k)} -F{x(t)} - F{x(t+k)}-F{x(t)} x(t +k) -x(t) · x(t+k)-x(t) k k _

60

FURTHER PROPERTIES

As h tends to 0, the second quotient on the right-hand side tends to x'(t) for almost all values of t. The first quotient on the right-hand side has the limit f{x(t)} for all x except a set of measure zero, say X. By Lemma 2 the subset of � for which x(t) is in X has measure zero. Hence, p. p. in �'

�{z(t)} = j{z(t)}z'(t).

Suppose first that j(x) is bounded, say I J(x) I :E; K. Then I F{x(t + h)} - F{x(t)} I So z'(t) = 0 implies that

� K{x(t + h) - x(t)}.

1f'{z(t)} = 0.

We have thus shown that, if f(x) is bounded,

:t'{z
p.p. in («,p).

It follows from Lemma 1 and the theorem of § 4·8 that

F(b) - F(iJ) =

J:!{z(t)}z'(t)dt.

To establish the result for an unbounded f(x), it is sufficient to give the proof for J(x) positive. Let

and let

/n (x) = f(x) if f(x) � n, if f(x) > n, /n (x) = n F,.(z) = =

J:t,.(z)dz.

lim {F,l. (b) - Fn (a)}, n-+oo by the definition of the integral of an unbounde
F(b) - F(a)

lim fPfn{x(t)}x' (t)dt. ft-+OOJQI

61

M ULTIPLE INTE GRALS

As n increases and tends to infinity f8{x(t)}x'(t) incr

eases

f{x(t)}x'(t). f{x(t)}x'(t) F{x(t)},

and

tends to But is integrable in («,P)-for it is integrable over the set 2}, being p.p. the derivative of the absolutely continuous function and it vanishes p.p. over the set comple­ mentary to T1• By Beppo Levi's theorem § 3·10 (A), lim

ft-+11)

J.P!n{x(t)}x'(t) dt = J.pf{x(t)}x'(t) dt, �

�

and the left-hand side has been proved equal to F(b) - F(a). 5-3.

Multiple integrals. The integral

Jj(:r:) tk was de­

fined as the plane-measure of an ordinate-set in two dimensions. In the same way the value of

JJJ<:r:, y)d:r:tly is defined to be

the three-dimensional measure of the appropriate ordinate-set erected on the plane set and so on for higher dimensions. The mode of evaluation of a double integral at the elementary level (e.g. when is continuous) is by repeated integra­ tion with respect to each variable in turn. We shall give a general theorem of this kind for a Lebesgue multiple integral, which shows that the existence of the multiple integral carries with it without furlher condition the existence p.p. of the repeated integrals with respect to the separate variables. The method will be clear if we give the detail when the number of variables of integration is two.

E;

f(x, y)

Let E be a m.eaBUrable plane set. Let Ez be tke linear &et whick is tke section of E by tke ordinate distant x frcnn tke origin. Then Ez is m.eaBUrable for almost all x and p.E = JmE,.Ik, tke limit8 of i.tegration being 8Uda. a8 to include the whole of LEMMA.

E.

62

FURTHER PROPERTIES

The lemma is plainly true if E is a rectangle, or a finite number of rectangles. We shall prove it for an open set 0. By the network construction (§ 1 ·5), 0 = lim G,, where G,cG,+1 and a. consists of a finite number of rectangles. Also Oz = lim (G3)z•

ft.-+IX) Then p.O = lim p.G3 (from the definition of p.O) �IX)

=

�Jm(G,.).,� = JmO.,tk, by Beppo Levi's theorem, §3·10(A).

Thus the lemma holds for open sets and similarly (or by complements) for closed sets. If now E is any �easurable set, take a decreasing sequence 01,02, of open sets containing E such that p,O,-+p.E; and an increasing sequence Q1 , Q1, of closed sets contained in E such that p.Qn�p.E. • • •

• • •

Then, for each

x,

(On)z =» Ez => (Qn)z,

and

Therefore

Since the integrand is a positive decreasing function of n, we have from § 3·9 (ii) and § 3·8 (ii),

lim {m(03).x - m(Q,)z} = 0 p.p.

ft-+CX)

J

Therefore E., is measurable for almost all x and mE.,tk, lying between two sequences of integrals-each tending to p.E, is equal to p.E. The lemma is proved.

F U B I N I ' S THBORBM

63

We could prove similarly a three-dimensional form of the lemma, and this is what will be needed in dealing with double integrals. If E is a measurable set in three dimensions and Ex is the plane set which is its section by a. plane perpendicular to the x-axis, distant z from the origin, then the measure of E is

J tk.

equal to ,.E.,

We are now ready to state Fubini's theorem on multiple integrals and in doing so we refer to the discussion of § 3·6. The domain of integration may be any measurable bounded set in the plane, or it may be the whole plane provided that the integral is absolutely convergent. On this understanding the most convenient notation is to leave the ranges of integration not explicitly specified.

