Real Variables

Uk1k. 4.19 Let A be an algebra of subsets of X. Then the u-algebra SeA) generated by A is the smallest family F of subsets of X that contains A and satisfies the following two conditions: (i) If En E F and En ~ En+l for n = 1,2, ... then U~=l En E F, and, (ii) If En E F and En ~ En+l for n = 1,2, ... , then n~=l En E F. Families that satisfy (i) and (ii) are said to be monotone. Prove that if the algebra A is a monotone family, then A is au-algebra.

4.

Problems and Questions

61

4.20 If Al is a a-algebra of subsets of Xl and A2 is a a-algebra of subsets of X2, show that Al x A2 is the smallest monotone family of subsets of Xl x X2 that contains all finite disjoint unions of "rectangles" El X E2 with El E Al and E2 E A 2.

x Bm = Bn+m. 4.22 Prove that S(El ) x S(E2 ) = S(EI

4.21 Prove that Bn

X

E2 ).

4.23 Let A be a a-algebra of subsets of X and E eX. If Al = A u {E}, show that SeAl) consists of those subsets of X of the form (AI n E) u (A2 n (X \ E)), where A}, A2 EA. 4.24 Let I be an extended real-valued function defined on X, and let J.L be the measure on (X, P(X)) given by J.L(E) L-XEE I(x). Find necessary and sufficient conditions, in terms of I, for J.L to be finite or a-finite.

=

4.25 Let (X,M,J.L) be a measure space, and let {Ek} ~ M. Show that if J.L (U E k ) < 00, and J.L(E k ) ~ 'fJ > 0 for infinitely many k's, then J.L (lim sup Ek) > O. By means of an example show that the condition J.L (U Ek) < 00 cannot be removed. 4.26 Let (X,M,J.L) be a measure space, and {Ek} ~ M. Show that J.L (lim inf Ek) ~ lim inf J.L (Ek), and that, provided that J.L (U Ek) < 00, lim sup J.L (Ek) ~ J.L(limsupEk). By means of examples show that we may have strict inequalities above. 4.27 Let (X,M,J.L) be a probability measure space. If J.L(lim sup An) = 1 and J.L(liminf Bn) = 1, prove that J.L(limsup(An n Bn)) = 1. What happens if we assume instead that J.L(lim sup Bn) = I? 4.28 Let J.L be a measure on (R, Bt) with the property that J.L( I) < every finite interval I, y E R, and put

J.L«y,xD

Fy(x)

={ 0

-J.L«x, yD

00

for

if x> y if x = y if x < y.

Show that Fy is a nondecreasing right-continuous function; Fy is called a distribution function induced by J.L. 4.29 Let (X,M,J.L) be a measure space, T be a mapping of X onto Y and set N = {A ~ Y: T-l(A) EM}. Furthermore, let v be the set function defined on N by yeA) = J.L(T-l(A». Prove that N is a a-algebra of subsets of Y and that (Y,N, v) is a measure space.

IV.

62

Abstract Measures

4.30 Let J.Ll,J.L2 be measures on (X,M), J.L2(X) < 00, and suppose that J.Ll(E) ~ J.L2(E) for all E E M. Show that there exists a (unique) measure J.L3 on (X, M) such that

J.Ll(E) = J.Li E ) + J.L3(E) , all E EM. Is the restriction J.L2(X) < 00 necessary? 4.31 Let {J.Lk} be a sequence of measures on (X,M) with the property that J.Lk( E) ~ J.Lk+1 (E) for all E EM. Is the set function J.L defined on (X,M) by J.L( E) = lim J.Lk( E) , k-+oo

necessarily a measure? 4.32 A useful concept in measure theory is that of semifiniteness, a notion weaker than that of a-finiteness. A measure J.L on (X, M) is said to be semifinite if for each A E M with J.L(A) = 00, there is a measurable set E C A, with 0 < J.L(E) < 00. Show that every 0'finite measure is semifinite, as is the counting measure, and give an example of a measure that is not semifinite. Also prove that if J.L is semifinite and M does not contain an uncountable, pairwise disjoint collection of measurable sets of positive measure, then J.L is a-finite. Show by means of an example that this does not hold without the assumption that there are no "infinite atoms." 4.33 Let J.L be a semifinite measure defined on (X, M). Show that every measurable set with infinite measure contains a measurable subset with arbitrarily large, finite, measure. 4.34 Let J.L be a measure on (X,M). Show that J.L can be written as the sum J.L(A) = J.Ll(A) + J.L2(A), A E M, where J.Ll is a semifinite measure and the measure J.L2 only assumes the values 0 and 00; the decomposition need not be unique. 4.35 Assume that (X, M, J.L) is a measure space, and for E}, E2 E M put

d (El, E2) = J.L(E1 b. E2). Discuss under what conditions endowed with this distance, (M, d) becomes a complete metric space. 4.36 Assume that (Xk,Mk,J.Lk) are measure spaces for k = 1,2, ... , and that the Xk'S are pairwise disjoint. Put X = Uk Xk, and let M be the class of subsets A of X of the form A = Uk Ak, where Ak E Mk for all k. Show that M is a a-algebra of subsets of X and that the set function J.L given on M by J.L(A) = Lk J.Lk(A k ) is a measure on (X, M). When is J.L finite? a-finite?

CHAPTER

v

The Lebesgue Measure

In this chapter we introduce the most important example of a measure on Rn, the Lebesgue measure. I learned this construction from A. Zygmund. 1.

LEGESGUE MEASURE ON Rn

In defining a measure on Rn we must first decide on the a-algebra of measurable sets; in the case at hand geometric considerations are also of importance. We call a closed parallelepiped, i.e., a closed bounded set of the form {(XI, ... xn): ak ~ Xk ~ bb 1 ~ k ~ n}, a closed interval. An open interval is the interior of a closed interval, i.e., a set of the form {(Xl, ... , Xn) : ak < Xk < bk , 1 ~ k ~ n}. Intervals I, open and closed, have volume vel) equal to IIk=1 (bk - ak). We require that the a-algebra of Lebesgue measurable sets contain all open and closed intervals and that the Lebesgue measure agree with the volume for intervals. Since each open set in Rn is the countable union of nonoverlap ping closed intervals, Lebesgue sets include all open sets, and consequently, also all closed and Borel sets. It is not intuitively clear what the Lebesgue measure of these general sets is; however, we expect the Lebesgue measure to satisfy two additional properties: It should be complete and translation invariant, i.e., if A is Lebesgue measurable and Ay = y + A = {y + x: x E A}, then Ay is also measurable and A and Ay have the same measure. Translation invariance plays an essential role in the determination of the nature of the a-algebra of Lebesgue measurable sets. In the early 1900's Vitali showed that if we accept the Axiom of choice, then not all subsets of R are Lebesgue measurable. In the early 1970's Solovay proved

63

v.

64

The Lebesgue Measure

that it is consistent with the usual axioms of Set Theory, excluding the Axiom of choice, for every subset of R to be Lebesgue measurable. One cannot conclude from Solovay's result, however, that the Axiom of choice and the existence of a subset of R which is not Lebesgue measurable are equivalent. Since the construction of a subset of R which is not Lebesgue measurable depends on general, rather than specific, properties of the measure, including translation invariance, we describe such a set at this point. Let 1= [-1/2,1/2] and let", be the relation defined for numbers X,Y in I by x '" y iff x - y is a rational number (between -1 and l)i it is not hard to check that '" is an equivalence relation. Let 'R, be the collection of all the distinct equivalence classes of "'i clearly 'R, can be indexed by a subset of I. Now, by the Axiom of choice, there is a set A C I which contains exactly one element from each equivalence class of "'i it is the set A we propose to show is not Lebesgue measurable. Suppose, to the contrary, that A is Lebesgue measurable, enumerate the rational numbers in [-l,l],ro = O,Tt, .•• , and let Ak = Tk + A,k 2: 0, be the rational translates of A. Since we assume that A is Lebesgue measurable, the Ak'S are also Lebesgue measurable and they all have the same measure. Note that translates corresponding to different Tk'S are disjoint: Indeed, iffor j i- k there is y E Aj n Ak, then there are x, x' E A such that y = x + Tj = x' + Tk. Now, since Tj i- Tk, then also x i- x', and x and x' correspond to two distinct equivalence classes of "'. But we also have that x - x' = Tk - Tj is a rational number (between -1 and 1), and consequently, x '" x'. However, since A contains one element from each equivalence class this cannot happen, and the Ak'S are pairwise disjoint. Next observe that I ~

U

Ak

~ [-3/2,3/2].

(1.1)

k

Suppose we have shown (1.1) to be true. Then the left-hand side inclusion implies that the Lebesgue measure of A cannot be 0, for otherwise Uk Ak would be a null set which contains a subset of measure 1, and the righthand side inclusion implies that the Lebesgue measure of A cannot be positive, for otherwise Uk Ak would be a measurable set of infinite measure contained in a set of measure 3. In short, (1.1) implies that A cannot be Le besgue measurable. To prove that (1.1) holds is not hard. For x E I, let x, be the representative of the equivalence class of x which belongs to A. Then there is a rational number Tk in [-1,1], such that x = x, + Tk. Whence x E Ab and the left-hand side inclusion in (1.1) holds. Moreover, since

1.

Lebesgue Measure on R n

65

each Ak ~ [-3/2,3/2], the same is true of the union, and the right-hand side inclusion in (1.1) also holds. Now that we are satisfied that not every subset of Rn can be Lebesgue measurable, how do we go about constructing the u-algebra of Lebesgue measurable sets? The idea is to introduce the Lebesgue measure as we go along. First we define a nonnegative u-subadditive set function on P(Rn ), the Lebesgue outer measure, and then select as Lebesgue measurable sets those subsets of R n which can be approximated, in a sense made precise by the outer measure, by open sets. Since intervals are the only sets we know how to measure at this point, it is natural that the definition of Lebesgue outer measure involve coverings with intervals. Given a subset A of R n , we define the Lebesgue outer measure IAle of A as the quantity

(1.2) where the infimum in (1.2) is taken over the family of all at most countable coverings of A by closed intervals. Now, a countable mesh of nonoverlapping unit intervals (actually n-dimensional cubes of sidelength one) covers Rn, and consequently it also covers every subset of Rn. Thus the inf in (1.2) is a well-defined quantity. By (1.2) it is clear that if A ~ B, then IAle ::; IBle (monotonicity). Moreover, as expected, the outer measure of an interval coincides with its volume. Proposition 1.1.

Let I be a closed interval in Rn. Then

IIle =

v(I).

Proof. Since I is a finite covering of itself, IIle ::; v(I) < 00. To prove the opposite inequality note that since IIle < 00, given c > 0, there is a family of closed intervals {Ik} such that I ~ UIk, and 2: v(Ik) ::; IIle + c. Furthermore, by extending the sidelengths of the Ik's and then deleting the edges, it is readily seen that given", > 0, there are open intervals I~ d Ik, such that v(ID ::; (1 + ",)V(Ik), all k. Since I ~ UI~, and since I is compact and the Irs are open, by the Heine-Borel theorem there is a finite subcovering, which for simplicity we also denote by {In, such that I ~ U:=1 I k. In this case it is intuitively clear, although involved to prove, that v(I) ::; 2::=1 v(I~), and consequently we also have v(I) ::; (1 + ",)

Lk

N =1

v(Ik) ::; (1

+ ",) L

v(Ik) ::; (1 + ",)(IIle

+ c).

(1.3)

Since c and", are both arbitrary, from (1.3) it readily follows that v(l) ::; IIle, and we have finished. •

66

V.

The Lebesgue Measure

Along the same lines it is not hard to see that the following result holds: Let {Ik}k=l be a finite c,ollection of nonoverlapping closed intervals in Rn, then

Corollary 1.2.

Let I be an open interval. Then Ille = vel).

Proof. Let I denote the closure of I, then we have Ille ~ v(1) = v(l), and one inequality holds. Also, if J is any closed interval contained in I, monotonicity implies that v(J) = IJI ~ Ille. But since v(I) = supv(J), where the sup is taken over the family of the closed subintervals J of I, we get that vel) ~ Ille, and the opposite inequality holds. • Next we show that the outer measure is q-subadditive. Proposition 1.3. Then

Let {Ek}k:l be any sequence of subsets of Rn. (1.4)

Proof. If IEkie = 00 for some k, then by monotonicity also the union has infinite outer measure, and we have equality in (1.4). Otherwise, suppose that IEkie < 00 for all k, and let c > 0 be arbitrary. For each k let {lk,j} be a covering of Ek by closed intervals with the property that

(1.5) Clearly

(1.6) and consequently, the closed intervals on the right-hand side of (1.6) form a covering of Uk Ek' Whence, by definition, we have

(1.7) and since the summands on the right-hand side of (1.7) are nonnegative we may interchange freely the order of summation and estimate that expression by

1.

Lebesgue Measure on R"

But since

E

67

> 0 is arbitrary, we also have

I Uk Ekle

~ L:k

IEkle. •

In contrast to the case of measures, it is interesting to point out that strict inequality may occur in (1.4), even ifthe Ek'S are pairwise disjoint. To see this observe that the Lebesgue outer measure is translation invariant, and that with the notation of (1.1) above, IAle > O. Then, again by (1.1), IUk Akl e ~ 3 < L:k IAkie = 00. Since open sets can be expressed as the union of closed intervals, it is reasonable to attempt to compute the outer measure of subsets of R n in terms of the outer measure of open sets. Specifically, we have Proposition 1.4. IE Ie

Let E be any subset of Rn. Then

= inf {IOle: 0

(1.8)

is open, and 0 ;2 E} .

Proof. If IEle = 00, by monotonicity every open set which contains E (and this class is nonempty since Rn is one such set) also has infinite outer measure and (1.8) holds. On the other hand, if IEle is finite, given E > 0, let {lk}k:l be a covering of E by closed intervals such that

Furthermore, for each k, let lfe be an open interval containing lk such that v(lfe) ~ v(lk) + E/2 k+1, and put 0 = Uklfe. By construction 0 is open and it contains E. Moreover, by Proposition 1.3 and Corollary 1.2 it readily follows that

IEle ~ IOle ~ ~

Lk Ilfele = Lk v(lfe)

Lk (v(lk)

+ E/2k+l)

= Lk v(lk)

+ E/2 ~ IEle + E.

Thus, for any E > 0, there is an open set 0 ;2 E such that IEle + E, and (1.8) holds. •

(1.9)

IEle ~ IOle

~

Proposition 1.4 is important in applications; to state some we need a definition. We say that a subset H of Rn is a G6 set if H is the intersection of an at most countable family of open sets. The complement of a G6 set is an Fu set, i.e.,.an at most countable union of closed sets. Corollary 1.5. Let E be an arbitrary subset of Rn. Then there is a G6 set H which contains E and such that IHle = IEle.

v.

68

The Lebesgue Measure

Proof. If IEle = 00, put H = Rn. If IEle < 00, by (1.9) there is a sequence {Ok}k:l of open sets containing E such that 10Ie ~ IEle + l/k. Let now H k Ok; by construction H is a Gs set, and E ~ H ~ Ok for all k. Thus by monotonicity it follows that

=n

(1.10)

Since k is arbitrary we get that IEle = IHle.

•

A closer look at inequality (1.10) for sets E with finite outer measure indicates that the following property is true: Given E > 0, there is an open set 0 ~ E such that 10Ie - IEle < E. This estimate does not make sense when IEle = 00, but a closely related result holds for any subset of Rn, to wit, there exists an open set 0 ~ E such that 10Ie ~ IEle+IO\Ele. This inequality is a simple consequence of the monotonicity and it hints that rather to seek to controIIOle-IEle, which is meaningless when IEle = 00, we may try to control 10 \ Ele. This control is, indeed, all that is needed. We say that E ~ Rn is Lebesgue measurable if for any E > 0, there exists an open set 0 ~ E such that

10 \ Ele <

E.

The class of Lebesgue measurable sets is denoted by Cn, or plainly by C. Notice that open sets, as well as sets with outer measure equal to 0, belong to C. In the case of open sets this is obvious, and for sets E of outer measure observe that by (1.10) there are open sets Ok ~ E with lOki ~ l/k for all k. Then, given E > 0, let k ~ l/E, and note that

°

10k \ Ele ~ 10kie ~ l/k ~

E.

In order to show that C is a O'-algebra we must verify that C is closed under countable unions, which is easy, and under complementation, which requires some work. We begin with the easier part. Proposition 1.6. UkEk E C.

Let {Ek}k:l be a sequence of subsets in C, then

Proof. For a given E > 0, we must find an open set 0 ~ Uk Ek so that 10 \ Uk Ekle < E. Since each Ek belongs to C, there are open sets Ok ~ Ek such that 10k \ Ekle ~ E/2 k for all k. Hence, 0 = Uk Ok is an open set which contains Uk Ek, and since as is readily seen (Uk Ok) \ (Uk Ek) ~ Uk (Ok \ E k ), by Proposition 1.3 we get that

10 \ UkEkle ~ Lk 10 k \ Ekle ~ Lk E/2 k = E.

1.

Lebesgue Measure on R n

Thus, the union of the Ek'S belongs to C.

69

•

To complete the verification that C is a u-algebra, we begin by showing that closed sets are indeed Lebesgue measurable. This requires some preliminary results. Lemma 1.7.

Let E 1 , E2 be subsets of Rn with the property that

Then, lEI U E21e = IEtie compact and disjoint.

+ IE2Ie. In particular, this is

true if Et, E2 are

Proof. If either lEt Ie or IE21e is infinite, then the same is true of lEt uE2le, and we are done. Now, if both are finite, since the outer measure is subadditive, it suffices to show that IEtie + IE21e ~ lEt U E21e. But in this case we also have that lEt UE21e < 00, and consequently, given £ > 0,

there is a covering {Ik} of E t U E2 consisting of closed intervals such that

There are three relevant kinds of Ik'S, to wit: (i) Those h's such that h nEt -::j:. 0, h n E2 = 0, call them Il's; (ii) Those Ik'S such that Ik nEt = 0, Ik n E2 -::j:. 0, call them I~'s; (iii) Those Ik'S which intersect both E t and E 2 • The intervals in the third class above may be subdivided into nonoverlapping closed subintervals of diameter less than or equal to d(Et, E 2 ). Each subinterval thus obtained either belongs to the first family (i), or to the second family (ii), or it does not intersect E t U E2 and it can be discarded. Therefore, we divide the Ik'S into a covering of E t , a covering of E 2 , and throwaway the rest. By definition we have

which implies, since

lEI U E21e.

£

is arbitrary, that, as asserted, IEti e + IE21e ~

•

We are now ready to prove Theorem 1.8.

Closed subsets of Rn are Lebesgue measurable.

V.

70

The Lebesgue Measure

Proof. Suppose first that the closed set F in question is bounded, and hence compact. Then, given c > 0, by inequality (1.10) there is an open set 0 :J F such that 10Ie :5 IFle + cj we would like to show that 10 \ Fie :5 c as well. Now, 0 \ F is also open, and consequently it can be expressed as the countable union of nonoverlapping closed intervals, Uk Ik, say. By Proposition 1.3, it follows that 10 \ Fie :5 ~k v(h). On the other hand, since

by monotonicity we get that

Furthermore, since F and Uf=l Ik are both compact and disjoint and the Ik's are nonoverlap ping, by Lemma 1.7 it readily follows that

In particular

But since this inequality holds with a bound independent of N, we have ~k v(Ik) :5 c, and consequently also 10 \ Fie :5 c. Thus, in this case, F is Lebesgue measurable. For general closed subsets F of Rn, let Fk = {x E F: Ix I :5 k}, and observe that each Fk is closed and bounded, and that F = Uk Fk. By the above argument each Fk is measurable, and by Proposition 1.6 so is their union, F. • Theorem 1.9.

Suppose E E C, then R n

\

E E C.

Proof. Let Ok :J E be a family of open sets such that 10k \ Ele :5 11k, k = 1,2, ... j each Rn \ Ok is closed, and hence measurable. furthermore, since Rn \ Ok ~ Rn \ E for all k, H = Uk(Rn \ Ok) is an Fq measurable subset of Rn \ E. Let A = (Rn \ E) \ Hj since Rn \ E = H U A, we will be done once we check that A is measurable. To do this we show that IAle = o. Indeed, since for each k, A ~ (Rn \ E) \ (Rn \ Ok) = Ok \E, it readily follows that

IAle :5 10 k \ Ele :5 11k,

all k.

Lebesgue Measure on R n

1.

Thus

IAle =

0, and we are done.

71

•

Theorem 1.9 completes the verification that C is a u-algebra and, at the same time, it provides a description of the Lebesgue measurable sets. Indeed, the argument of Theorem 1.9 applied to the (measurable) complement R n \ E of a measurable set E gives that E = R n \ (Rn \ E) = HUN, where H is an FO' set and INle = 0. It also gives that the Lebesgue measurable sets are precisely those subsets E of R n which satisfy the following property: Given c > 0, there is a closed set F ~ E such that IE\ Fie ~ c. Next we construct the Lebesgue measure on (Rn, C)j it is the restriction of the Lebesgue outer measure to C. More precisely, we have Theorem 1.10. The set function I . Ie restricted to C is a measure. We call this measure the Lebesgue measure on Rn and denote it by 1·1. Proof. The proof amounts to showing that the set function I . Ie is u-additive on C. For this purpose, let {Ek}~l be a sequence of pairwise disjoint measure sets, and let E denote its union. Note that, by Proposition 1.3, lEI ~ L:k IEkl. As for the opposite inequality, assume first that the Ek'S are bounded. Given c > 0, let Fk ~ Ek be a sequence of closed sets such that IFk \ Ekl ~ c/2 k for all k. Since Ek = Fk U (Ek \ Fk) we also have IEkl ~ IFkl + c/2 k . Furthermore, since the Ek'S are pairwise disjoint, the sequence of Fk'S is composed of pairwise disjoint compact subsets of Rn. Fix N, and note that by (a simple extension of) Lemma 1.7, I Uf=l Fkl = L:f=lI Fkl. Thus, since Uf=l Fk ~ E for all N, it follows that L:f=l IFkl ~ lEI, all N, and consequently also L:k:llFkl ~ lEI. Whence 00

00

k=l

k=l

L IEkl ~ L (IFkl + c/2k) ~ lEI + c,

and, since c is arbitrary, we are done in this case. In the general case, fix an increasing sequence {Ij} of bounded intervals so that Uj Ij = Rn, 10 = 0, and put Sj = Ij \Ij-bi = 1,2, ... Then, the sets Ek,j = Ek n Sj are measurable, pairwise disjoint and bounded, and for each k we have Uj Ek,j = Ek. Thus, on the one hand,

IUj,kEk,j1 In other words,

IUj,k Ek,jl

=

I Uk Ekl, and, on the other hand,

= Lk,jIEk,jl = LkLjlEk,jl = LkiEkl.

1·1 is u-additive on C, and we have finished.

•

V.

72

The Lebesgue Measure

The following characterization of C, due to CaratModory (1873-1950), highlights the interplay between the Lebesgue measurable sets and the Lebesgue measure, and it is very interesting since it can be used to define C, and more general a-algebras of sets, d. 3.40 below. Theorem 1.11 (CaratModory). A subset E of Rn is Lebesgue measurable iff for every subset A of Rn we have

IAle = IA n Ele + IA \

Ele .

(1.11)

Proof. We begin by assuming that E is measurable and A is any subset of Rn. Since A = (A n E) U (A \ E), by the sub additivity of the Lebesgue outer measure it follows that

IAle

~ IA U Ele

+ IA \ Ele .

To prove the opposite inequality, note that by Corollary 1.5 there is a Gs measurable set H ~ A such that IAle = IHI. Now, since H is also measurable, we have IHI = IH n EI + IH \ EI. Whence, by monotonicity

and (1.11) holds. Next assume that (1.11) is true for every subset A of Rnj we distinguish the cases IEle < 00 and IEle = 00. In the former case, by Corollary 1.5 there is a Gs set H ~ E such that IHI = IEle. By (1.11) we have

IHI = IH n Ele + IH \ Ele = IEle + IH \ Ele.

(1.12)

(1.12) gives at once that H \ E is a measurable set of measure 0, and that E = H \ (H \ E) is also measurable. As for the latter case, let Ek = {x E E: Ixl ~ k} be so that IEkl < 00, and let Hk be a Gs set containing Ek such that IHkl = IEkie for all k. By (1.11), with A = Hk there, we get

Since IHkl = IEkle, we have IHk \ Ele = 0 for all k. Whence, setting H = Uk Hk, it readily follows that H is a measurable set which contains E, and that H \ E = Uk(Hk \ E) is a measurable set of measure o. Thus, E = H \ (H \ E) is also measurable. •

2.

73

The Cantor Set

2. THE CANTOR SET

°

It is easy to see that there are uncountable sets of measure in Rn, n ~ 2; indeed, the boundary of any interval is such a set. How about R? The Cantor set is such an example, and we construct it next. Consider the closed interval Co = [0,1]. The first stage of the construction is to trisect Co and to remove the interior of the middle interval, (1/3,2/3). Each successive step is essentially the same. Let C 1 = [1,1/3] U [2/3,1]; C 1 is the union of 21 = 2 closed disjoint intervals. At the second stage we subdivide each of the closed intervals of C1 into thirds and remove from each one the middle open thirds, (1/9,2/9) and (7/9,8/9). Suppose that Cn has been constructed and that it consists of 2n closed disjoint intervals, each of length 3- n • Subdivide each of the closed intervals of C n into thirds and remove from each one of them the interior of the middle intervals. What is left from C n is C n+!; note that C n+! is the union of 2n+1 closed intervals, each of length 3-(n+1). The Cantor set C is now defined as C = n~=o Cn. Some of the elementary properties of C are the following: It is closed, it contains the endpoints of all intervals in Cn, and any point of C is the limit of a nondecreasing (and a nonincreasing) sequence of endpoints of the intervals of the Cn's. It is not hard to give an analytical description of the elements of C. Let x = L:~=1 a n 3- n be the tryadic expansion of an arbitrary x E C. We observe that since x ~ (1/3,2/3),a1 ~ 1; similarly, since x ~ (1/9,2/9) U (7 /9,8/9),a2 ~ 1, and so on. In other words, by induction we see that an ~ 1 for all n, and C consists precisely of those points with an = 0,2 in their tryadic expansion. For example, the number 1/4 = L:~=1 2· 3- 2n is in C, but is not an endpoint of any of the intervals of the Cn's.

o o

1/3

1/9

2/9

2/3

2/3

1/3

-

1

7/9

8/9 I

Figure 2

1

...... \

v.

74

The Lebesgue Measure

As for the cardinality of C, we have Proposition 2.1.

C is uncountable.

Proof. The idea is to show that C '" 2N ,2 = {O,l}. If (xn) E 2N , let Yn = 2x n , and put f((xn)) = E~=l Yn3-n. Since Yn -::j:. 1 for all n, f maps 2N into Cj we want to show that f is one-to-one and onto. Suppose that (xn) -::j:. (x~), and let m = min{n: Xn -::j:. x~}j we may assume that Xm = 0 and x~ = 1. Since 2 E~=m+1 3- n = 3- m, it follows that 00

m-l

f((x~)) = 2 Lx~3-n ;::: 2 L n=l

x n3- n + 2· 3- m

n=l

00

> 2 L xn3- n = f((xn)) ' n=l

and

f is one-to-one.

Since given x onto. •

= E~=l an3-n in C,

we have f((a n /2))

= x,

f is also

Is C measurable, and if so, what is its measure? Since C is covered by the intervals in any C n we have ICle ::; 2n3- n for all n, and consequently, ICI = o. Thus C is an example of an uncountable set of measure 0 in the line.

3. PROBLEMS AND QUESTIONS 3.1 Suppose A, B are not Lebesgue measurable, is the same true of AUB? 3.2 Suppose IAle = 0 and show that for every subset of B of R n we have IB U Ale = IB \ Ale = IBle . 3.3 Let A, B

~

Rn. Show that

3.4 Suppose {Ek} is a nondecreasing sequence of subsets of R n and let E = Uk Ek. Is it true that limk-+oo IEkie = IE Ie ? 3.5 Does the notion of outer measure change if we replace the coverings by intervals by coverings with balls? How about parallelepipeds with a fixed orientation?

3.

Problems and Questions

3.6 Show that if

75

L: IEkie < 00, then

3.7 Suppose A,B and IAI = IBI·

~ Rn ,

Ilim sup Ekle

= O.

A E £, and IA ~ Ble = O. Show that BE£'

3.8 Assume {Ek} is a sequence of pairwise disjoint Lebesgue measurable sets and let A be any set. Is it true that

IA n (U

OO

n=I

Ek)

I= e

",00

L....tk=I

I An Ek Ie ?

3.9 Consider the transformation ¢>( x) = 71X + 6 from R into itself, where 71 f; 0 and 6 are real numbers. Show that: (i) For any set E, 1¢>(E)le = 171IIEle. (ii) E is Lebesgue measurable iff ¢>(E) is Lebesgue measurable, and in this case I¢>(E)I = 171IIEI. Can you think of extensions of this result to R n ? 3.10 A mapping ¢> from R into itself is said to be an isometry if for any x,x'in R we have I¢>(x) - ¢>(x')1 = Ix - x'i. Show that if ¢> is an isometry and E E £', then ¢>(E) E £, and I¢>(E)I = lEI.

3.11 Assume that INI = 0 and show that {x 3 : x E N} is a null Lebesgue set. 3.12 Suppose IEle < 00 and show that E E £, iff for any E > 0, we can write E = (A U AI) \ A 2 , where A is the union of a finite collection of nonoverlap ping intervals and IAIle, IA21e < E. 3.13 Is the set of irrational numbers in the line a Gs set?

3.14 Show that E E £, iff E = H \ N, where H is a Gs set and INI = O. 3.15 Does there exist a function f: R --+ [0,1] such that the set D of its discontinuities has IDI = 0 and D n I is uncountable for every interval I of R?

3.16 Assume A is a Lebesgue measurable subset of R of finite measure and put ¢>(x) = IA n (-00, xli. Show that ¢> is continuous at each x of R.

3.17 Let A be a Lebesgue measurable subset of R and let 0 < 71 < IAI. Show that there exists a Lebesgue measurable set B so that B ~ A and IBI = 71. 3.18 Given E > 0, show that there exists a dense open subset 0 of [0,1] with 101 < E so that its boundary 80 satisfies 1801 ~ 1 - E.

3.19 Let A = {x E [0,1]: x = .aIa2 ... , an f; 7, all n}. Prove that IAI = O. Generalize this result to different configurations of an's and to dyadic, tryadic expansions.

V.

76

The Lebesgue Measure

3.20 Let A = {x E [0,1]:x = .a1a2 ... ,an = 2 or 3, all n}. Show that A is measurable and compute IAI. 3.21 Let A = {x E R: there exist infinitely many pairs of integers p, q such that Ix - p/ql ~ 1/q3}. Show that IAI = o.

=

3.22 Suppose II, ... ,In are open intervals in R, so that if Q1 Q n [0,1] denotes the rational numbers in [0,1], then Q1 c Uj=l Ij. Prove that 2:j=l IIj I ~ 1. Is the conclusion true if the Ii's are measurable sets rather than intervals? What if we allow the collection of intervals to be infinite rather than finite? 3.23 Let

T1, T2, .•.

be an enumeration of Q. Show that

On the other hand, also show that

may, or may not, be empty. 3.24 Suppose E is bounded measurable subset of R, there exist Xl, X2 E E so that Xl - X2 E Q. 3.25 Show that if B is a Hamel basis for R, then

lEI> o.

Prove that

IBI = O.

3.26 Construct Lebesgue null subsets B 1 , B2 of R such that

3.27 Construct a Cantor-type subset C n of [0,1] by removing at the nth stage a "middle" interval of length (1 - 7] )3- n , 0 < 7] < 1. Show that C n enjoys all the properties of C, but it has Lebesgue measure

ICnl

= 7].

3.28 If -1

~ T ~

1, show there exist x, y E C such that y -

X

= T.

3.29 Does the Cantor set contain a Hamel basis for R? 3.30 Construct a Cantor-like subset of [0,1] which consists entirely of irrational numbers. 3.31 Let (an) be a fixed decreasing sequence of real numbers such that ao = 1, and 0 < 2a n < an-I, and define the sequence (d n ) by dn = a n-1 - 2a n , n ~ 1.

3.

Problems and Questions

77

= [1 - at,l], h,2 = [al - a2,al], Ia,3 = [al - a2,al - a2 + a3], 13,4 = [al - a3,aI], and so on; this definition can be made precise by induction. Now put Now let 10,1 = [0,1], h,l = [O,al],

Fn =

U

2"

k=l

In k '

and

h2

P =

n°on=l Fn.

Show that P E £ and IPI = lim n ..... oo 2n a n • Moreover, if 0 ~ "I < 1, the an's can be chosen so that IPI = "I. Also, if Tn = an-l - an, then the elements of P are precisely those real numbers of the form 2:k=l ckTk, ck = 0 or 1. 3.32 Show that the dinality 2 c • Consider now say that El rv relation on £ cardinali ty c.

class of Lebesgue measurable subsets of R n has carthe following relation on £: Given EI, E2 in £, we E2 if lEI l:::. E21 = o. Show that rv is an equivalence and that the family of the equivalence classes has

3.33 Prove that there is no Lebesgue measurable subset A of R such that alII ~ IA n II ~ bill for all open intervals I of the line. More precisely, prove the following two assertions: (a) If IA n II ~ bill for all open intervals I C R and b < 1, then IAI = 0, and, (b) If alII ~ IA n II for all open intervals I C R and a > 0, then IAI = 1. 3.34 Prove there exists a Lebesgue measurable set E C R such that

o < IE n II

< III, all bounded intervals I CR.