If Jff(x,y)tkdy e:rist8, then Jf(:r:,y) dy e:rist8for almost all x, i8 an integrablefunction of:r: and Jff(:r:.y)tkdy ftkJf(:r:.y)dy. Similarly, Jff(:r:,y)tkdy Ja11Jf(:r:,y)tk. For the proof, suppose first that f is positive. The three­ dimensional form of the lemma just stated, applied to the ordinate-set of f, gives the result. If I takes values of either sign, we have only to write f f+ -f_, where f+ and f_ are positive, and apply the lemma to the ordinate-sets of I+ and f_. S-4.

Fubioi's theorem.

=

=

=

The operation involving multiple integrals which one most often wishes to justify in practice is the interchange of the order of integration in a repeated integral. The condition of validity of Fubini's theorem which necessitates a direct investi­ gation of the multiple integral may not be easy to establish. The following simple variants are analogous to § 3·1 0 (B) and (F) for integration of series.

•

64

FURTHER PROPERTIES

If f(x, y) is a rable function of (x, y) and iff� 0 for all (x, y) in the domain of integration,, then tke existence of any of m,eaau

one

implies the existence and eq'U4lity of all three. follows: Suppose that fd f!dy exists. Define /,. (x,y) a: In = / if l x l � n, I Y I � n and f� n, In = n if I x I � n, I y I � n and f> n, fn = 0 if I x I > n or I y I > n. Then /,. ill bounded urable function and ffj,. tkdy exists and is equal to either of the repeated integrals of fn · From the hypothesis, fJdy exists for almost all x. fJ,.dy inc with n and tends to fJdy. 88

a.

meas

reases

By § 3·10 (A), But the left-hand side ill

ffJ,.tkdy and its limit,

88 n-+oo,

Jfftkdy. Hence ff!tkdy = fd ffdy, and the rest follows. a: Ifj(x,y) i8 a mbk juncti&n of (x,y) and if ftkfi fl dy ea:iat8, then ftkf!d71 = fayfJtk. From the last result, ffil l tkdy exists, therefore so does ff!tkdy and the conclusion follows from Fubini's theorem.

ill

meJUU

THB

S·S.

CLASS V

65

Dift"erendation of multiple integrals. This is an

operation of theoretical interest rather than one which occurs in day-to-day analysis, and we only mention the main result (for double integrals). 8 is

If a Bq'U4re of side h, containing (x, y), then p.p. · • lim � J f f(x, y) dxdy f(x, y). A-+oh Js =

A proof can be based on Vitali's covering theorem § 4·3 (note Corollary 2).

f(x)

is in the Lebesgue 5·6. The class LP. \Ve say that class LP (where > 0) for a given set of values of if is

p x f(x) measurable and l f(x) I P is integrable in the set. For example: (1) x-l is in P, for p 2, in (0, 1). (2) In a finite interval (a, b), a function in liP is also in Lq for 0 � q p; a bounded function is in J;P for every p. in (0, co), but not (3) The function x-l(I + f log x f )-1 is in in LP in (0, oo) for any value of p other than 2. The most interesting case is p � 1 and we shall assume this. <

<

L•

The integrals will be supposed to be taken over a given set E of finite or infinite measure. is defined by If > 1, the index

conjugate p' p. 1 1 . 1 I.e. p, = p + p p' 1 The classes LP, LP' will be called conjugate; L2 is self-conjugate. We define N ( f) , the of j, by P P Np(f) (ti / IP tkr . HoLDER's INEQUALITY. If, in E, f is in V and g ia in LP', then with equality only when A I J IP B I g IPP' p.p. for A , B not both zero (or, as we m.ay 8ay, IJ I and I g IP' are effectively proportional). p

=

_

norm =

=

scnne con8lant8

66

FURTHER PROPERT IES

Write

( j (P

=

-p1 = <X, p1

cf>,

,

Then we have to prove that

=

fl

.

By the inequality of the arithmetic and geometric means, if 0, 0, 1),

a> b >

a« bP �a<X+ bfJ, (« + fl = with equality only when a = b. In the last inequality write

4> a = -, t.,s and integrate over E.

cx +fJ

The right·hand side is integrable and its value is I. The left·hand side, being measurable, is integrable and the result follows. MINKOWSKI'S INEQUALITY.

=

Iff and g are in LP, then

with equality only if f and g are effectively proportional. For fi!+u I" � fi!l-lf+ul"-1 + fig 1-lf+g 1"-1 ' /p ll/p p �� /p' } } } { { { { 1 � 1/ � 1 � � J lf lP J IJ+g I" + J I g fP J l f+g (P by Holder's inequality. ' 1/p } Dividing each side by { �l f+ g ( P we have the result. J ,

67

T H E M E TRIC SPACE LP

5·7. The metric

space LP. The members of a set are said

to be the elements of a metric space if, for every pair x, y, there is defined a distance-function d(x, y) with the properties:

d(x, y) > 0

(i) (ii) (iii)

if x =1= y;

d(x, y)

=

d(x, x) = 0.

d(y, x).

d(x, z) � d(x, y) + d(y, z)-the triangle inequality.