3.35 Does there exist a measurable subset E of R such that

o < IE n II ,

0

< II \ EI , all intervals I

C R?

3.36 A measurable subset A of R is said to have a well-defined density, if the limit D(A) = lim IA n (-A,A)I ~ ..... oo

2A

exists. In this case D( A) is called the density of A. Give an example of a measurable set whose density is defined, and one whose density is not defined. Further, prove that if Al and A2 are disjoint and have well-defined density, then Al U A2 also has a well-defined density, and D(AI U A 2) = D(Ad + D(A 2). 3.37 Prove that the Lebesgue measure enjoys the following property, known as regularity: Given a measurable set A we have

IAI = sup{IKI : K is compact, and K

~

A} .

V.

78

The Lebesgue Measure

3.38 It is difficult to approximate sets that are not Lebesgue measurable with measurable ones. Specifically, suppose E C R n is not Lebesgue measurable. Show there is .,., > 0 such that if E ~ A and R n \E ~ B, and if A and B are Lebesgue measurable, then IA n BI ~ .,.,. 3.39 Decide whether the following statement is true: A ~ Rn is Lebesgue measurable iff for every open subset G of Rn we have

IGI = IG n Ale + IG \ Ale. 3.40 Suppose p.* is a nonnegative u-subadditive monotone set function defined for all the subsets of a set X such that p.*(0) = O. We say that E ~ X is measurable with respect to p.* if for every subset A ~ X we have p.*(A) = p.*(A

n E) + p.*(A \

E).

Let M be the class of subsets of X which are measurable with respect to p.*. Show that M is a u-algebra of subsets of X and that the restriction of p.* to M defines a measure on (X, M). This construction is known as the Caratheodory extension of an outer measure.

CHAPTER

VI

Measurable Functions

In this chapter we introduce the class of measurable functions, for which the integral will be defined, and discuss some of its basic properties.

1. ELEMENTARY PROPERTIES

OF MEASURABLE FUNCTIONS Let M be a u-algebra of (measurable) subsets of X and suppose f is an extended real-valued function defined on X; by this we mean that, in addition to real values, f may also assume the values ±oo. We say that f is measurable if for any real number~, {x EX: f(x) > ~} = {f > ~} E M; that is to say, all the level sets of f are measurable. For instance, for any M, f = XA is measurable iff A EM. If M = {0,X}, only constant functions are measurable, and if M = P(X), all functions are measurable. We begin by exploring some simple properties of measurable functions. Proposition 1.1. Suppose M is a u-algebra of subsets of X and let f be an extended real-valued function defined on X. Then, the following statements are equivalent: (i) f is measurable. (ii) For any real ~,{f ~ ~} E M. (iii) For any real ~,{f < ~} E M. (iv) For any real ~,{f =:; ~} E M. Proof. (i) implies (ii). Fix~, and for n ~ 1 let An = {f > ~-l/n}; by assumption An E M, all n. Now, since {f ~ ~} is the intersection of the An's, it also belongs to M, and (ii) holds.

VI.

80

Measurable Functions

(ii) implies (iii). {f < A} = X \ {f ~ A}. (iii) implies (iv). {f ~ A} = n~=l {f < A + lin}. (iv) implies (i). {f > A} = X \ {f ~ A}. • In working with measurable functions it is essential to know whether certain sets are measurable. Since these sets are readily obtained from those introduced in Proposition 1.1 we merely indicate how their measurability is established.

{f

= co} = n~=l {f > n},

{f < co} = U~=l {f < n},

= -co} = n~=l {f < -n}.

(1.1)

{f> -co} = U~=l {f > -n}.

(1.2)

{f

{ -co < f < co} = {f > -co}

n {f < co} .

(1.3)

Also, for any real numbers A, T/, we will have the occasion to deal with the measurable sets

P < f < co} =

{f > A} n {f < co} , {-co < f < A} = {f < A} n {f > -co}.

(1.4)

P < f < J.L} = {f > A} n {f < J.L}, P < f ~ J.L} = {f > A} n {f ~ J.L}. P ~ f < J.L} = {f ~ A} n {f < J.L},

(1.5)

P~f~J.L}={f~A}n{f~J.L}.

(1.6)

{f = A} = {f ~ A} n {f ~ A}, {f t A} = X \ {f = A}.

(1.7)

Our next result indicates how to handle the infinite values of a measurable function. Proposition 1.2. Let M be a u-algebra of subsets of X and let f be an extended real-valued function defined on X. Then, f is measurable iff {f = -co} EM and for each real A, {A < f < co} E M. Proof. The necessity has been established in (1.1) and (1.4). As for the sufficiency, first observe that

{f < co}

= {f = -co} u (U:=-oo {n < f

< co})

1.

Properties of Measurable Functions

81

belongs to M by assumption. Whence {I in M, and since for each real A we have

{I> A} the level sets of

I

= oo} = X \

{I < oo} is also

= {A < 1< oo} U {I = oo} EM,

are measurable and

I

is measurable.

•

In fact, a more general statement is true.

Proposition 1.3. Let M be a cr-algebra of subsets of X and suppose is an extended real-valued function defined on X. Then, I is measurable iff {I = -oo} E M and for each open subset 0 ~ R, 1-1(0) E M.

I

Proof. Since for each real A, (A, 00) is open, the sufficiency follows from Proposition 1.2. As for the necessity, suppose 0 is an open subset of R and write 0 = Uh, where the Ik's are an at most countable collection of pairwise disjoint open intervals, one or two of which are possibly unbounded. By 5.8 in Chapter I

1-1(0) =

U1-1(Ik).

(1.8)

k

Now, by (1.3), (1.4) and (1.5), the sets in the union on the right-hand side of (1.8) all belong to M and 1-1(0) is measurable. Since {I = -oo} E M whenever I is measurable, we have finished. • So far the role of measures on (X, M) is not apparent, but in dealing with measurable functions sets of measure 0 are important and the following concept essential. Given a measure space (X, M,J.L), we say that a property P(x) is true J.L-almost everywhere on a measurable subset E of X, and denote this by J.L-a.e. on E, if 1'( {x E E: P( x) is not true}) = o. For instance, we say that a measurable function I is finite J.L-a.e. on E if 1'( {x E E : I( x) = ±oo}) = o. It is natural to expect that measurable functions that coincide J.L-a.e. on X be, in some sense, equivalent. A more precise statement is

Theorem 1.4. Let I' be a complete measure on (X,M), and let I,g be extended real-valued functions defined on X. If I is measurable and 9 I J.L-a.e., then 9 is also measurable and

=

1'( {g > A}) = 1'( {f > A}) ,

all real A .

(1.9)

VI.

82

Measurable Functions

=

Proof. Let N {g "I J}; by assumption N is a null, measurable, set. Now, for each real A we have

{g > A} u N = {f > A} UN.

(1.10)

Since f is measurable, the set on the right-hand side of (1.10) is measurable, and so is the set on the left-hand side there. Moreover, since J.l is complete and No = {x E N :g(x) ~ A} ~ N, then No is also a null, measurable set and consequently,

{g> A}

= ({g > A} U N) \ No

(1.11)

is also measurable. Next observe that since N is null, we have J.l( {f > A} U N) = J.l( {f > A}) for all real A. Whence, by (1.11) and (an argument similar to) 3.2 in Chapter V, we get

J.l( {g > A})

= J.l( {g > A} UN) -

J.l(No) = J.l( {f > A} U N) = J.l( {f > A}). •

Theorem 1.4 states that functions that coincide J.l-a.e. are roughly interchangeable; this property is essential in operating with extended realvalued functions. Consider, for instance, addition: f( x) +g( x) is undefined for those x's where f and 9 assume infinite values of opposite sign. The idea is to work with functions j and 9 which are closely related to f and g, and for which the sum makes sense. We proceed as follows: Let the (bad) set

B = {x E X:f(x) = oo,g(x) = -oo}U{x E X:f(x) = -oo,g(x) = co}. Since B is measurable, MX\B = {E n (X \ B): E E M} is a u-algebra of subsets of X \ B. Observe that j = fl(X \ B) and 9 = gl(X \ B) are also measurable on (X \ B,Mx\B), and that lex) + g(x) is defined for any x E X \ B. In fact, as we shall prove below, j + 9 is also measurable. To avoid having to go through various technical considerations each time we discuss an operation involving measurable functions, we sort the functions out into equivalence classes and operate at the level of classes. Let (X,M,J.l) be a measure space. We consider the collection :F consisting of those functions f that satisfy the following properties: (i) f is an extended real-valued function defined on X \ N, where N is a null subset of X.

1.

Properties of Measurable Functions

83

(ii) I is measurable, as a function on (X \ N,MX\N). Note that we only require functions in F to be defined J1,-a.e. on X. Next we identify those measurable functions which coincide J1,-a.e.j more precisely, given I, 9 E F, we say that I '" 9 iff there is a null subset N of X such that I(x) = g(x) for x E X \ N. It is clear that", is an equivalence relation on Fj the only property that offers any difficulty is the transitivity, and this follows at once from the fact that the union of null sets is null. We return to the addition: Given equivalence classes ],y E F corresponding to the finite J1,-a.e. functions I,g, by removing the bad set B we readily see that hex) = I(x) + g(x) is a finite quantity for x E X \ N, N null, and we put ] + Y = h. It is straightforward to verify that h is well-defined, i.e., it is independent of the representatives of the classes ],y, and this completes our discussion. Now that we know how things should be done we agree to denote the equivalence class] of a function I once again by I, and to operate with the classes as if they were functions. This should cause no undue stress, and the reader should keep in mind that a statement such as "a function I defined on X" actually means "an equivalence class ] of a function I defined J1,-a.e. on X." To deal with the usual arithmetic operations we need a preliminary result.

Lemma 1.5. Let (X,M,J1,) be a measure space, and suppose I,g are extended real-valued measurable functions defined on X. Then

{I> g} EM. Proof. Let {rk} be an enumeration of the rational numbers and observe that by Proposition 1.1

Ek={/>rk}n{g g} = Uk Ek is also measurable. • We are now ready to prove

Theorem 1.6. Let (X,M,J1,) be a measure space and I,g be extended real-valued measurable functions defined on X. Then I ± 9 is also measurable. Proof. We only do the addition. Observe that for any real A, A - 9 is measurable. Since

{/+g>A}={/>A-g},

real A,

VI.

84

the conclusion follows at once from Lemma 1.5.

Measurable Functions

•

The other operations of interest to us are covered by the following result. Theorem 1.7. Let (X,M,J.L) be a measure space, assume I is a measurable, finite J.L-a.e. function defined on X and let 4> be a real-valued continuous function defined on R. Then the composition 4> 0 I is measurable. Proof. Since {I = ±oo} is a null set we may assume that 4> 0 I is well-defined and that {4> 0 I = -oo} = 0. By Proposition 1.2 the measurability of 4> 0 I will be established once we show that (4)0 l)-l((A,oo))

= 1-1 (4)-l((A,oo)))

EM,

all real A.

(1.12)

But this is not hard; indeed, since 4> is continuous, 4>-1 (( A, 00)) = 0 is an open subset of R, and, by Proposition 1.3, 1-1(0) E M. Thus (1.12) holds and we have finished. • Theorem 1.7 shows that the composition 4>0 I of a measurable function with a continuous function 4> is measurable; it is not intuitively apparent that the composition 104> should also be measurable. In fact, it is not, as the following example shows. Let {J(n} be a sequence of Cantor-like sets, lJ(nl = 1 - lin, n 2,3, ... , and let A = Un J(n' Since

I

1[0,1] \

AI

~ 1[0,1] \

it readily follows that [0,1] = any subset B of [0,1] we have

J(nl

~

lin,

Un J(n U Z,IZI

n = 2,3 ... = 0, and consequently for

In particular, if B is not Lebesgue measurable, there is an index N so that B n J(N is not Lebesgue measurable. Referring to the construction of the Cantor set, let Dn = [0,1] \ Cn, where as usual C n denotes the union of the intervals remaining after n steps. Dn consists of 2n - 1 open intervals, I! say, ordered from left to right by k, removed in the first n steps of the construction of C. Since J(n is a Cantor-like set, there also is a sequence of open intervals, J! say, 1 ~ k ~ 2n - 1, ordered from left to right by k, removed in the first n steps of the construction of J(N.

1.

Properties of Measurable Functions

85

/

/ /

/

/

/ / Figure 3 We define now a function h from [0,1] onto [0,1] as follows: Construct KN in the interval [0,1] corresponding to the domain of h, and C in the interval [0,1] that corresponds to the range of h. Then h is the function that maps the left-end point of J~ into the left-end point of the rightend point of J~ into the right-end point of I~, and is extended to [0,1]\KN by continuity. It is not hard to check that h is well-defined, one-to-one (if this were not the case KN would contain an interval, and this is not possible) and onto [0,1]. Let BnKN be a non-Lebesgue measurable subset of KN and put A = h(B n KN) ~ C. Then A is null, and consequently measurable; in other words the image ofthis non-Lebesgue measurable set by a continuous function is Lebesgue measurable. Another way to express this situation is the following: If = h- 1 is the continuous inverse of h, then (A) = B n KN, and the image of a Lebesgue measurable set by a continuous function is not necessarily Lebesgue measurable. Also <1>( C) = KN, and takes a null set onto a set of positive Lebesgue measure. Returning to the question at hand, let f = XA; since A is null f is Lebesgue measurable. Consider now the composition f 0 <1>. The inverse image (J 01/;)-1((1/2,3/2» is readily seen to equal

I!,

which is not measurable. Thus, as asserted, f

0

is not measurable.

VI.

86

Measurable Functions

The measurability of several expressions involving I follow at once from Theorem 1.7, with an appropriate choice of
I(x) ± A ,A real AI(x), A real l(x)P,p ~ 1 I/(x)IP,p ~ 0 I+(x) = v(O,/(x)) I-(x) = V(O, - I(x)) and when


I i- 0 J.L-a.e.,

l/l(x) I/( x )1 71 , 'fJ real


We also have Theorem 1.8. Let (X,M,J.L) be a measure space and assume I,g are measurable, finite J.L-a.e. functions defined on X. Then Ig is measurable, and, if 9 i- 0 J.L-a.e., also 1/9 is measurable. Proof. By Theorem 1.6, I ± 9 are measurable, and by the above remarks so are (J ± g? Whence, once again by Theorem 1.6,

is also measurable, and so is

Ig.

Since 1/ 9 is measurable, it readily follows that measurable.

•

1/9 = I . !

is also

9

Corollary 1.9. Let (X,M,J.L) be a measure space, and suppose I is a "simple" function defined on X, Le., I( x) = E~=l CkXE/c (x), where the Ck'S are real constants and the Ek'S form a measurable partition of X. Then I is measurable. Next we consider whether measurability is preserved under limiting operations. We begin by discussing the inf and sup of a sequence {In(x)} of measurable functions, which exist for any x E X.

1.

Properties of Measurable Functions

87

Lemma 1.10. Let (X,M,J.L) be a measure space, and let {In} be a sequence of extended real-valued measurable functions defined on X. Then the functions inf In{x) ,

and

sup In{x) ,

x EX

are measurable. Proof.

Since for arbitrary sequences {gn} we have

infgn{x)

= -sup{-gn{x)),

all x EX,

we only need prove one statement, the one with the sup, say. But this is not hard; indeed given any real A, note that {sup In

> A} = U{/n > A}.

(1.13)

n

By assumption the set on the right-hand side of (1.13) is the countable union of measurable sets, and so it is measurable. Thus the set on the left-hand side is measurable, and the sup of the In's is measurable. • Next we consider the limsup and liminf of a sequence of functions; these limits exist for every x EX, even if the sequence does not converge. Theorem 1.11. Assume (X,M,J.L) is a measure space and let {In} be a sequence of extended real-valued measurable functions defined on X. Then lim inf In and lim su PIn are measurable. In particular, if I (x) = lim In( x) exists for all x EX, then I is measurable. Proof.

Since liminf In(x)

= sup (inf 1m (X)) , k~l

x EX,

m~k

the measurability of the liminf follows at once from Lemma 1.10. Similarly, since lim sup In(x) = inf (sup k~l

m~k

1m (x )),

x EX,

also the lim sup is measurable. Finally, if the sequence converges, I{ x) coincides with both the lim sup In and Hminf In, and it is measurable as well. • We close this section with an interesting and important result; it shows how to approximate arbitrary functions by functions that assume finitely many values.

VI.

88

Measurable Functions

Theorem 1.12. Let (X,M,J.L) be a measure space, and I be an extended real-valued function defined on X. Then there is a sequence {In} of simple real-valued functions defined on X, i.e., kn

In(x) = LCiXEi(X) ,

Ci real,

Ei pairwise disjoint,

i=l

so that lim n -+ oo In(x) = I(x),x EX. Furthermore, (i) If I is measurable, so are the In's. (ii) If I is nonnegative, then the sequence {In} is nondecreasing, and

o ~ In(x)

~

I(x) ,

all x EX,

n

= 1,2, ...

(iii) If I is bounded, i.e., I/(x)1 ~ M for all x E X, then the In's converge uniformly to I. Proof. The In's are defined by looking closely at the level sets of Suppose first that I is nonnegative, fix an integer n ~ 1, and consider the n2n pairwise disjoint subintervals of [0, n) given by

I.

Put now In(x) = {n(k - 1)2- n

if (k - 1)2- n ~ I(x) otherwise.

< k2- n , 1 ~ k

~ n2n

Clearly In( x) ~ I( x) for all x E X and n = 1,2, ... Also, each In assumes a finite number of values; more precisely, if Ak,n

= {(k -

we have

1)2- n ~

1< k2- n }

and

An

= {I ~ n},

n2n

In(x)

= L(n -1)2- n XA",Jx) + nXAn(x).

(1.14)

k=l

Observe that In+1(x) is obtained from In(x) by dividing each interval [(k - 1)2- n , k2- n ) in half, and then only increasing In(x) to In+1(x) at those x's where In(x) is changed; this proves the remarks in (ii). We claim that lim fn(x)

n-+oo

= I(x),

x EX.

(1.15)

2.

Structure of Measurable Functions

89

Now, if I(x) = 00, then In(x) = n for all n, and In(x) ~ 00. On the other hand, if I( x) is finite, by the definition of In( x) it readily follows that

I(x) - In(x)

~

2- n ,

all n> I(x),

(1.16)

and (1.15) holds. Moreover, if I is measurable, by (1.6) the Ak,n'S are measurable, and, by (ii) in Proposition 1.1, the An's are measurable. Whence In is a simple measurable function and (i) is proved. Furthermore, if I is bounded, (1.16) is true for n > M, uniformly for x in X, the convergence is uniform and (iii) holds. In short, we have obtained the desired conclusion for nonnegative functions. To complete the proof recall that each function I is the difference of two nonnegative functions, I = 1+ - 1-, and apply the first part ofthe proofto 1+ and 1- separately. Note that in this case, the In's also have the property that IIn(x)I~I/(x)l,

allxEX,

n=1,2, . . . •

2. STRUCTURE OF MEASURABLE FUNCTIONS What does a measurable function I look like? In case there is a topology defined on X, how far is I from being continuous? First an example: Let I be the characteristic function of the set of irrational numbers [in [0,1]. I is Lebesgue measurable, yet discontinuous at every point of [0,1]. There is, however, another way to interpret this situation: Since I is constant on [, it is continuous there in the relative topology in [with respect to the usual topology of R. Now, 1[0,1]\[1 = and consequently, we have that I is continuous on a subset [ of [0,1] of full measure in the relative topology in [. This observation points to a general fact concerning measurable functions. In order to avoid technical difficulties, and since the nature of the question is already apparent there, we restrict our attention to the Lebesgue measure on X = [0,1]. We begin by considering the simpler question of the boundedness of measurable functions.

°

Proposition 2.1. Assume I is an extended real-valued Lebesgue measurable function defined on X such that I{III = 00}1 = o. Then, for

VI.

90

Measurable Functions

any E > 0, there exist a measurable subset B of X and a constant M with the following properties: (i) IBI < E. (ii) II(x)1 ~ M for x E X \ B. Proof. Let Bn = {III> n},n = 1,2, ... ; {Bn} is a nonincreasing sequence of measurable subsets of X. Suppose that for all n, IBnl > E. Since Bn = {III = oo} and IXI = 1, by (3.4) in Chapter IV we get

n

lim

n-+oo

IBnl = I{III

=

00}1

~ E,

which is impossible. Thus, there exists m such that IBml ~ desired conclusion obtains with B = Bm and M = m. • As for the continuity of provides the answer.

E,

and the

I, the following result of Lusin (1883-1950)

Theorem 2.2 (Lusin). Suppose I is an extended real-valued measurable function defined on X with the property that I{III = 00}1 = 0. Then, given E > 0, there is a closed subset F of X such that (i) IX \ FI < E • (ii) IIF is continuous on F, in the relative topology in F. Proof. Since I = 1+ - 1- is the difference of two nonnegative measurable functions, we may assume that I ~ 0. Now, given E > 0, let B be a measurable subset of X corresponding to the choice E /2 in Proposition 2.1. The restriction II(X \ B) of I to X \ B is measurable and bounded, and with no fear of confusion we also denote it by I. By Theorem 1.12 there is a sequence {In} of simple functions defined on X \ B that converges uniformly to I there. Let n=I,2, ...

where the Ci,n'S are real numbers and the Ei,n'S form a pairwise disjoint partition of X \ B. By the regularity properties of the Lebesgue measure, given TJn > 0, there exist closed subsets li,n of Ei,n with the property that

IEi,n \li,nl ~ TJn

i

= 1, ... ,in, n = 1,2, ...

Put now Fn = U:~l Fi,n; since each Fi,n is closed, Fn is a closed subset of X. Moreover, since the Fi,n'S are pairwise disjoint and compact, they

3.

Sequences of Measurable Functions

91

can be separated by pairwise disjoint open subsets of X. Whence, the restriction of In to F n, or any of its subsets for that matter, is continuous in the relative topology in Fn. Next observe that

I (X \ B) \

Fnl = I U~~lEi,n \ U~~lFi,nl ""in :$ L..Ji=l I Ei,n \ Fi,nl :$ i n17n.

(2.1)

A good choice for 17n is c/i n2n+1 for then, by (2.1), we have

Put now F = n~=l Fn; F is a compact subset of X \ B and {/nIF} is a sequence of continuous functions in the relative topology in F which converges uniformly to I on F. Since the uniform limit of a sequence of continuous functions on a compact set is continuous, IIF is continuous in the relative topology in F. It only remains to estimate IX \ Fl. First observe that I(X \ B) \ FI equals

I (X \ B) \

nnFnl

= IUn((X \ B) \ Fn)1 :$ Ln I(X \ B) \ Fnl :$ Ln c/2n +1

= c/2. (2.2)

Now, since X \ F = ((X \ B) \ F) U B, by (2.2) and the choice of B it readily follows that

IX \ FI :$ c /2 + c /2

=c .

•

3. SEQUENCES OF MEASURABLE FUNCTIONS Let (X,M,Il) be a measure space. The pointwise convergence of a sequence {In} of real-valued functions defined on X is a well-defined notion. Here, and unless otherwise specified, the term "convergence" means that the convergence is to a finite limit. Actually, a more relevant concept is that of Il-a.e. convergence; sets of measure 0 playa special role in this setting. For instance, on X = [0,1], the sequence In(z) = 0 if Z '" 1/2 and In(1/2) = 1 + lin tends to

92

VI.

Measurable Functions

I(x) = 0, Da-a.e., a i- 1/2, but In f+ I D1 / 2 -a.e. Other important notions of convergence involve uniformity, and we shall discuss them shortly. How can we express the notion of p.-a.e. convergence in terms of familiar concepts? First an observation: Since the limiting function I is finite p.-a.e., the statements In(x) - I(x) p.-a.e. and Iln(x) - l(x)1 - 0 p.-a.e. are equivalent. So, we may restrict our attention to sequences of nonnegative functions which converge to 0 p.-a.e. For such sequences we have Proposition 3.1. Let (X,M,p.) be a measure space and let {In} be a sequence of measurable nonnegative functions defined on X. Then, In - 0 p.-a.e. on a subset of M E M iff for each 'fJ > 0, the sets Bn.,., = {x E M :/n(x) > 'fJ} satisfy p. (lim sup B n.,.,) = o. Proof. Let In - 0 p.-a.e. on M and suppose that for some 'fJ > 0 we have p. (lim sup B n.,.,) > O. Then each x E lim sup Bn.,., belongs to infinitely many of the Bn.,., 's, and consequently, there is a sequence nk - 00 such that Ink (X) > 'fJ. Whence, lim sup In(x) ~ 'fJ > 0, and the In's do not converge to 0 on a subset of M of positive measure; this is a contradiction. Conversely, given c > 0, pick 0 < 'fJ < c, and consider a point x in M \ (lim sup B n.,.,). Since x belongs to at most finitely many of the Bn.,., 's there is a no such that In(x) ~ 'fJ < c for all n ~ no. In other words, In(x) - 0 for x E M \ N,p.(N) = 0, which is precisely what we wanted to show. • Since lim sup Bn.,., = nk:l (U~=k Bn.,.,), if P.(U~=k Bn.,.,) < 00 for some k, by (3.4) in Chapter IV the conditions of Proposition 3.1 are satisfied iff OO (3.1) lim p. B n .,.,) = 0 , all'fJ>O. k-+oo n=k

(U

In particular, if p.(X) < 00, (3.1) describes convergence p.-a.e. The relation (3.1) points to a possible limitation of the concept of p.-a.e. convergence, namely, we require the control of all the Bn.,., 's, from one index on. To illustrate this point, consider the sequence of (dyadic) subintervals of I = [0,1] defined as follows: 10 = I, II = [0,1/2], 12 = [1/2,1], 13 = [0,1/4], and so on. In other words, the sequence consists of successive blocks of 2n nonoverlap ping intervals, each of length 2- n , and the union of the intervals in each block is I. Let {In} be the sequence consisting of the characteristic functions of the In's. Clearly {In} does not converge to 0 anywhere on I, yet in some sense the In's are getting

3.

Sequences of Measurable Functions

close to

93

o. Specifically, lim 1{ln

n ..... oo

> TJ}I = 0,

all real TJ>

o.

(3.2)

Notice that in contrast to (3.1), we are dealing with one Bn,T/ at a time. Motivated by this remark we introduce the following definition. Given a measure space (X,M,p) and a sequence {In} of measurable nonnegative extended real-valued finite p-a.e. functions defined on X, we say that In converges to 0 in p-measure iff lim p( {In> TJ})

n ..... oo

=0,

all real TJ

> o.

(3.3)

If p(X) < 00 we refer to convergence in p-measure as convergence in the sense of probability, or convergence in probability. Thus, p-a.e. convergence implies convergence in probability, but the opposite is not true. Also, p-a.e. convergence does not, in general, imply convergence in p-measure. To see this consider (R,.c, 1·1), and observe that the sequence In(x) = X[n,oo)(x) tends to 0 everywhere, but 1{ln > 1/2}1 = 00 for all n. Nevertheless, a closer look at the first example indicates that there is a subsequence {Ink} of the In's, specifically that consisting of the characteristic functions of the intervals [0,1/2 n ], n ~ 1, with the property that limk ..... oo Ink (x) = 0 for all x E (0,1]. The remarkable fact is that this property is true for arbitrary sequences; before we prove this we need a bit of information concerning convergence in p-measure. Proposition 3.2. Let (X, M,p) be a measure space and suppose {In} is a sequence of measurable nonnegative extended real-valued finite p-a.e. functions defined on X. Then, In -+ 0 in p-measure iff for any e,8 > 0, there exists a constant N = Ne,s such that p( {In> 8})

< e , all n

~ N .

(3.4)

Proof. The necessity of the condition is obvious. As for the sufficiency, if In 0 in p-measure, then by (3.3) there exist TJ > 0 and a sequence nk -+ 00 such that

r

L = limsupp({lnk > 77}) >

Thus, (3.4) cannot hold for c completes the proof. •

=

L and Ii

=

o.

77, and this contradiction

Next we show that convergence in probability implies p-a.e. convergence along a subsequence.

VI.

94

Measurable Functions

Proposition 3.3. Let (X,M,J.t) be a finite measure space, and assume {In} is a sequence of measurable, nonnegative, extended real-valued, finite J.t-a.e. functions defined on X. If In --+ 0 in probability, then there is an increasing sequence nk --+ 00 such that lim Ink = 0 J.t-a.e. on X.

k-+oo

Proof. By Proposition 3.2 it follows that for each n we may find an index nk < nk+1 with the property that (3.5) Let Bk = {Ink > 1/2k} and consider the (bad) set B = lim sup B k. Since by (3.5) L:~1 J.t(Bk) < 00, by the Borel-Cantelli Lemma we have J.t(B) = O. Now, it is not hard to see that lim Ink (x)

k-+oo

Indeed, if x (3.6) holds.

ft

=0 ,

xEX \ B.

(3.6)

B, then x belongs to at most finitely many of the Bk'S and •

Sometimes we have to deal with questions of convergence when no limit is in evidence. For J.t-a.e. convergence this can be reduced to the numerical case, where the Cauchy criterion is available. Specifically, let (X,M,J.t) be a measure space, and let {In} be a sequence of measurable extended real-valued finite J.t-a.e. functions defined on X. We say that {In} is Cauchy J.t-a.e. if for J.t almost every x EX, given e > 0, there is an integer no = no( x) such that

I/n(x) - In,(x)1 ~ e,

all n,n' ~ no.

By the Cauchy criterion of convergence of numerical sequences, if {In} is Cauchy, then lim n -+ oo In( x) = I( x) exists J.t-a.e. The same is true for convergence in probability. Let (X,M,J.t) be a finite measure space, and let {In} be a sequence of measurable, extended real-valued, finite J.t-a.e. functions defined on X. We say that {In} is Cauchy in probability if given e, 6 > 0, there is an integer no such that

J.t({l/n - In' I > 6}) < e,

all n,n' ~ no.

Sequences which are Cauchy in probability converge in the sense of probability, cf. 4.30 below, and convergence in probability corresponds to a notion of "metric" convergence, cf. 4.31 below.

3.

Sequences of Measurable Functions

95

Next we discuss the concept of uniform convergence. Let (X,M,J.t) be a measure space and assume {in} is a sequence of measurable nonnegative extended real-valued finite J.t-a.e. functions defined on X. We say that in -+ 0 almost uniformly if given c > 0, we can find a measurable subset B of X such that J.t(B) < c and lim in(x) = 0,

uniformly for x E X \ B .

n--+oo

Proposition 3.4. Let (X,M,J.t) be a measure space, let {in} be a sequence of measurable nonnegative extended real-valued finite J.t-a.e. functions defined on X, and suppose that in -+ 0 almost uniformly. Then in -+ 0 J.t-a.e. Proof. For every positive integer k there is a measurable subset Bk of X such that J.t(Bk) < 11k and in(x) -+ 0, uniformly for x E X \ Bk. Clearly in -+ 0 pointwise on the (good) set G = Uk:l(X \ Bk). It only remains to check that J.t(X \ G) = 0; this is not hard. Since X \ G equals

and J.t(Bt)

< 00, it readily follows that

How about the converse to Proposition 3.4? To decide whether it is true we investigate the rate at which arbitrary sequences converge pointwise to o. First we show that the convergence occurs at a fairly rapid rate. Theorem 3.5. Let (X,M,J.t) be a finite measure space, and suppose {in} is a sequence of measurable nonnegative extended real-valued finite J.t-a.e. functions defined on X so that in -+ 0 J.t-a.e. Then there exists a non decreasing sequence of integers An -+ 00 with the property that lim Anin

n--+oo

= 0 J.t-a.e. on X .

(3.7)

Proof. Redefining the In's if necessary on a set of measure 0, we may assume that the In's are finite everywhere and that In(x) -+ 0 for every x E X. Let 9n(X) = sup A(x), x EX. k~n

VI.

96

Measurable Functions

Clearly the 9n'S are measurable, 9n(X) 2: In(x), all n,x E X, and lim 9n(X) = O.

n-+oo

In other words, working with the 9n'S instead, we may also assume that, in fact, the In's decrease to 0 everywhere. The first step is to construct the An's. Let nl = 1, and note that since In -+ 0 in probability, for each integer k = 2,3, ... , there exist integers nk > nk-l such that (3.8) Put now for nk ~ n

An = k,

< nk+1, k =

1,2,. . .

(3.9)

In other words, the sequence of An's is defined in blocks: The first (n2 - nl) entries are l's, the next (n3 - n2) entries are 2's, and so on. Furthermore, since nk -+ 00 as k -+ 00, also An -+ 00 as n -+ 00. Next we deal with the convergence of the sequence {Anln}. Let

Bm =

U {An In > 11m},

m = 1,2, ...

(3.10)

n~nm

It is not hard to estimate J.L(B m ). First observe that since the An's are constant on blocks we have 00

nk+l- 1

00

nk+l- 1

U U {Anln > 11m}

Bm =

=

U U {kIn> 11m}.

(3.11)

k=m n=nk

Furthermore, since the sequence {In} is nonincreasing and since k 2: m, the innermost union in (3.11) is contained in {Ink> 1/k 2}, and 00

Bm ~

U {Ink> 1/k2}.

(3.12)

k=m Whence, by (3.12) and (3.8) 00

J.L(Bm) ~

L

k=m

J.L( {Ink> 1/k2}) ~ 2- m+1

•

3.