Functions of JiP are elements of a metric space if we take

d(f, g) = Np (f-g).

Two functions differing only in a set of measure zero are indistinguishable as elements of the metric space. With this convention, the properties (i), (ii), (iii) of the distance function are satisfied, (iii) being Minkowski's inequality. The reader will recognize in the following discussion an extension to the space of functions in JiP of ideas such as limit­ point in the theory of sets of points. and NP (fn - /)-+0 as n-KO, we say that If/, and f are in J,�f(LP) or alternatively that f. converge8 strongly to f (with index A nece88ary and sufficient C01Ulition that fn�f(LP) is that NP (f - /,.. �0 as m and n tend to infinity. Two such limit

LP,

p). m

functi0'1UJ f ran differ only in a set of measure zero. Moreover, there i& a sub-sequence n,. suck that /,r�f We first prove sufficiency. Given E, there is n0(E) such that for m � n0, n � n0, /,. I P lkt: < Ep+l ,

p.p.

film 1

and so the set in which /m -/, I > e has measure less than E. Replace E by e/2, . . . , ef2k, . . . successively and let n1, , nk, . . . be the indices corresponding to the n0 of the last paragraph. Then • • •

FURTHER PROPERTIES

68

except in a set of measure at most E/2"-1 . Since the measure of the exceptional set tends to 0 as r-+eo, it follows that the series

� {!.J:+,(x)- /n1(x)} ,

k-o

is absolutely convergent p.p., that is to say, there is a function / ) defined p. p. to which the suh·sequence J ) converges as

. (x

(x

,

r�.

We shall prove that /, converges strongly to J with index p. By Fatou's lemma (§ 3·9),

i.e.

Jl!-f,. i�'dx �!,.�{11.., -In i�' dxfO �t>+l

if n ;> no,

J,�J(LP).

To prove the uniqueness of the limit function (ignoring differences in sets of measure zero), suppose that /.,. also con· verged to g(LP). By Minkowski's inequality, Np(/-

g)� Np(/-/,) + Np(/, - g),

and the right·hand side tends to zero as 1H-OO The necessity part of the theorem also follows at once from Minkowski's inequality. For •

�

Np(/m - /.,.) Np(/m -

.

f) + Np(/-/,).

E X A M PL E S O N CHAPTER V

( 1) Investigate the question of existence and equality of the double and repeated integrals of the following functions over the square 0 � � 1 , 0 � y � 1 ,�a: ' xtt+-ytti ' (i) (ii) (x y ) ( 1 -xy (iii) /(x,y) (x-1 !)* for 0
I.

=

=

=

69

EXAMPLES

j(x) is integrable, then /,. (x) I';ccJ:(x-t)«-lf(t)dt (« > 0) exists p.p. and is integrable. (The function /« (x) is the �tk integral of /(x). The definition (2) Prove that, if =

agrees with the ordinary usage when and is valid when � is not an integer.) is the fJtk integral of (3) If

fa.fJ (x)

f«.fJ (x) f«+fJ(x), =

ex

is a positive integer,

j« (x), prove that

(� > 0, fJ > 0)

if the right-hand side exists. (4:) If are integrable, the function

J,g

r(x) Jf(x -y)g(y)dy is called the re8'Ultant off and g (sometimes convolution, German Faltung). Taking two cases, (i) f and g periodic, and the integrals taken over a period, (ii) integrals over (- oo, oo), prove that r(x) is integrable, and that J1 r(x) I dx � Jlf(x) I dxJ1 g(x) I dx. (li) If /,.-+/(LP), then Jl /,. 1"-+Jill"· (6) If F(x) is the integral of a function of LP, then, as h-+0, F(x + h) - F(x) = o( 1 h 11/P'). (7) Prove that, if f is in LP for all p, then, as p�, NP (/)-+max. ! /I, =

where max. l /1 is the 'essential upper bound' of 1/1, i.e. the smallest number for which 1/f p.p. This gives an interpretation of p 1, p' = oo as conjugates, where LP' is the class of functions which are bounded (except in a set of measure zero).

M

�M =

CHAPTBR VI

THE L E B E S G UE - STIELTJES INTEGRAL 6·1. Integradon with respect to a funcdon. The idea of

replacing the varia.ble of .integra.tion

x in

a.n

J!(x)th

integra.l

q,(x) of the variable is due to Stieltjes ( 1894). The simplest definition of a. Stieltjes integra.! J:f(x)d
a.a

limit of approximative sums (1)

x,.

x,._1

x,..