Sequences of Measurable Functions

97

Let the (bad) set B = limsupBm. Since Lp(Bm) < 00, by the Borel-Cantelli Lemma we have pCB) = O. It only remains to check that for any x E X \ B, we have lim n -+ oo Anln(x) = o. But this is not hard: Given c > 0, let m be so large that l/m ::; c and x rt. Bm; such a choice is always possible since x belongs to at most finitely many of the Bm's. Then by (3.10) there exists nm so that

Our next result is an interesting interpretation of Theorem 3.5. Theorem 3.6. Let (X, M,p) be a finite measure space, and let {In} be a sequence of measurable nonnegative extended real-valued finite p-a.e. functions defined on X such that In -4 0 p-a.e. Then there exist a measurable nonnegative finite p-a.e. function I defined on X and a sequence of real numbers 'f/n

-4

0 such that

In ::; 'f/nl p-a.e. on X . Proof.

(3.13)

In the notation of Theorem 3.5, let

I(x) = sup {).n In (x)} ,

x EX.

n

Clearly I is measurable and nonnegative, and since Anln also finite p-a.e. Put now 'f/n = 1/ An, and note that

In ::; 'f/nl p-a.e., with 'f/n as asserted.

-4

-4

0 p-a.e., I is

0,

•

We are now ready to show that convergence p-a.e. implies almost uniform convergence; this result is due to Egorov (1869-1931). Theorem 3.7 (Egorov). Let (X,M,p) be a finite measure space, and let {In} be a sequence of measurable nonnegative finite p-a.e. functions defined on X such that In -4 0 p-a.e. Then In -4 0 almost uniformly.

98

VI.

Measurable Functions

Proof. We must show that given e > 0, there exists B E M such that J-L(B) < e and In(x) - 0 uniformly on X\B. Let I be the J-L-a.e. finite function corresponding to the sequence {In} constructed in Theorem 3.6. By Proposition 2.1 there is a constant M such that I( x) ~ M for x E X \ B, J-L(B) < e. By Theorem 3.6

In(x)

~

'I]nM,

x E X \ B, J-L(B) < e,

and In(x) - 0 uniformly for x E X \ B.

•

The measurability of the In's is essential to the validity of Egorov's theorem, cf. 4.38 below, as is the assumption J-L(X) < 00. Indeed, in the measure space (R,C, 1·1) the sequence In = X[n,oo) tends to 0 everywhere on R, but not uniformly on any unbounded subset of R.

4. PROBLEMS AND QUESTIONS The setting of the first thirteen problems and questions is the following: M is a u-algebra of (measurable) subsets of X and I is an extended real-valued function defined on X.

{I> A} EM for each rational number Aj is I measurable? Suppose I is a measurable real-valued function defined on X, and

4.1 Suppose 4.2

put g( x) = 0 if I( x) is rational and g( x) 9 measurable?

= 1 if I( x) is irrationalj is

I is measurable and B E Bl is a Borel subset of Rj does it follow that 1-1 (B) E M?

4.3 Suppose

I is a measurable real-valued function defined on X, and let be a real-valued Borel measurable function defined on R. Show that the composition 0 I is measurable.

4.4 Suppose

4.5 Suppose I is measurable and show that for each real r, s truncations

>

0, the

r if I(x) > r Ir,s(x) = { I_(sx) if - s ~ I(x) ~ r if I(x) < -s, are measurable. 4.6 If I,g are measurable real-valued functions defined on X and

U,g) is measurable.

4.

Problems and Questions

99

4.7 Suppose X = R, and show that Bl is the smallest u-field with respect to which all continuous functions are measurable. More precisely, show that every continuous function is measurable iff Bl ~M. 4.8 Let {In} be a sequence of measurable functions defined on X, and let A = {x EX: lim n -+ oo In(x) exists}. Show that A E M. 4.9 An elementary function is one which assumes at most count ably many values. Show that if I is an everywhere finite real-valued measurable function, then it is the uniform limit of a sequence of elementary functions. Also, unless I is bounded, it is not the uniform limit of a sequence of simple functions. 4.10 Let I' be a measure on (X,M). Show that the following assertions are equivalent: (a) (X,M,I') is a complete measure space, and, (b) If {In} is a sequence of measurable functions defined on X which converges I'-a.e., and if I is any extended real-valued function defined on X such that I = lim n -+ oo In I'-a.e., then I is measurable. 4.11 Suppose (X,M,I') is a u-finite measure space and let

I be a mea~

surable function defined on X. Show that the function I'({I/I > A}),

A> 0

is nonincreasing and right-continuous. Furthermore, if I, It and h are nonnegative and measurable and "'1, 'f/2 are nonnegative real numbers so that I ~ ",til + 'f/2h I'-a.e., then for any A > 0

4.12 With the notation of 4.11, if {In} is a non decreasing sequence of measurable functions and I lim n -+ oo In, then for any A > 0,

=

lim 1'({l/nl

n-+oo

> A}) = I'({I/I > A}).

4.13 With the notation of 4.11, suppose that I'({I/I A --+ 00, and let

ret)

r

= inf{'\:I'({1/1 > A} ~ t},

Show that satisfies the following properties: (i) It is nonincreasing and right-continuous.

t

> A}) ~ O.

--+

0 as

VI.

100

(ii) If

°< 7]

~ p,( {III> A})

Measurable Functions

< 00, then

f*(p,({1/1 > A})) ~ A < f*(p,({1/1 > A})-7]).

(iii) If /* is continuous at t = p,( {III> A}), then /* (p,( {III> A})) = A. (iv) I and /* are equimeasurable, i.e., for A > 0, p,({1/1 > A}) =

I{/* > A}I· Because of property (iv), /* is called the nonincreasing equimeasurable rearrangement of I. The next six problems and questions deal with the Lebesgue measure on

R. 4.14 Suppose I is defined a.e. on [0,1] and it is continuous a.e. there. Is I Lebesgue measurable? What if I is right-continuous instead? 4.15 Suppose

I

is differentiable on [0,1]. Is

f'

Lebesgue measurable?

4.16 Show that I is Lebesgue measurable iff there is a Borel measurable function 9 such that I = 9 Lebesgue-a.e. 4.17 Suppose I(x, y) is a Lebesgue measurable real-valued function defined on R2 with the property that I(x,.) is Lebesgue measurable as a function of x E R, and 1(·, y) is continuous as a function of y E R. Let now 4>( x) = maxc is Lebesgue measurable. - 4.18 Suppose that I is Lebesgue measurable and 4> is real-valued, continuous, and has the following property: For any null set N, 4>-l(N) E £,. Show that 104> is Lebesgue measurable. 4.19 Suppose {An} is a sequence of Lebesgue measurable subsets of R, and let In = XA n' n = 1,2, ... Find necessary and sufficient conditions for the sequence {In} to: (a) converge Lebesgue-a.e., and, (6) converge uniformly.

In the next three problems and questions we assume that (X, M , p,) is a probability measure space.

I defined on X is said to be bounded in probability if given c > 0, there exists a finite real number Me such that p,( {III ~ Me}) ~ 1 - c. Prove that I is bounded in probability iff I is finite p,-a.e.

4.20 An extended real-valued measurable function

4.21 A sequence of extended real-valued measurable functions {fn} is

said to be bounded inl probability iff sup Ifni is bounded in probability. The sequence {fn} is said to diverge to 00 in probability iff

4.

Problems and Questions

101

for each M > 0 and c > 0, there exists a finite integer no(M,c) such that if n > no, then J-t( {l/nl > M}) > 1- c. Prove that if {In} diverges to 00 in probability and {9n} is bounded in probability, then the sequence {In + 9n} diverges to 00 in probability. 4.22 Suppose that sup In = 00 J-t-a.e. Does there necessarily exist a subsequence {In,,} that diverges to 00 in probability? In the next nine problems and questions we assume that (X,M,J-t) is a measure space. 4.23 This result concerns the approximation Theorem 1.12. Suppose the function I there enjoys the property that {x EX: I( x) =F O} is O'-finite. Then prove that the approximating sequence {In} may be constructed with the additional property that J-t( {In =F O}) < 00 for each n. Conversely, if each of the simple functions In satisfies J-t( {In =F O}) < 00 and for each x E X we have lim n-+ oo In( x) = I( x), then the set {I =F O} is O'-finite. 4.24 Let J-t(X) < 00, and suppose {In} is a sequence of extended realvalued measurable functions defined on X that satisfies 00

L J-t( {l/nll An > 1}) <

00.

n=l

Prove that lim sup(l/n II An) ~ 1 J-t-a.e. 4.25 Suppose J-t(X) = 1. Show that the sequence of extended real-valued measurable functions {In} converges J-t-a.e. to the measurable function I iff for any c > 0 and N we have

J-t({l/n - II

~ c, for all

n ~ N})

= 1.

4.26 Suppose J-t(X) < 00 and let {In} be a sequence of extended realvalued measurable functions defined on X which converges to I in probability. Show that I is measurable and that I: - t 1+, I; - t 1- and Ilnl - t III in the sense of probability. Are the analogous statements for convergence in measure valid? 4.27 In the setting of 4.26, suppose that {gn} is a sequence of extended real-valued functions defined on X which converges to 9 in probability, and {An} a sequence of real numbers, An - t A. Does it follow that In + An9n - t 1+ A9 in probability? Is it also true that Inl9n - t I I 9 in probability? How about the corresponding statements for convergence in measure?

VI.

102

Measurable Functions

Consider further the following statement: If 4>: R2 ~ R is continuous, then 4>(Jn,gn) ~ 4>(J,g) in probability, or in measure. Is it true? 4.28 In the setting of 4.26, assume that the sequence {In} is nondecreasing. Show that in this case we have limn_oo In = I p,-a.e. 4.29 In the setting of 4.26, suppose that p, is a probability measure and, for t =I 0 real, let D(t) = 1 if t < 0 and D(t) = 0 if t > O. Show that In ~ 0 in probability iff lim p,( {In> t}) = D(t) ,

n-oo

<

all t

=I O.

and let {In} be a Cauchy sequence in the sense of probability, i.e., for any c > 0 there exists an integer N = Ne such that

4.30 Suppose that p,(X)

00,

P,({l/n - Iml > c}) < c,

all n,m ~ N.

Show that there exists a measurable function = I in probability.

I such that limn_ oo In

< 00, and for extended real-valued measurable functions I,g defined on X put

4.31 Suppose that p,(X)

d(J,g) = inf {c > 0:p,({11 -

gl > c}) ~ c}

.

Observe that d(J,g) = 0 iff I = 9 p,-a.e.; whence (at the level of classes) upon identifying functions which agree p,-a.e. we have d(J,g) = 0 iff 1= g. Show that d(J,g) is a distance function and that In ~ I in probability iff d (In, J) ~ O. Furthermore, show that endowed with the metric d, the space of measurable functions is a complete metric space. Is a similar result, involving now convergence in measure, true for arbitrary measure spaces? 4.32 Show that if p, is the counting measure on the integers X = Z, then convergence in measure is equivalent to uniform convergence. 4.33 Suppose (X,M,p,) is a finite measure space, and let {In} be a

sequence of measurable real-valued functions defined on X. Show that given c > 0, there exist a (bad) set B with p,( B) < c and a constant M such that

I/n(x)I::5M,

allXEX\B,

alln.

4.

Problems and Questions

103

Is the conclusion true if we assume instead that the in's are finite J.L-a.e. ? 4.34 Suppose (X,M,J.L) is a finite measure space, and let {in} be a sequence of extended real-valued measurable functions defined on X. Show that {in} converges to a finite limit J.L-a.e. iff given e > 0, there exists a finite constant Me such that

4.35 Suppose (X,M,J.L) is a finite measure space, and let {in} be a sequence of measurable real-valued functions defined on X. Show that there exists a sequence (an) of positive real numbers with the property that lim anin = 0 J.L-a.e. n-+oo

4.36 Show that the proof of Lusin's theorem may be adjusted to give the following result: Suppose i is an extended real-valued finite Lebesgue-a.e. measurable function defined on X = [a,b]. Then, given e > 0, there is a continuous function 4> defined on X such that

I{x EX :i(x) ~ 4>(x)} I < e. 4.37 In the setting of 4.36 show that the following is true: There is a sequence of continuous functions {in} defined on X such that limn -+ oo in = i Lebesgue-a.e. 4.38 By means of an example show that the conclusion of Egorov's theorem does not necessarily hold if the in's are not Lebesgue measurable. 4.39 Construct a sequence of Lebesgue measurable functions {in} defined on X = [a,b] with the following properties: The in's converge at every point of X, but they do not converge uniformly on any Lebesgue measurable subset E of X with IX \ EI = O. 4.40 Let {in} be a sequence of Lebesgue measurable functions defined on X = [a,b] and suppose that limn -+ oo in = i exists Lebesgue-a.e. on X. If i ~ 0 Lebesgue-a.e. and in ~ 0 Lebesgue-a.e. on X, prove that given e > 0, there exist c > 0 and a sequence {En} of Lebesgue measurable subsets of X such that

lin(x)1

~ c, x E E,."

and

IX \ Enl ~

E,

n = 1,2, ...

VI.

104

Measurable Functions

4.41 Prove the following variant of Egorov's theorem: Let X = [a,b] and suppose I, In, n = 1,2, ... , are Lebesgue measurable functions defined on X such that lim sup In ::; I on X. Then, given € > 0, there is a Lebesgue measurable set E ~ X with lEI < € such that for each 17 > there is N with the property that

°

In(x)::;/(x)+l7,

allxEX\E,n~N.

4.42 Prove the following extension of Egorov's theorem due to Lusin. Let (X, M ,J.L) be a 0'- finite measure space, and I, In be extended real-valued measurable functions defined J.L-a.e. on X, n = 1,2, ... , such that lim n -+ oo In = I J.L-a.e. Then there exists measurable sets N, EI, E 2 , ••• such that X = NUU k Ek, J.L(N) = 0, and the sequence {In} converges uniformly to I on each Ek. 4.43 Let M be a O'-algebra of (measurable) subsets of X, and let I be a complex-valued function defined on X. We say that I = ~I + i';Sl is measurable iff the (real-valued) functions ~I and ';SI are measurable. This is an open ended question: Discuss the properties of complex-valued measurable functions. For instance, show that I is measurable iff for any open subset U in the complex plane, l-l(U) E M.

CHAPTER

VII

Integration

In this chapter we introduce the integral and establish its basic properties, including taking limits under the integral sign. The relation between the Riemann and the Lebesgue integrals of a bounded function is elucidated.

1. THE INTEGRAL OF

NONNEGATIVE FUNCTIONS Suppose (X,M,J.L) is a measure space and let

ak real,

k=l

where the Ak'S form a measurable pairwise disjoint partition of X. The integral of
Ix

Ix

(1.1)

Ix

The usual convention 0 . 00 = 0 is in effect, so that
105

VII.

106

Integration

(i) It is well-defined, Le., if also 4>( x) = Lh=l bhXBh (x), then n

m

k=l

k=l

L: akJ.L(Ak) = L: bhJ.t(Bh). (ii) If 4> and 'I/J are nonnegative simple functions and 4> = 'I/J J.L-a.e., then

Ix

4> dJ.L

=

Ix

'I/J dJ.L .

(iii) It is positively linear, i.e., if 4>, 'I/J are nonnegative simple functions and A > 0, then

Ix

(4) + A'I/J)dJ.L = lx4>dJ.L+A lx'I/JdJ.L.

(iv) It is monotone, i.e., if

°: ; 4> ::;

Ix

(1.2)

'I/J are simple functions, then

4> dJ.L ::;

Ix

(1.3)

'I/J dJ.L .

(v) For each nonnegative simple function 4>, the set function v given by

veE) =

Ix

4>XE dJ.L = t4>dJ.L,

E EM,

is a measure on (X, M). Proof. Since the proof of (i) follows along the lines to that of (iii), we only do (iii). Let, then, 4> = L anXAn and 'I/J = L bmXBm be nonnegative simple functions; 4> + A'I/J is then the simple function that takes the value an + Abm on the set An n Bm E M. Note that the an + Abm's are not necessarily distinct, but that the An n Bm's are pairwise disjoint. Thus, by the definition of the integral, we have

If the sum in (1.4) is infinite there are indices n, m such that an + Abm =I 0, J.L(An n Bm) = 00, and either anJ.L(An) = 00 or bmJ.L(Bm) = 00. In the former case we have 4>dJ.L = 00, in the latter case it follows that 'I/J dJ.L = 00, and in either case (1.2) holds. On the other hand, if the sum in (1.4) is finite, since the summands there are nonnegative they may be rearranged freely and we obtain at once that the sum equals

Ix

Ix

n

m

m

n

1.

The Integral of Nonnegative FUnctions

107

By the additivity of J.L this expression is

and (iii) holds. We prove (ii) next. Since 4> = "p J.L-a.e., there are nonnegative simple functions (,4>' and "p' and a null set A such that 4>' and "p' vanish off A and 4> = ( + 4>', "p = ( + "p' . By (i) it readily follows that

and (ii) is true. As for (iv), let 4> = L: anXA n , "p = L: bmXBm' and observe that if J.L(An n Bm) 0, then an :::; bm . Whence, since Un,m(An n Bm) = X, we have

t

and (iv) is true. (v) is a useful property and, among other things, it gives new examples of measures on (X, M). Clearly 11 is a nonnegati ve set function and 11( 0) = 0; only the a-additivity requires some work. First observe that if 4> is a nonnegative simple function and E EM, then 4>XE is also a nonnegative simple function and

Suppose, then, that {Ek} is a sequence of pairwise disjoint, measurable subsets of X, and let E denote its union. Now, since J.L is a measure, the right-hand side of the above expression equals

Whence

11

is a measure on (X,M).

•

VII.

108

Integration

How about the integral of arbitrary nonnegative measurable functions? By Theorem 1.12 in Chapter VI, these functions are limits of nondecreasing sequences of simple functions, and this fact suggests a way of defining the integral. Let (X,M,JL) be a measure space, I a nonnegative measurable function defined on X, and set

F J = {: is simple, and 0 ~ ~ f} . The integral of lover X with respect to JL is denoted by or simply I dJL, and it is defined as the quantity

Ix

Ix I( x) dJL( x), (1.5)

Ix

By Theorem 1.12 in Chapter VI, }='J "# 0, and consequently, I dJL is a well-defined nonnegative real number or 00. (1.5) is similar to the definition of the lower Riemann integral of a nonnegative function I, but with a crucial difference: Rather than considering partitions of the "domain of integration" X, we work with partitions of the "range" of I, in a manner compatible with each individual function I. More precisely, }='J contains the 's constructed in Theorem 1.12 in Chapter VI, and these simple functions are closely related to the level sets of I. Before we go on we must check that if I is simple, then the definitions in (1.1) and (1.5) coincide. If I dJL denotes the expression given by (1.1), since I E }='J it readily follows that

Ix

(1.6) On the other hand, if E }='J' then by (1.3),

Ix

dJL

~

Ix I

dJL,

all E }='J,

the inequality opposite to (1.6) holds, and the definitions given by (1.1) and (1.5) coincide. We are now ready to consider some elementary properties of the integral.

1.

The Integral of Nonnegative Functions

109

Theorem 1.2. Assume (X,M,p.) is a measure space, and let I,g be nonnegative measurable functions defined on X. We then have (i) If 1= 9 p.-a.e., then Ix I dp. = Ix 9 dp.. (ii) IxU + g) dp. ~ Ix I dp. + Ix 9 dp.. (1.7) (iii) If I ~ g, then Ix I dp. ~ Ix 9 dp.. (iv) If A ~ B are measurable, then

(1.8) Proof. (i) follows at once since for any E F, there exists a'IjJ E Fg such that = 'IjJ p.-a.e., and, by property (ii) in Theorem 1.1, the integrals of and 'IjJ over X with respect to p. are equal. As for (ii), observe that since for any E F, and 'IjJ E Fg we have + 'IjJ E F1+g, by (1.2) it follows that

Whence taking the sup of the left-hand side in the above inequality over E F, and 'IjJ E Fg gives (1.7), and (ii) holds. As for (iii), note that in this case we have F, ~ F g , and consequently,

L ~L dp.

9 dp.,

all E F, .

Taking the sup over the 's above gives (iii). To verify (1.8) it suffices to note that in this case FhA ~ F'XB' Thus (iv) holds, and the proof is complete. • It is of interest to determine whether equality holds in (1.7). To address this question, and to investigate the behavior of the integral with respect to limits, we consider the following result.

Theorem 1.3 (Beppo Levi). Let (X,M,p.) be a measure space and assume {In} is a nondecreasing sequence of nonnegative finite p.-a.e. measurable functions defined on X. Then, lim n -+ oo In(x) = I(x) exists everywhere on X, I(x) is nonnegative and measurable and

lim f In dp. (= sup f In dP.) Jxf I dp. =n-+oo Jx Jx

.

(1.9)

VII.

110

Integration

Proof. That I is nonnegative and measurable is clear. By monotonicity, the numerical sequence In dJ.L, n = 1,2, ... , is nondecreasing, and consequently, it has a limit L, say. Also, by monotonicity, In dJ.L ~ I dJ.L for all n, and

Ix

Ix

Ix

L = sup n

f In dJ.L ~ f I dJ.L . Jx Jx

(1.10)

If L = 00, the right-hand side of (1.10) is also 00, and (1.9) holds in this case. On the other hand, if L is finite we must show the inequality opposite to (1.10), and this requires some work. Given 0 < TJ < 1 and E F" let En = {In> TJ}; {En} is a sequence of measurable sets and since the In's are non decreasing and ~ I, it readily follows that En ~ En+! for all n, and that lim En = X. Consider now the measure veE) = IE dJ.L, E E M, and observe that by monotonicity we have

(1.11) By (1.10) and (v) in Theorem 1.1 both sides of (1.11) have a finite limit as n --+ 00. Whence, taking limits there we obtain at once L

~ n~oo lim TJv(En) =

Now, (1.12) holds for each



TJv(X) = TJ

Jxf dJ.L.

(1.12)

E F" and taking sup over F, we get

TJ

L

I dJ.L

~ L.

Since TJ < 1 is arbitrary it is clear that (1.13) is also true with TJ the inequality opposite to (1.10) holds. •

(1.13)

= 1 and

This result of Beppo Levi (1875-1961), also known as the Monotone convergence theorem, or MCT, has many important applications; before we discuss them we present a simple extension of MCT, also useful in applications. Corollary 1.4 (J.L-a.e. version of MCT). Let (X, M, J.L) be a measure space and {In} a J.L-a.e. nondecreasing sequence of nonnegative finite 1'a.e. measurable functions defined on X. Then limn~oo In(x) = I(x) exists J.L-a.e on X, I( x) is nonnegative and measurable and

lim f n~oolx

In dJ.L = lx f I dJ.L •

1.

The Integral of Nonnegative Functions

111

Proof. Let N be the null set outside of which the In's increase to I, and put 9n = In on X \ N, and 9n = 0 on N, and 9 = Ion X \ N, and 9 = 0 on N. The point is that now the 9n'S converge to 9 everywhere, and they coincide with the In's and I at the level of integrals. More precisely, by property (i) of Theorem 1.2,

Ix I

dl'

=

Ix

9 dl' ,

and

Ix In = Ix 9n dl'

dl' ,

all n,

and consequently

lim n-+oo

Jxr In dl' =

lim n-+oo

Jxr9n dl' = Jxr9 dl' = Jxr I dl' .

•

As for the applications, we do the additivity and u-additivity of the integral first. Proposition 1.5. Suppose (X,M,I') is a measure space and let 1,9 be nonnegative extended real-valued measurable functions defined on X. Then (1.14) Proof. Let {
Ix (
dl'

dl' ,

all n.

(1.15)

Ix(J +9) dl' as n -+ 00, Ix I dl' + Ix 9 dl' as

By MCT the left-hand side of (1.15) converges to and also by MCT the right-hand side converges to n -+ 00. Thus (1.14) holds. • We are now ready to prove

Theorem 1.6. Assume (X,M,I') is a measure space, suppose {In} is a sequence of nonnegative extended real-valued measurable functions defined on X and let I = 2:n In. Then I is nonnegative, extended realvalued and measurable, and

{ I dl' = L:

Jx

n

{ In dl'.

Jx

(1.16)

VII.

112

=

Integration

=

Proof. Let Sk E:=l In, k 1,2, ... , and observe that the Sk'S form a non decreasing sequence of nonnegative, extended real-valued measurable functions defined on Xj moreover, limk-+oo Sk = I, and I is nonnegative, extended real-valued and measurable. By MeT

{ I dJ.L = lim { Sk dJ.L .

Jx

(1.17)

Jx

k-+oo

Now, by (a simple extension of) Proposition 1.5, we have

1

k

Sk dJ.L =

X

~

n=l

1

In dJ.L,

X

and consequently,

k~~

Ix

k

SkdJ.L =

l~~~ n=l

Ix

00

IndJ.L =

~

n=l

Ix

IndJ.L.

Whence, replacing this expression in the right-hand side of (1.17), (1.16) follows. • An interesting consequence of Theorem 1.6 is

Proposition 1.7. Suppose (X,M,J.L) is a measure space and let I be a nonnegative extended real-valued measurable function defined on X. Then the set function

v( E)

=

l

I dJ.L ,

E EM,

is a measure on (X,M).

Proof. That v is nonnegative and v(0) = 0 is clear. As for the a-additivity, let { En } be a sequence of pairwise disjoint measurable sets, and put In IXEn' n 1,2, ... The sequence {In} satisfies the hypothesis of Theorem 1.6, and consequently, by (1.16) we have

=

=

LV(En) = L { IndJ.L = { LlndJ.L n n Jx Jx n

= Jx { ILXEn dJ.L = Jx ( IXUEn dJ.L = v(UEn)' In other words, v is a-additive.

•

It is natural to consider whether Theorem 1.3 can be extended to include more general classes of functions. First we discuss a result due to Fatou (1878-1929); since its statement involves the liminf it applies to arbitrary sequences.

1.

The Integral of Nonnegative Functions

113

Thereom 1.8 (Fatou's Lemma). Suppose (X,M,JL) is a measure space and let {In} be a sequence of nonnegative extended real-valued measurable functions defined on X. Then

Ix Proof. spot. Let

liminf In dJL

~ liminf

Ix

IndJL.

(1.18)

The idea of the proof is to invoke MCT at the appropriate 9k

= inf In, n~k

k = 1,2, ...

The sequence {9k} satisfies the following properties: (i) The 9k are nonnegative extended real-valued measurable functions. (ii) The sequence is nondecreasing. (iii) 9k ~ In, all n ~ k ~ 1. (iv) limk-+oo 9k = sUPk~l (infn~k In) = lim inf In. By (iv) and MCT we have lim

f

k-+ooJx

9k dJL =

f

lim 9k dJL =

Jx k-+ oo

f

Jx

liminf In dJL.

(1.19)

On the other hand, by (iii) and monotonicity we get

Ix

9k dJL

~

Ix

In dJL ,

all n

~ k,

and consequently for each fixed k we have

L 9k dJL

~ Hminf Lin dJL.

(1.20)

Combining now (1.19) and (1.20) it readily follows that (1.18) holds, and we have finished. • It is not hard to see that strict inequality may occur in Fatou's Lemma. Indeed, for the Lebesgue measure on [0,1] and the sequence Un} given by In(x) = nX(O,l/n)(X), n = 1,2, ... , we have Hminf In(x) = 0 for all x, and consequently, the left-hand side of (1.18) is 0, but itO,l] In dJL = 1 for all n, and the right-hand side there is 1. Fatou's Lemma is very important in applications; Fatou discovered it while investigating the convergence properties of the Poisson integral, a problem that lies at the heart of Harmonic Analysis. A related result will be discussed in Theorem 3.1 in Chapter XVII.

VII.

114

Integration

2. THE INTEGRAL OF ARBITRARY FUNCTIONS As interesting a result as Theorem 1.8 is, it still does not address the question of interchanging limits with integrals. We discuss this general question in the context of integrals of functions of arbitrary sign. Let (X,M,JL) be a measure space, and let f be an extended realvalued measurable function defined on X; we can write f = f+ - f-:as the difference of two nonnegative functions. In particular, both of the integrals f+ dJL and f- dJL exist, and if either of these integrals is finite we define the integral f dJL of f over X with respect to JL as

Ix

Ix

Ix

(2.1) Observe that if

f and 9 are measurable and f = 9 JL-a.e., then

f + -- g+, f- -- 9 - JL-a.e. and consequently, by (i) in Theorem 1.2, the integral of f over X with respect to JL exists iff that of 9 exists, and in this case they are equal. Some of the properties of the integral are readily obtained, for instance (2.2) and (2.3) Indeed, (2.2) follows at once from (2.1), and (2.3) is a simple consequence of this observation: For A > 0 we have (AJ)+ = Af+ and (AJ)- = Af-, and for A < 0 we have (AJ)+ = -Af- and (AJ)- = -Af+. An interesting theory may be developed for those functions for which both of the integrals on the right-hand side of (2.1) are finite. This class of functions is denoted by L1(X,JL), or plainly L(X,JL), L(X), or L(JL), and it is called the Lebesgue class Ll; the functions f in Ll are said to be integrable. Note that since (2.4) and for functions I in L(JL) the expression Ix (1+ + 1-) dJL = Ix III dJL is finite, it is built into the definition of L(JL) that I E L(JL) iff the integral

2.

The Integral of Arbitrary Functions

Ix I dJ.L

115

IIx

is defined and I dJ.L1 < 00. On the other hand, as simple examples show, it is possible for III to be integrable, and yet for I not to be measurable. Thus the measurability of I is essential in the definition of L(J.L). The following estimate, in the spirit of (3.1) in Chapter III, is quite useful. Proposition 2.1. Let (X,M,J.L) be a measure space, and let I be an extended real-valued measurable function defined on X for which the integral over X with respect to J.L is defined. Then we have

Ii

I dJ.L1

~

i

(2.5)

III dJ.L.

Proof. If the right-hand side of (2.5) is infinite there is nothing to prove. On the other hand, if IE L(J.L), then by (2.4) and Proposition 1.5 the integrals of 1+ and 1- over X with respect to J.L are also finite and we have

Ii

I dJ.L1 =

~

Ii

1 +i

1+ dJ.L -

i 1+

dJ.L

1- dJ.L1

1- dJ.L =

i (1+ +1-) = i dJ.L

III dJ.L. •

When I is integrable, (2.5) may be interpreted as a statement concerning its "size." A more precise estimate was proved by Chebychev (1821-1894); the result is known as Chebychev's inequality. Theorem 2.2. Let (X,M,J.L) be a measure space and suppose I is an extended real-valued measurable function defined on X. Then for any real A > 0 we have AJ.L({I/I > A})

~

i

I/ldJ.L.

(2.6)

Proof. Let A..\ = {III > A}; A..\ is measurable and AXA>. < III J.L-a.e. Thus, by Theorem 1.2 (iii), it readily follows that

VII.

116

Integration

Corollary 2.3. Let (X,M,JL) be a measure space and let Then I is finite JL-a.e. Further, if I is nonnegative and I dp 1= 0 JL-a.e.

Ix

Proof.

I

E L(JL).

= 0, then

By Chebychev's inequality we have

JL( {III> n})

:::;!n lxfill dJL -+ 0

as n

-+ 00 •

Whence JL({I/I

= oo}) = nlim JL({I/I ..... oo

> n})

= 0,

and I is finite JL-a.e. Moreover, if I is nonnegative and its integral over X vanishes, then, also by Chebychev's inequality, JL({I > A}) = 0 for all A > 0, and I vanishes JL-a.e. • How does the integral behave with respect to addition? Proposition 2.4. Let (X,M,JL) be a measure space and suppose A is a real number and I, 9 E L(JL). Then I + Ag is integrable and

Ix (J +

Ag) dJL

=

Ix I

dJL + A

Ix

9 dJL .

(2.7)

Proof. The integrability of 1+ Ag follows at once from the estimate II + Agi :::; III + IAllgl· Now, since h = 1+ Ag is integrable, the integral of h over X with respect to JL is a well-defined finite number. Furthermore, we have h+ - h- = 1+ - 1- + (Ag)+ - (Ag)- ,

and consequently, also h+

+ 1- + (Ag)- = h- + 1+ + (Ag)+ .

(2.8)

All the summands in (2.8) are nonnegative, and by Corollary 2.3 finite JL-a.e. By Proposition 1.5 then, it readily follows that

fx

h+ dJL+

fx

1- dJL+

fx

(Ag)- dJL =

fx

h- dJL+

fx

1+ dJL+

fx fx

(Ag)+ dJL,

and since all the integrals are finite we may move them freely and obtain

L L h+ dJL -

h- dJL =

Thus (2.7) holds.

L L L 1+ dJL -

1- dJL+

(Ag)+ dJL -

(Ag)- dJL.

•

The following variant of Proposition 2.4 is important in applications since it allows us to consider arbitrary functions for which the integral is defined.

2.