� �r � where the are points of subdivision of (a, b) and The existence of the integral can be proved if suitable assump­ tions are made about the functions of which the most natural are that is continuous and monotonic (or of bounded variation). As an illustration let be a function which is discontinuous at isolated points with + 0) and 0) whose graph consists otherwise of horizontal stretches. It is

f, �' �(x)

f(x)

q,(x) x Xn,

q,(xn- kn, easy to see tha.t the va.lue of the integra.! Jfd· One of the =

q,(xn

=

advantages of the Stieltjes integral is that it thus includes as special cases series of discrete terms. It is often �onvenient to use a physical mode of speech. An increasing function determines a 'distribution of mass' along the x·axis, which can include continuous distributions and also masses concentrated at certain points (see Ch. IV, Ex. 6 for the detailed analysis of

�(x)

�(x)).

THB VARIATION OF A .F U N C TION

71

In this chapter we shall set up a 'Lebesgue·Stieltjes' integral. Our first task is to develop a theory of variation of
I I,

I)

vi = tfo(b - 0) - t/J(a + O). If 0 is an open set, vO is defined to be !.vi summed over the intervals of 0. With this beginning, a theory of variation can be set up orl the model of Chapter II. In the first place, variation over a closed set is defined 'by complements '. We observe that, for a closed interval,

I

v(�, [a, b]) = �(b + 0) -
v(t/J, x) =
• • •

72

THE LEBBSGUB-STIELTJBS I N T E G RAL

discll88ion of a general plane mass-distribution, containing as it may masses concentrated at points and along curves involves complications of detail. The concept of measurability (B) (see p. 23) is here of some importance. In general, a measurable set may or may not be measurable for a particular . A set which is measurable (B) is however measurable for every . For open and closed sets are measurable (), and so are sets obtainable from them by elementary operations including that of taking a limit (§ 2·9). A function l(x) is said to be measurable () if the set of values of x for which l(x) > A is measurable () for every A . From the last paragraph, a function which is measurable (B) is measurable with respect to every increasing . 6·3.

The Lebesgue-Stieltjes integral. We assume increasing, and, for the present, I positive. It is useful to have both a geometrical representation of the integral by means of ordinate-sets and an analytical expression 88 the limit of approximative sums. We start with the former. Let f = (x). Then to an interval (a, b) of Ox corresponds an interval (cx, /J) of oe. The correspondence between X and e is one-one, except that (i) if (x) is constant in an interval (c, d), the same value of f corresponds to every value of X in (c, d), (ii) if (x) has a discontinuity at x0, we agree that every value of f between (x0- 0) and (x0 + 0) shall correspond to X0• The function X = +-l(f), or X = x(f) say, inverse to e = (x) is uniquely defined except for the values of e specified in (i) which form at most an enumerable set. Throughout this chap­ ter we shall reserve the notatibn x = x( f) for the inverse of e = (x) . To a given function l(x) corresponds a functidn of f, l{x(f)}. If we wish to make the definition of this function unique we may agree that, for values of e of type (i), 1 shall be equal to the lower bound of the aggregate of values of the corresponding f(x); for purposes of integration, however, indetermiMcy in an

73

OR D I N ATE ·SBTS

enumerable set will not matter. We observe that a discontinuity of tf>(x) corresponds to an interval of constancy of x(e) and so of l{x( �)}. We can now adopt as the definition of the Lebesgue· Stieltjes

(LB) integral

J:f{x(f)}de,

ffdt/>

the value of the Lebesgue integral

tha.t is to s"a.y, the measure of the ordina.te-set off

having as its base the set of values of e, i.e. of t/>(x). In the same wa.y

J/t¥

is defined to mean

JJ<xW}de

where I is the set of f corresponding to the set E of x. · If I takes both positive and negative values and I = f+ - 1 , as in § 3·1, then fdt/> is defined to be f+dt/> - f

J

�

� I!_ _, _ _, __ -.,.. __ .,... --r--. r-

J

J -dtl>·

-- - - - ---

t =-;(x) I I I I I I I I I

f(x)

0

a

b

X

Fig. 2.

Text4fig. 2 illustrates (in the first quadrant) an increasing function e = tf>(x) having a horizontal stretch and a discontin­ uity. The graph of it is 'projected' on to the g axis, and ordinates of a function f(x) (shown as positive and increasing) are set up on the projection. The shaded area represents

t¥· i f

•

74

THE L E B E S G U E· STIELTJES INTEGRAL

The altemative method of definition of the integral of a bounded is as the common limit of approximative sums

f

8

=

8 =

ft.

L l,.+1 e,., 1

ft.

L l,. e,.,

1

where e, is the variation of q, over the set E,. for which l,. �f(x) < l,.+1 (cf. § 3·4). Extension to unbounded is made as in § 3·5. If + is a function of bounded variation, then q, q,1- �� where 1 and 2 are increasing functions (they are taken to be the positive and negative variations of � ). We then define

f

=

Ifdl/> I11lfh - I1dll>l· =

If the integrals on the right exist, so does this is appropriately written

I11 dl/> 1.

Iftllh Ifd.h and +

It is easy to adapt the arguments of Chapter III to yield theorems about the LS integral. A set of x of measure zero is to be replaced by a set over which the total variation of is zero. A property which holds except possibly in a set over which � has zero variation is said to hold p.p. (�). We give two illustrations of useful results derived from those of Chapter III. THEOREM OF DOMINATED OONVBRGENCE.

If, for

aU

n,

I.p I dl/> I exi818, and /,.-+/ p.p. (4>), then IJdl/> limII diJ>. The theorem of bounded convergence is a special case of this. INTEGRATION OF SERIES. If either I<:E I u,.l) ldl/> I or :EI I u,. I I dl/> I la finite, then I<:E u,.) d!/> :EI•,.dll>·

I /,.(z) I � ,P(x), wkt:te

=

..

=

I N T E GRATION BY PARTS

E:t:amplu

75

( 1) Establish the equivalence of the 'geometrical' and the 'approximative sum' definitions of the LB integral. ( 2) Prove that

Ifds = Ijd(
A �J � B, then

E) <; IId4> .;; B tl(f., E). (4) If 1 / I E; M, then Ijd <; MI I d4> I · E A tl(f>,

6·4. Integration by parts. This theorem for Stieltjes inte­ grals takes the elegant form

(I)

I:Jd + rf>d/

=

/(b) f>(b) -j(a) f>(a).

We shall investigate under what hypotheses on f and � this holds. If the integrals are defined in the original 8ert8e of BtWtjes (§ 6· 1 ( 1)), then, if either integral on the left·hand 8ide exist!, so doe8 the other and the farmula is true.

I

Suppose that fd exist-a. Let a = x0 � �1 � x1 � . . . � X3_1 � �. � x. = b be any dissection of (a, b). Define €0 = a, €a+I = b. Then if • � � f ) {/(xr ) -f(x,._1)}, '1' 1 ( ,. =

and

we have identically,

T+� = /(b) �(b) -f(a) �(a).

76

THE L E B E SG U E - S TIELTJES INTEGRAL

Observe that if either max (x,. - Xr-d Ol' m.a.x (€,. - �,_1) tends to 0, so does the other.

[!d4>

Since

exists, it is the limit of

T1

as max (z, - z,._1)

J

0 . Therefore T tends to a limit, and so df exists and relation (I) holds.

tends to

We now link this up with the LS integral.

bounded

If j,tfi have variation and there is no value of x fur which they are both diacontinluouB, the:n, if a, b are points of con­ tinuity off, cp, the formula (I) holiJB. (If a, b are discontinuities of f or q,, the right-hand side of ( 1 ) is to be replaced by

f(b - 0) t/J(b - 0)-j(a + 0)
or

according as the interval is open or closed.) The general case can be built up from monotonic functions f, tfi, and it will be sufficient to give the detail for increasing J, ;. Since

f, tfi

have no common discontinuities, given take a subdivision

(x,_1, x,)

so fine that in any

�,

we can

at least one of /(x,.) -f(x,._1) and

f/J(x,) - f/J(x,_1) is less than £. The Stieltjes sum �/(f,){tfi(x,) - f/J(x,_1)} lies between U u

and

=

�/(x,.){f/J(x,) - f/J(x,_1)}

�/(x,._1){t/J(x,.) - f/J(x,_1)},

U - u < E{t/J(b) - f/J(a) +/(6) -/(a)}.

where Moreover

==

U

decreases and

u

�

increases (in the wide sense)

when further·points of subdivision are inserted. So by the argu­ ment of § 3·4, U and u have a common limit as max (xt' - x,_1) tends to 0, and the Stieltjes integral exists.

C H A N G E OF VARIABLE

77

There is a pair of approximative sums for the

Jfd.l>

lying between

U

LS

integral

and u, a.nd the result follows.

The theorem has a simple geometrical interpretation. The reader should draw a diagram taking, say, I and


both to be

positive and increasing. Then, with values off and ' measured along two perpendicular axes, the ordinate-sets defining the

integrals fill up the part of the plane between the two rectangles whose opposite vertices are the origin and the points {l(a),
{j(b),
Example Develop this geometrical interpretation into a proof of the theorem.

6·5. Change of variable. Second mean-value theorem.

J:

If t/>(x) is of bounded variation and lll(x) = gd.f>, then

J:fdlll J:fgd.f>. =

We may suppose that + is increasing and that I and

g are

of constant sign (say both positive). Write �

==

tions be


and E

=

(f)(x),

and let their inverse func­

respectively. Then, by definitions of the integrals as ordinate·sets, we have

Jg{xCe)}dg, Jfdlll = Jf{X(S: } dE, JJgd.f> Jf{xCe>l g{xCe>l de, :e: =

)

=

the integrals being taken between the appropriate limits.