The Integral of Arbitrary Functions

117

Proposition 2.5. Let (X,M,JL) be a measure space and suppose j, 9 are extended real-valued measurable functions defined on X which satisfy the following conditions: The integral of j over X with respect to JL is defined, and 9 is integrable. Then the integral of j + 9 over X with respect to JL is defined and (2.9) Proof. By Proposition 2.4, (2.9) is only novel when f is not integrable. If this is the case one of the quantities, f+ dJL or I~ f- dJL, is infinite and the other one is finite. To fix ideas suppose that Jx f- dJL = 00, and observe that with the notation of Proposition 2.4, recalling that h = f + 9 and setting>. = 1 in (2.8), we have

Ix

(2.10)

Ix

Ix

Ix

Since f+ dJL, g+ dJL < 00, from (2.10) it follows that h- dJL = 00. Furthermore, since h- = 0 when h+ 'I 0, by (2.10) we also get that h+ ~ f+ + g+, and consequently h+ dJL < 00. Thus the integral of h over X with respect to JL exists and it equals -00. Whence the left-hand side of (2.9) equals -00, and so does the right-hand side. Thus (2.9) holds, and we are done. •

Ix

We are now in a position to explore what Fatou's Lemma says in the general context of functions of arbitrary sign. Theorem 2.6 (Fatou's Lemma). Let (X,M,JL) be a measure space and suppose {fn} is a sequence of extended real-valued measurable functions defined on X which satisfy the following property: There exists an integrable function 9 such that 9 ~ fn

(2.11)

all n.

Then the integrals of lim inf fn and fn over X with respect to JL exist, n = 1,2 ... , and we have

Ix

lim inf fn dJL

~ lim inf

L

fn dJL •

(2.12)

VII.

118

Integration

Proof. Since by (2.11) In - 9 ~ 0, the integral of the functions In - 9 over X with respect to p, is well-defined for n = 1,2, ... ; similarly for liminf In - g. Now, by Fatou's Lemma for nonnegative functions we have

Ix

lim inf(Jn - g) dp, =

~

Ix

(lim inf In

lim inf

- g) dp,

Ix (J

n -

g) dp, .

(2.13)

First we consider the left-hand side of (2.13). By Proposition 2.5 with I = liminf In - 9 and 9 = 9 there, we get that 1+ 9 = liminf In has a well-defined integral over X with respect to p, which satisfies

Ix

lim inf In dp,

=

Ix

(lim inf In

- g) dp, +

Ix

(2.14)

9 dp, .

Since 9 is integrable, by (2.14) it readily follows that the left-hand side of (2.13) equals

Ix

lim inf In dp, -

Ix

(2.15)

9 dp, .

A similar argument gives that the integral of In over X with respect to p, exists, n = 1,2, ... , and that the integral that appears on the right-hand side of (2.13) is equal to

Ix In

dp, -

Ix

9 dp, ,

(2.16)

n = 1,2, ...

Thus combining (2.15) and (2.16) we may rewrite (2.13) as

Ix

lim inf In dp, -

fx

9 dp,

~ lim inf

Ix In

dp, -

fx

9 dp, .

Ix

Since 9 is integrable we may now cancel 9 dp, in the above inequality. Whence (2.12) holds, and we have finished. • There is a version of Theorem 2.6 with the inequality (2.12) reversed, but with the lim inf replaced by the lim sup. Corollary 2.7. Let (X,M,p,) be a measure space and suppose {In} is a sequence of extended real-valued measurable functions defined on X

2.

The Integral of Arbitrary Functions

119

which have the following property: There exists an integrable function 9 such that (2.17) In:5 g, all n. Then the integrals of lim sup In and n = 1,2, ... , and we have

In

over X with respect to J.l exist,

(2.18) Proof.

Observe that (2.17) is equivalent to -g

:5 - In,

all n,

and that 9 is integrable iff -g is integrable, in other words, the hypotheses of Fatou's Lemma are satisfied by {- In} and -g. Since lim inf( - In) = -lim sup In, by Theorem 2.6 we have

-Ix

lim sup In dJ.l

=

Ix

lim inf ( -

:5 liminf

In) dJ.l

Ix (- In) Ix In

= -lim sup

dJ.l

= liminf

(-Ix In

dJ.l.

dJ.l ) (2.19)

Now, (2.19), whether involving finite quantities or not, is equivalent to (2.18), and we have finished. • Some remarks concerning these results: Clearly we may assume that (2.11) and (2.17) hold J.l-a.e. and obtain the same conclusion; also strict inequality may occur in estimates (2.12) and (2.18). For instance, for the Lebesgue measure on 1= [0,1] and the sequence In = X[O,3/4)' n odd, and In = X[3/4,1],n even, we have

and lim sup

1In

dJ.l = 3/4 <

1

lim sup In dJ.l = 1.

We close this section with the Lebesgue dominated convergence theorem, or LDCT, which describes under what conditions we may pass to the limit under the integral sign. \

VII.

120

Integration

Theorem 2.8 (LDCT). Let (X,M,JL) be a measure space and suppose {In} is a sequence of extended real-valued measurable functions defined on X such that (i) lim n -+oo In = I exists JL-a.e. (ii) There is an integrable function 9 so that for each n, IInl ~ 9 JL-a.e. Then I is integrable and lim f IndJL. Jxf IdJL = n-+oo Jx

(2.20)

Proof. By (i) it readily follows that I is measurable and by (ii) that III ~ 9 JL-a.e., thus I is also integrable. As for the In's, by (ii) they are integrable and for any n we have -9 ~

In

~ 9

JL-a.e.

So, Fatou's Lemma and its corollary apply, and since lim sup In JL-a.e., it readily follows that

I

= liminf In

L I dJL ~ liminf L In dJL

~ lim sup L IndJL ~ LIdJL . Whence, all four quantities in the above inequality are equal (and finite), and (2.20) holds. • By the way, the example following Theorem 1.8 shows that, in the absence of an integrable majorant, the conclusion of LDCT may fail.

3. RIEMANN AND LEBESGUE INTEGRALS Suppose 9 is a Riemann integrable function over an interval I. Does the integral of 9 over I with respect to the Lebesgue measure exist? If so, do both integrals coincide? In other words, we would like to know whether the notion of Lebesgue integral extends that of Riemann integral. First some notations. Lebesgue measurable functions will be called measurable functions, Lebesgue integrable functions will be called integrable, Lebesgue a.e. will be denoted plainly by a.e. and the integral of 9 over I with respect to the Lebesgue measure is denoted by 11 9 dx. We then have

3.

Riemann and Lebesgue Integrals

121

Theorem 3.1. Let 9 be a bounded real-valued function defined on 1= [a,b] and suppose that 9 E n(I). Then 9 E L(I) and

1b

(3.1)

g(x)dx = i9dX.

Proof. Let Pn = {a = Xl,n < ... < Xkn,n = b},n = 1,2, ... , be a sequence of finite partitions of I such that P n +1 is a refinement ofPn , n ~ 1, and so that the norm of P n ~ 0 as n ~ 00. If h,n = [Xk,n,Xk+1,n] , 1 ~ k ~ k n - 1, are the intervals induced by P n , and

mk,n = inf g, I",n

Mk,n

= supg I",n

note that the functions

=L

Ln(x)

mk,nXI",JX),

Un(x)

and

= LMk,nXI",JX) k

k

are bounded and measurable, and hence integrable over I. Now, the sequence {Ln} is non decreasing and the sequence {Un} is nonincreasing, and consequently, the limits

L(x) = lim Ln(x) , n~CX)

U(x) = lim Un(X) ,

and

n~oo

(3.2)

exist and are finite everywhere on I. Furthermore, Land U are measurable, and L(x) ~ g(x) ~ U(x), x E 1. (3.3) Next observe that for n

= 1,2, ... we have (3.4)

and

f Un dx = L

11

Mk,nI1k,nl

= S(g, Pn).

(3.5)

k

Since 9 is bounded the Ln's and the Un's are uniformly bounded on I, and consequently, by LDCT we get that

f L dx =

JI

lim f Ln dx , n-ooJI

f U dx =

JI

lim f Un dx . n_oo JI

(3.6)

VII.

122

Integration

Moreover, since 9 E 'R,(I), by (3.4), (3.5) and (3.6) we have

i

L dx =

i

(3.7)

U dx = lb g(x) dx .

From (3.7) it readily follows that the integral of the nonnegative function U - L over I vanishes, and by Corollary 2.3 we have that U = L a.e. Hence, by (3.3) we obtain that 9 = U a.e., and since the Lebesgue measure is complete, by Theorem 1.4 in Chapter VI we get that 9 is also measurable. By (3.3) it now follows that (3.1) holds, and we have finished. • The converse to Theorem 3.1 is false, to wit, there are bounded integrable functions 9 defined on I which are not Riemann-integrable; the characteristic function of the rationals in I will do. The notion of Riemann-integrability incorporates unbounded functions by means of the so-called "improper" convergence methods. For instance, suppose 9 is unbounded on I, but it satisfies the following properties: (i) For each 0 < c < b - a, 9 E 'R,([a + c,b]). (ii) lime.... o+ J:+e g( x) dx = J:+ g( x) dx exists.

J:-

J:;

Of course there are similar definitions for g( x) dx and g( x) dx. 1 2 For instance, the function g(x) = x- / is unbounded on (0,1], but

Functions for which the improper Riemann integral exists are also integrable, as our next result shows. Theorem 3.2. Suppose that the nonnegative function 9 is defined and finite on I = (a,b] and that J:+ g(x) dx exists. Then 9 E L([a,b]) and

f gdx =

Jf

lb g(x)dx. a+

Proof. Let 0 < en < b-a be a sequence which tends to 0 as n - 00. By Theorem 3.1 the functions gn = gX[a+En,b), n = 1,2, ... , are integrable on I and f gn dx = fb g( x) dx. Jf Ja+En

3.

Riemann and Lebesgue Integrals

123

Moreover, since the gn's increase to 9 on I (the value of 9 at a is irrelevant) by MeT it readily follows that

t g( x) dx = n-+oo lim f gn dx = f 9 dx . Ja+ JI JI

•

A similar result holds for the other improper integrals. On the other hand, it is also possible to consider the Riemann integral of a function 9 defined on R by simply letting

1

g( x) dx = lim

00

-00

N-+oo

IN g( x) dx , -N

whenever the limit exists. In this case the function g( x) = sin x / x is Riemann integrable over R, but not integrable there since, as is readily seen, iRlg(x)ldx = 00. With the aid of the Lebesgue measure we are also able to identify those functions which are Riemann integrable on finite intervals. Theorem 3.3. Suppose 9 is a real-valued bounded function defined on I = [a,b]. Then 9 E 'R(I) iff 9 is continuous a.e. on I. Proof. First assume 9 E 'R(I), and fix a sequence of partitions {Pn } of I as in Theorem 3.1. If L and U are defined by (3.2), let

N = {x E I:L(x) < g(x) or g(x) < U(x)}. In Theorem 3.1 we actually proved that INI = O. Now let N' be the set consisting of all those points in I which belong to some P n ; N' is countable and hence null. We claim that 9 is continuous off the null set NUN'; we prove this by contradiction. Given an interval J ~ I, consider the oscillation osc (g, J) of 9 over J given by osc (g, J) = sup 9 - inf 9 . J

J

Now, if 9 is not continuous at x ~ NUN', there exists e > 0, such that the oscillation of 9 on any interval containing x is at least e. Since x ~ N', x is an interior point to one of the intervals of each Pn , and consequently,

Un(x) - Ln(x)

~

e

for all n.

Fromfoll (3.2) it follows that if this is the case, then U(x) - L(x) and x EN, which is the desired contradiction.

~

e,

VII.

124

Integration

Conversely, suppose 'Pn is an arbitrary sequence of partitions of J such that limn-+oo(norm 'Pn) = 0 and observe that by assumption, lim Ln(x) = lim Un(x) = g(x) a.e.

n-+oo

n-+oo

By (3.4), (3.5) and LDCT we get that lim 8(g, 'Pn) = lim S(g, 'Pn) = [g dx .

n-+oo

h

n-+oo

Whence, by Proposition 2.1 in Chapter III, 9 E R(J).

•

4. PROBLEMS AND QUESTIONS The first four problems describe how the concept ofintegral was viewed by different mathematicians. For simplicity we assume that Jl is the Lebesgue measure defined on (X, C), where X is a compact interval of Rn, and that I is a nonnegative measurable function defined on X. 4.1 Lebesgue defined the integral of a bounded measurable function as follows: If 0:::; I:::; M a.e. on X, put

1 x

I dx

mM

I

k

= m-+oo lim L -I{klm:::; 1< (k + l)lm}l· k=l m

Show that this definition coincides with the one given in the text. 4.2 de la Valh~e-Poussin (1866-1962) extended the definition of Lebesgue

to include unbounded functions I as follows: Let 1m = AU, m) be the truncation of I at level m. {1m} is a sequence of bounded measurable functions, cf. 4.5 in Chapter VI, and we put

[ I dx =

Jx

lim [ m-+oo Jx

1m dx.

Show that this definition is equivalent to the one given in the text. 4.3 Saks (1897-1942) defines the integral in a manner reminiscent ofthe Riemann-Stieltjes integral. More precisely, let 'P = {E1 , • •• ,Ek} be a measurable partition of X, let mk = infEIc I, and put

4.

Problems and Questions

125

Show that this definition is also equivalent to the one given in the text. 4.4 Finally, there is the notion of integral as the "area under the graph." CaratModory defines the integral of a bounded measurable function f as follows: Let AU) = ((x,y) E X X R:O ~ y ~ f(x)}; show that AU) is a Lebesgue measurable subset of R n +1 and put

Lf

dx

= IA(J)I·

(4.1)

Show that (4.1) is equivalent to the definition given in the text. It is also interesting to interpret results such as MCT in terms of (4.1), as their meaning is quite apparent. ( 4.5JTrue or false: If f is a nonnegative function defined on Rand IR f dx < 00, then limlxl-+oo f(x) = O. 4.6 Suppose f is integrable on R n and for a fixed h E R n let g(x) = f(x + h) be a translate of f. Show that 9 is also integrable and that

f

iRn

gdx

= f

iRn

fdx.

(4.2)

(4.2) is a restatement of the translation-invariance of the Lebesgue measure. 4.7' Let (X,M,JL) be a measure space and {fn} a sequence of mea- surable functions such that L:n I~ Ifni dJL < 00. Show that L:n fn converges absolutely JL-a.e. and Jx (L:n fn) dJL = L:n Ix fn dJL. In particular, also lim n -+ oo fn = 0 JL-a.e. 4.8 Let Tl, T2, ••. ,Tn, .•. be an enumeration of the rational numbers in 1= [0,1], and let f(x) = 2- n •

L

{n :X>rn}

Compute IIf(x)dx. 4.9 Prove that the sum L:~=o ito,1r](1 - "'sin x)n cos x dx converges to a finite limit, and find its value. 4.10 Let f be a nonnegative measurable function defined on R. Prove that if L:~=-oo f( x + n) is integrable, then f = 0 a.e. On the other hand, if I is integrable, then 4>( x) = L:~=-oo 1(2nx + 1/ n) is finite a.e. and integrable, and JR 4>( x ) dx = JR I( x ) dx.

VII.

126

Integration

4.11 Let (X,M,JL) be a measure space and IE L(JL). Show that the set {I i: O} is O'-finite, i.e., the at most countable union of sets of finite measure. 4.12 Referring to Proposition 1.7, decide when the measure v introduced there is: ( a) finite, and (b) 0'- finite. 4.13 Referring to Proposition 1.7 again, suppose that 9 is a nonnegative 9 dv = gl dJL. measurable function defined on X. Show that

Ix

Ix

4.14 Suppose that the assumptions of 4.27 in Chapter IV hold and let I be a nonnegative real-valued measurable function defined on Y. Prove that the following "change of variable" formula holds:

4.15 Show that Fatou's Lemma is also true for functions that depend on a continuous parameter. More precisely, under the relevant assumptions, the following is true: lim infiEI Ii dJL ~ lim infiEI Ii dJL.

Ix

Ix

4.16 Prove the following variant of Fatou's Lemma: If {In} is a sequence of nonnegative measurable functions which converges to I JL-a.e. and In dJL ~ M < 00 for all then I is integrable and I dJL ~ M.

Ix

n,

Ix

4.17 Decide whether the following Fatou-like statements are true: (a) If {In} is a sequence of nonnegative measurable functions and In converges to I in probability, then I dJL ~ liminf In dJL, and, (b) Same result with convergence in probability replaced by convergence in measure.

Ix

Ix

4.18 Show that the following extension of Fatou's Lemma is true: Rather than assuming 9 E L(JL), we may assume that g- dJL < 00 in g+ dJL < 00 in Corollary 2.7. Theorem 2.6, and that

Ix

Ix

4.19 Describe the relation of 4.26 in Chapter IV to Fatou's Lemma. 4.20 Let (X,M,JL) be a measure space and {
¢>k'S, call it ¢>, so that

iR" (j - ¢» dx < e/2.

We have thus reduced the problem at hand to one of approximating simple integrable functions by continuous functions in the metric of L(Rn).

VIII.

134

More About £1

Suppose = E~=l CkXEk' where for each k, Ck > 0 and Ek is a measurable set of finite measure, 1 ~ k ~ m. It suffices now to approximate each summand that appears in the definition of , or equivalently, the characteristic function of a measurable set E, say, of finite measure. By the regularity of the Lebesgue measure, for any "I > 0, there is an open set 0 of finite measure such that

10\EI<"I,

O~E.