78

THE LEB ESOUE·STIELTJES I N T E G RAL

The result will now follow from the theorem of change of variable for Lebesgue integrals (§ 5·2), if we satisfy ourselves that the integrals are unaffected by the many-valuedness of the inverse functions x and X. We find that may be un­ defined in an enumerable set of �, and f in an enumerable set of E, corresponding to intervals of constancy of q,(x); J may also be undefined in an interval in which (so that S: is constant); both the integrals vanish over such an interval. We now prove the second mean-value theorem (for the L integral). Iff

g

g=0

is mmwtonic and g integrable, then f:Jgdz =/(a)f�gdz +J(b)L6glk for some � satisfying a�;� b, where a, b are values for which J is co-ntinuous. (H a,b discontinuities ofj, J(a) and f(b) on the right-hand side are to be replaced by J(a + O) and j(b-0).) Let G(z) = J:glk. Then the L8 integral J!dG exists and we have by the last ...heorem, are

J:Jglk = J:!dG = [!GJ:-J:odj, by integration by parts, =/(b) G(b) -G(�){/(b)-/(a)}, by the first mean-value theorem (§ 6·3, Ex. 3), and this is equal to

\

\

EXAMPLES

79

The brief account that we have given of Stieltjes integrals should enable the reader to manipulate them with confidence. Differential properties (depending on the notion of differ· entiating with respect to a function tfo) do not often come into question. EXAMPLES O N CHAPTER VI

(I) If tfo"'�' state sufficient conditions for

£!# �J:fd.f>. ..

(2) Construct a Stieltjes-Fubini theorem (§ 5•5). (3) Prove that

� (n )J.1xm(I - x)"'-m�(x). J.l�X(x) n=m m 0

==

0

SOLUTIONS O F SOME EXAMPLES Hints a.re given for the solutions of all but the easiest examples, and more detailed solutions of those which are most important. CHAPTER

I

P. 9, Ex. 6. Let E11 be the set of x such that /(f)
where the a 's are either 0 or 2 (not 1). There is a one·one correspondence between these numbers and

·b1 b1 b3

• . •

b, . . .

i n the scale of 2, defined by b, 0 or 1 according as a,1 These include all the numbers of (0, 1 ) . =

CHAPTER

=

0 or 2.

II

P. 25, Ex. I. The fraction of the interval (0, 1) left after the

nth opera.tion is (2/3)"'. P. 25, Ex. 2. Consider rational values of y. P. 25, Ex. 3. We construct first a. step-function t/J(x) (i.e .P is constant in each of a finite number of interval which together make up (a, b)), which satisfies the inequalities postu­ lated for
81

SOLUTIONS

Define .;,. = l,. in 8,. and 0 elsewhere. Then .p

=

function which differs from f by less than "

"

� .p,. is a step-

1

€

except in

Eg + � (e; + e;) and this set has measure less than €.

1

Any step-function can be modified into a continuous function differing from it in a set of arbitrarily small measure. For, if e is a ty.pical point of discontinuity of .p, with ifi(f- 0) = k and ifi(e + 0) = l, define � in the interval ( e -1'], e + 1])-where 1'J is arbitrarily small-as the linear segment joining (f - TJ, It) to ( f + TJ, l); outside intervals ( f -,.,, f+'J1) , keep ifJ = .p. si n Further conditions could be imposed on 4>, e.g. the posseso of a derivative of any order; we could satisfy them by 'rounding off the comers' of the ifJ already constructed. P. 25, Ex. 4. Write f/-/,. I · Let €,. be a decreasing sequence tending to 0. Let E., ,. be the subset of E in which g, < E,. for v � 11.. Then E,., ,.cEn+l, ,. and lim E.... ,. = E (since,

g,.

=

given r, every point of E belongs to some E,.,,.). Hence, given r, we can find n(r) such that 8 ft-+GO

m(E - E,.,c,.,, r) < 2""

Write E0 En
g.�

g,. <

€r

00

m(E - E0) � � m (E - En(r), r) • r-1 CHAPTER III

§ 3·5, p. 33, Ex. 3.

J(X) == 2xSID. xt - X2 COS xtl l

(x =F 0),

/(0) = 0.

The first term of /(x) is continuous and integrable. For the second term, let /" be the interval of x defined by 1T

1

1T

2nm - < xz < 2n7T + s · 3

82

SOLUTIONS

n yn .

I I In 1,, cos 1 � -and

A

Its length > -,-

Jflo

x

2

f

1•

2 ll

A

- cos 1 dx > - . If

x

x

n.

I;Jf . 1.

I � cos t dx existed, it would be > X X 1 § 3·7, p. 37. ( 1 ) g bounded or (2) r, g integrable. If (1), then fg is measurable and, if I g I � K, integrability of fg follows 1 from that of K l f l · If (2), use 2 1fg l �r+g • Ex. 2. !, (x) = nxe-nz' in (0, I). Ex. 3. Let H be a subset, with mH > mE - 8, in which j,_,.j uniformly. Then there is n.0(8) such that I f-fn I < 8 in H for n > n.0•

P. 42, P. 42,

fE-H JH

f.it-f I = t + IB-H• ..

� {upper bound of I !-!, l } m(E - H) < 2m, � II mE for n > no.

Jl !-/,. l-+0 as �. P. 42, Ex. 4. (1 - :r < and-+ e..... Use theorem of domin­ Hence

e-<"

ated convergence. (Observe how the use of the Lebesgue integral avoids the tiresome inequality work needed to discuss uniformity of convergence.) 43, Ex. 9. Take an enumerable sequence �
P.

=

Define 4>, (x) M;n) in 8�,, for all r of �
=

<1>, (x}-+M(x).

Cl>,(x) is measurable. Let bounded convergence.

�

and apply the theorem of

83

SOLUTIONS

P. 43, Ex. 10. M (x) = m(x) p.p. P. 43, Ex. 1 1 . See Ch. II, Ex. 3. P. 43, Ex. 12. Use Ex. I I.

CHAPTER IV

P. 57, Ex.

I. Let f(x), g(x) be absolutely continuous in

and l: (x,. ,x,. + h,.)

a.

set of disjoint intervals in

(a, b).

(a, b)

l: l f(x,. + h,.)g(x,. + h,.) -f(x,.) g(x,.) I

= l: l f(x,. + h,.) {g(x,. + h,.) - g(x,.)} + g(x,.) {f(x, + h,.) - f(x,.) } I

� M l: I g(x,. + h,.) - g(x,.) I + N l: I f(x,. + h,.) - f(x,.) I , where

M, N

are the upper bounds of I f I , I g I in last two sums tend to zero with l:h,..

(a, b).

These

P. 57, Ex. 2. I = /(I) -/(0) = �1 f1f+l: zfl f, where �1 is ta.ken

over the intervals of lengths

�, :1 ,

• • •

,

:

k

specified in the con·

struction and �z over the intervals complementary to them.

= 0 and so �� = 1 . But the intervals of �� have total length (2/3)k and this tends to 0 as proof (using Vitali's theorem ) . Suppose P. 57, Ex. 3. that there is a. sub-set E1 of E with mE1 > 0 in which E has not density I . Then there i s an a < 1, such that the set E1 at points m( ) of which lim !H < a:, where H is one of the intervals By definition �1

First

(x - k, x) , (x, x + h) , has mEz = e1 (say) > 0. Given E, enclose E1 in 0, mO < e1 + E.

k-+00.

is a. set I( = 11 + . . . + /") of dis· joint intervals with ml > ez - E in each of which m(l,E) < a.ml,.. The sub·set of Es in 8 has measure < a.ml; the sub·set of Ez in 0 - I has measure < 2£. Hence e1 < (X(e1 + E) + 2£, which is false if E is small enough. 8ecun,d proof. This, due to Lebesgue, starts from scratch By Vitali's theorem, there

and oonta.ins its own 'covering' argument.

84

SOLUTIONS

LEMMA. Let 0 be an open aet. The set of left-hand points of intervals in which 0 1148 average de?UJity greater than 1/K 1148 meaBure KmO, where K > 1. Let (a1, 61), , (a", b*') ' . . . be the intervals of 0. Extend the interval (a1, b1) to the left by taking (�1, b1) such that • • •

b1 - �� = K (b1 - a1). Do the same to (a1, b1), giving (a1, b1). If (�1, b1), (a1, b1) have an interval in common, move this to the left so as to give an interval (�, p.) containing (a1, 61) and (cxt, b1) of length equal to (b1 - a1) + (b1- a1). We have thus either one or two intervals, say (A1 , B1), (At, B1). Carry out the same construction with (as, b3) giving (as, b3). If (a3, b3) has an interval in common with either (A1, B1) or (At, B1), carry it to the left ..as before. We then have either

one, two or three intervals. This construction leads to an enumerable set n of non­ overlapping intervals (�1, p.1 ) , , (�, of total measure KmO. The set such that, for any (�, p.) inside (�, p.,.), the measure of the set of points of ( �' not in 0 is at least (K- 1 ) times the measure of the set common to 0 and (�, p.). There­ fore, in an interval (�, of which the left-hand end-point does not belong to n, the average density of 0 is at at most 1/K. To prove the density theorem, we first show that an open set 0 has zero density p.p. in CO. Take p intervals of 0, say ()P, with miJP > mO - �. Given K, construct the set n of the lemma for the set 0- 0P. Then

0 is

• . •

p.)

p.)

p.*')'

. • .

�op + P> � mfJP + Km(O - Op )

� mO + (K - l)E. All the points of CO at which the right-hand density of 0 is

greater than 1/K lie in a set of measure at most (K - 1)E. Since E is arbitrary they form a set of measure zero. Let K--+«.>; the right-hand density of 0 zero p.p. in CO. Similarly so is the left-hand density. Let now E be any measurable set; we prove that its density is zero p.p. in CE. Enclose E in 0 with m(O - E) < E. The set

is

85

SOLUTIONS

of points of CO in which E has not zero density is contained in the set in which 0 has not zero density, i.e. has measure zero. Since £ is arbitrary, E has density zero p.p. in CE. Interchange the roles of E and CE. CE has density zero p.p. in E and so E has density 1 p.p. in E. P. 57, Ex. 4. Let I l l < K. Let E,,., = set where u <:.j
F'

I

- Km(l . CE,.. • ) + um(l . E.,.,) �

=

L1

� Km(l . CE,, .,) + vm(J . E,,.,). Divide by ml and let ml�o. Using the density theorem

ff� v. ml 1 P. 57, Ex. 5. Let I �'(x) I < K. Then

we have u � lim

-

1

J

�(�+h) - f>(x) h

=

l �'(x+ 8h) I < K.

Let h take a sequence of values tending to 0. . h)1- f>(x) -f>(x - +-·--� ,(x) boundedly. Argue as tn § 4·7.

P. 57, Ex. 6. Provef1 = /- x is (1) continuous, (2) increasing. Then � =

J!i· CHAPTER V

P. 68, Ex.

I.

(i) Repeated integrals unequal, (ii) double

integral exists if .. < 2,

P. 69, Ex. 2.

J.af(Y.(x)dx 0

JdyJu exists; JuJdy does not. J.a J.x r<�> dx (x - t)«-l f(t)dt.

(iii)

=

1

0

0

86

SOLUTIONS

By Fubini, this

� J:dtJ,"(x - t)"-1/(t)tk I J.a I (a- t)« dt = o: /(t)

=r �

=r (o:) 0

0:

(a), r ( + 1 ) /a+l

since the latter repeated integral exists when the integrand is replaced by its modulus.

P. 69, Ex. 3.

!..,, (z) = I'(�:run J:(z - t)ll-l dt£
In the inner integral put t - u = v(%- u) and it becomes

P.

J.

1 r (o:) r (p) . (x - u)«+P-1 ( 1 - v).8-l v«-l dv = (x - u)a+P-l r(o:+ fJ) o 69, Ex. 4.

Jr<x) tk = Ju(y) dyJf(x - y) tk = Ju(y)dyJf(x) tk, J I r(z) I J1 g(y) l dyJl f(z) l tk, tk �

P.

the inversions of the repeated integrals being justified by Fubini. 69, Ex. 5. Np (/) - Np (/-/,) � Np (/, ) � Np (/) + Np (/- /n).

�+,.

P. 69, Ex. 6. Apply Holder to rz f(t) dt.

J (f) � M is clear.

f J f > M - E in E1 where 1 mE1 > 8, and so Np(/) � 8 '" (M - E) and this �M -:- E as p-+«J. P. 69, Ex. 7. Np

CHAPTER VI

§ 6·4, p. 77.

Consider the ordinate-set defining

J:!d

(diagram of § 6·3); their tops are f = cp(x), TJ = f(x). The function

,

'

I

'

� 1J l

87

S O LUTIO N S

1

=

/{x(f)} is a continuous function of e except

q,,

(I) for e

correa­

ponding to an interval of constancy of or (2) for e �where is discontinuous. or 2) there is a linear segment defined For each f of type by 71<e - o> � , � 11<� + o). Adjoining these segments to the set of points forming tops of o•dinates, we have a continuous curve (in general with hori­ zontal and vertical stretches). This same curve is obtained

f(x)

q,(x),

(I) (

fro� the tops of ordinates of J:
P. 79,

=

q,")

I. f

Ex. continuous, total variation of (q, in (a, b.) tends to as �oo or obtain a set of conditions by inte­ . grating by parts.

P. 79,

Ex. 2.

0

Jl(x)If(x, y) dofo(y) = Id./1(1/)If(x, y) d(x),

provided that one of

II d(x) I Il f(x,y) I I dofo(!l) I I

if \finite. From § 5·4. ' �. 79, Ex. 3. The terms of i: (n)xm(I-x)"-m are positive : .,._ m ' l

I

ip 1(0,

,

j.

l

I) and the sum is I/x.

m

REJ'ERENCES J'OB §2·7

.

LITTLEWOOD, Element& of the Theory

Publications, 1954).

of Real Functions (Do,·er

RoSSER, Logic for Mathematicians (McGraw-Hill, 1953). )v. SrERPINSKI, .Le9ons sur les nombres transfinis (Gauthier-Villars, t 1928). . (

, . a.

,

..