This estimate, in particular, implies

and since "I above is arbitrary, we may assume that the set E in question is actually open. It is at this point that the geometry of the situation plays a role. Let 0 = Uk:l h be an open set of finite measure; here the Ik's are nonoverlap ping closed intervals. Since 101 = Ek:llhl < 00, it readily follows that

and consequently it suffices to approximate the characteristic function of U~=l Ik, all finite m. But then it is enough to consider XI' where I is a closed interval of Rn. Suppose first that n = 1 and I = [0,1], and given "I> 0, let t/J." be the continuous function

0

1

1 + xj"l 1 ~"I+1-x)j"l

t/J.,,(x) =

Then clearly t/J."

~ Xb

if x < -"I if -"I ~ x < 0 if 0 ~ x ~ 1 if 1 < x ~ 1 + "I if x> 1 + "I.

and

L

(t/J." - XI) dx = "I ~ 0 with "I.

Now, if 1= [a,b], the function t/J." «x - a)j(b - a» does the job.

2.

Metric Structure of L1

135

=

As for the general case, observe that if I [at, b1 ] X ••• X [an, bn), then XI can be approximated in the metric of L( Rn) as close as we want by the continuous function

7/J1/

(Xlb - a1) X ••• X 1/;1/ (xn b 1 -

n

al

an), an

TJ small . •

Theorem 1.3 indicates that simple functions are dense in L(J.L). As for the Lebesgue integral, one of its important applications concerns the continuity of the translates of integrable functions. Proposition 1.4. Integrable functions are continuous in the metric of L(Rn). More precisely, for any IE L(Rn) we have lim

f

Ihl--+O iRn Proof.

I/(x + h) - l(x)1 dx = 0.

We show that given c

f I/(x + h) iRn

l(x)1 dx

> 0, there exists 6 > 0, such that

~ c,

whenever Ihl

~ 6.

(1.5)

This is not hard. First let 9 E Co(Rn) be such that dU,g) ~ c/3, and observe that I/(x + h) - l(x)1 may be estimated by

g(x + h)1 + I g(x + h) - g(x)1 + Ig(x) - l(x)l. (1.6) Whence, integrating (1.6) over Rn, we get that the integral on the lefthand side of (1.5) does not exceed I/(x

+ h) -

f I/(x+h)-g(x+h)ldx+ f Ig(x+h)-g(x)ldx+ f I/(x)-g(x)ldx iRn iRn inn =A+B+C, say. By 4.6 in Chapter VII we have A = C for all h, and, by our choice of g, A,C ~ c/3. As for B, note that 9 is actually a uniformly continuous function that vanishes off a bounded interval of Rn. Thus, given TJ > 0, there exists 6 > such that Ig(x + h) - g(x)1 ~ TJ for alllhi ~ 6 ,X E Rn. Moreover, since for any fixed h, Ihl ~ 1, also g(x + h) - g(x) vanishes off a bounded interval I of Rn, it is clear that

°

B ~

III TJ,

whenever Ihl ~ min (1,6). Thus, by choosing TJ small enough, we also have B ~ c/3 whenever sufficiently small, and A + B + C < c. •

Ihl is

VIII.

136

More About Ll

2. THE LEBESGUE DIFFERENTIATION THEOREM Given x = (xt, ... ,xn ) E R n and r > 0, let I(x,r) = {y: IXi - Yil < r, i = 1,2, ... ,n} denote the open interval of sidelength 2r centered at x. The question we address in this section is: If I is an integrable function, for what x's does lim

f

1

r-+O II(x, r)1 JI(:c,r)

I dy = I( x)

?

(2.1)

At those points x where (2.1) holds we say that the (indefinite) integral of I differentiates to I(x). In case n = 1, the question is whether lim .!.

j

I dy = I( x ) ,

.!. f

Idy

r-+O 2r (:c-r,:c+r) which is equivalent to lim

h-+O,ht=O h J[:c,:c+h)

= I(x).

(2.2)

If we set F(x) = J[O,:c)ldy, (2.2) reads precisely F'(x) = I(x), and this justifies the terminology. In fact, there are two questions implicit in (2.1): When does the limit exist, and, if it exists, when does it equal I(x). For instance, when 1= [0,1] and I = XI' we have

.!. f

2r J(-r,r)

I dy = 1/2, all r > 0.

(2.3)

Thus the limit of the left-hand side of (2.3) exists and it equals 1/2 =I1(0) = 1. Some observations are in order. First, the question we posed is "local" in nature, i.e., since we take limits as r -+ only the values of I near x are relevant. Thus, we may assume that x E 1(0,1) and that I vanishes off 1(0,2). Next, since the example given in (2.3) is not very reassuring, we consider an instance where (2.1) is true. Suppose, then, that I is continuous at x and note that

°

11(:r)1 f IdY-/(x)=II(:r)1 f (f-/(x))dy. , JI(:z:,r) , JI(:z:,r)

(2.4)

2.

The Lebesgue Differentiation Theorem

137

Given c > 0, let r be so small that I/(x)- l(y)1 we may assume r < 1. By (2.4) it follows that

II( 1 )1 x,r

r

JI(x,r)

Idy-/(x)

~

II( 1 )1 x,r

< c for y

r

JI(x,r)

E I(x,r)j clearly

1/-/(x)ldy~c,

and (2.1) is true in this case. Since continuous functions are dense in the metric of L(Rn), we expect that the good behaviour of continuous functions will somehow translate into an a.e. good behaviour of integrable functions. The idea of Hardy (1877-1947) and Littlewood (1885-1977) is to seek the control of all the averages of I. They devised this procedure to study the convergence of Fourier series and were inspired by the averages in the game of cricket. To control the averages of I we introduce the so-called HardyLittlewood maximal function. Specifically, suppose I is an integrable function which vanishes off 1(0,2), and for x E R n put

M(J)

= sup II( 1 )1 r i l l dy . r>O x, r JI(x,r)

(2.5)

What can we say about M I? We claim that M I is a nonnegative lowersemicontinuous, and hence measurable, function, which tends to as Ixl 00 at the rate of Ixl-n. To show that M I is lower-semi continuous we must verify that for each oX > 0, the set {M I > oX} is openj this is not hard. Working with complements we show that for each oX > 0, {M I ~ oX} is closed. Fix oX > 0, then, and suppose {Xk} is a sequence of points in {M I ~ oX} such that Xk Xj we show that x E {M I ~ oX} as well. In other words, we check that all the averages of III about x are less than or equal to oX. First observe that since Xk - x,

°

lim I(xk, r)

k~oo

~

I(x, r) = 0,

all r.

Therefore, if Xk denotes the characteristic function of I(xk,r) and ik = IXk' it follows that

I/k(y)1

~

I/(y)1 and

lim

k~oo

!k =

~

I(x,r)

°

a.e.

Thus, by LDCT

(2.6)

VIII.

138

Next consider the average II(~,r)1

fI(x,r)

More About L1

If I dy. Since

the average in question does not exceed

r

1

11(x,r)1 JI(x,r)~I(xk,r)

+

If I dy

1

11(xk,r)1

r

If I dy = A

+ B,

JI(xk,r)

say. By (2.6), A -+ 0 as k -+ 00. As for B, since M f( Xk) ::; A for all k, we also have B ::; A. Thus all averages of If I about x are less than or equal to A, and Mf(x) ::; A. Next we show that Mf(x) Ixl- n for Ixllarge. To see this take x E Rn with Ixllarge, x E Rn \ 1(0,10) will do, and observe that unless r> clxl, where c is a dimensional constant independent of x, the average of If I about x vanishes. Thus, with a dimensional constant c which may differ at different occurrences even in the same chain of inequalities, we have f'V

1

11(x,r)1

r

If I dy ::;

JI(x,r)

~ rn

r

and consequently,

Mf(x)::;

If I dy ::; _c_

Ixl n

JI(O,l)

ICln x

r

r

If I dy ,

JI(O,l)

(2.7)

Ifldy.

JI(O,l)

Since there is a dimensional constant c such that for Ixl large we have 1( x ,clx I) ~ 1(0,1), it readily follows that

Mf(x) >

1 - 11(x,clxl)l

r

JI(x,clxl)

Ifldy> _c_ - Ixl n

r

Ifldy,

JI(O,l)

the inequality opposite to (2.7) holds, and, as asserted, Mf(x) Ixl- n for Ix I large. It is then apparent that M f is not integrable, cf. 4.46 in Chapter VII, but just barely. As for the function Ixl- n, it satisfies a weak integrability condition reminiscent of Chebychev's inequality. More precisely, there is a constant c such that f'V

AI{lxl- n

> A}I

::; c,

all A >

o.

(2.8)

Indeed, if Ixl-n > A, then there is a dimensional constant c so that x E 1(0, CA- 1 / n ) , and

2.

The Lebesgue Differentiation Theorem

The class of those measurable functions

AI{I/I > A}I::; c,

all

139

I

which satisfy the estimate

A> 0,

(2.9)

was studied by Marcinkiewicz (1910-1940). It is called the weak-Ll class of Marcinkiewicz and it is denoted by wk-L(Rn). By Chebychev's inequality, L(Rn) ~ wk-L(Rn), and for integrable functions I, (2.9) is true with c = I/(y)1 dy. On the other hand, Ixl- n E wk-L(Rn) \ L(Rn). The remarkable fact that Hardy and Littlewood proved is that although for I integrable M I is not necessarily integrable, it belongs to wk-L(Rn); in a sense this gives the next best result.

JRn

Theorem 2.1 (Hardy-Littlewood). Suppose I is an integrable function which vanishes off 1(0,2). Then M IE wk-L(Rn), and for any A > 0 we have (2.10) AI{MI> A}I ::; 3n I/ldy.

iR!'

Proof. Given A > 0, let (h = {M I > A}; we want to show that the open set V;, has finite measure and that (2.10) holds. Since by (2.7) MI(x) - 0 as lxi- 00, V;, is a bounded set of finite measure; to show that (2.10) holds requires some work. The following line of reasoning is a prototype of the so-called "covering arguments" and it is due to Wiener (1894-1964). If V;, = 0 there is nothing to prove. Otherwise, let x E V;" and observe that by the definition of M I(x) there exists T = Tx such that 1 Il(x, Tx)1 Clearly

V;, ~

1

I(x,r.,)

III dy > A.

U lex,

Tx).

(2.11)

(2.12)

xED>.

Although the set on the right-hand side of (2.12) appears to be quite cumbersome, with the overlaps and all, things are not as complicated as a first impression might indicate. Since by the regularity of the Lebesgue measure, cf. 3.37 in Chapter V, IV>.I

= sup{IJ(I: J( C V>., J(compact},

(2.13)

VIII.

140

More A bout L 1

it suffices to estimate !KI for each compact subset I( of 0.". Now, for each such a compact set of K we also have

U I(x,Tx)

K ~

xeO",

and, since the I( X, T x) 's are open, by the Heine-Borel Theorem there exist finitely many intervals I(xt, Td, ... ,1(xm' Tm ), say, so that m

I( ~ UI(xi,Ti).

(2.14)

i=l

Clearly we may assume that no interval that appears on the right-hand side of (2.14) is contained in the union of all the others, that is to say, each interval there contributes something to the union. Let T = ma.x{ Tt, ... , Tm} and, by renaming the intervals if necessary, suppose that TI = Tj if more than one Ti equals T just choose any. At this juncture of the argument the geometry of the situation takes over: Observe that if

I(xt,TI)nI(xj,Tj)

=10,

then

I(xj,Tj)

~

I(Xt,3Tl)'

Because ofthis property we discard all the intervals I( Xj, Tj), j =11, which intersect I( xl, TI), and repeat the same procedure with the remaining intervals, i.e., the family ofthose intervals I(xj, Tj) which are disjoint with I( Xl, Td. In other words, we separate an interval with largest sidelength, and then discard all the intervals which intersect it. Since the original family of intervals is finite, after a finite number of steps we are left with a pairwise disjoint family of open intervals I(xt,Tt), ... ,I(Xk,Tk), say, which by (2.14) has the property that k

K ~

UI(xi' 3Ti). i=l

Whence

k

k

IKI ~ LII(xi,3Ti)1 i=l

= 3n LII(xi,Ti)l.

(2.15)

i=l

But the intervals that appear on the right-hand side of (2.15) are special: They are pairwise disjoint and they all satisfy (2.11). Therefore the sum on the right-hand side of (2.15) is less than or equal to

L:\11

k i=l

I(Xi,ri)

IfI dy =

11

:\

k

U i =l I (xi,ri)

IfI dy ~

11 IfI

-

..\

R!'

dy ,

2.

The Lebesgue Differentiation Theorem

141

and consequently, (2.16) Taking the sup of the left-hand side of (2.16) over those K CO)., by (2.13) it follows that (2.10) holds. • We are now ready to address the questions raised in (2.1), to wit, the existence of the limit there, and its precise value. Theorem 2.2 (Lebesgue Differentiation Theorem). integrable function which vanishes off 1(0,2). Then lim Il( 1 )1 ( fey) dy = f(x) x,r JI(x,r)

Suppose

a.e. on 1(0,1).

f is an (2.17)

r-+O

Proof. First we show that the limit on the left-hand side of (2.17) exists a.e. For this purpose consider the function

~(f,x) = limr-+Osup Il( X,1 r )1 JI(x,r) { f dy -liminf lIe 1 )1 ( f dy. r-+O X, r JI(x,r) Although ~ is measurable, we need not make use of this fact to proceed with the proof. However we point out that ~ is a well-defined nonnegative function, and show that ~(f,x) = a.e. on 1(0,1). The idea is to control ~ by the Hardy-Littlewood maximal function. Now, if 9 is continuous it is clear that ~(g, x) = everywhere and consequently, ~(f, x) = ~(f - g, x). Whence it readily follows that for any continuous function 9 we have

°

°

~(f,x)

= ~(f -

g,x)

~

2M(f - g)(x) ,

all x.

Let now A > 0, and note that the above inequality gives

Thus, by the monotonicity of the Lebesgue outer measure and Theorem 2.1, we obtain I{~(f,·)

> A}le

~ I{M(f - g)

> A/2}1

2 ·3n

~ -A-

JR( If -

gl dy.

(2.18)

VIII.

142

More About Ll

Now, since continuous functions are dense in the metric of L(Rn), the integral on the right-hand side of (2.18) may be made arbitrarily smail, and consequently,

I{x E Rn:i)(j,x) > A}le = 0, all A > O. This can only be true if i)(j,x) = 0 a.e., in other words the lim sup is equal to the lim inf and the limit on the left-hand side of (2.17) exists a.e. Next we show that the limit equals

W(j,x)= lim 11(1 )1 r-+O

x, r

f

f a.e. Let

JI(x,r)

f(y)dy-f(x).

W is an a.e. well-defined nonnegative function, and it is apparent that for continuous 9 W(j,x) = W(j-g,x), all x. Whence it readily follows that W(j,x) ~ M(j - g, x) + If(x) - g(x)l, and consequently, for A > 0 we have

{x E R n : W(j,x) > A}

~

{x E R n : M(j - g)(x) > A/2} U{x E R n : If(x) - g(x)1 > A/2}.

Thus by the monotonicity of the Lebesgue outer measure, the HardyLittlewood maximal theorem and Chebychev's inequality, it follows that

I{w(j,·) > A}le ~ I{M(j - g) > A/2}1 + 1{lf - gl > A/2}1 2·3 n f 2 f ~ -A- JR'" If - gl dy + X JRn If - gl dy. Since 9 is an arbitrary continuous function, we get that

I{x E R n :W(j,x) > A}le = 0, Whence W(j, x) = 0 a.e., (2.17) holds. •

all A > o.

Although we have only discussed a "local" version of (2.1), it is not hard to obtain the "global" version as well. One way to go about this is to use the general Hardy-Littlewood maximal theorem, cf. 3.23 below, but a simpler way to proceed is this: First observe that Rn = U~l 1(0, 2k). Now, an argument entirely analogous to Theorems 2.1 and 2.2 gives that if I E L(I(O,2k», then the integral of I differentiates to f(x) a.e. on I(O,k). Given an arbitrary integrable function I note that !k = IXI(O,2k) is integrable and it vanishes off I(O,2k), and consequently there exists a null set Nk such that the integral of !k differentiates to !k(x) = I(x) a.e. on I(O,k), k = 1,2, ... Clearly the integral of I differentiates to I(x) off the null set N = U~l Nk, and the global version of the differentiation theorem also holds.

3.

Problems and Questions

143

3. PROBLEMS AND QUESTIONS In what follows (X,M,JL) is a measure space. 3.1 By means of an example show that if f, 9 E L(JL) it is not necessarily true that f 9 is integrable. However, if p, g2 E L(JL), prove that

fg E L(JL) and

Ix

If gl dJL

~

(Ix

f2 dJL ) 1/2

(Ix

g2 dJL) 1/2

(3.1)

(3.1) is known as the Cauchy-Schwarz inequality and it has many interesting applications. One of them is the following: Show that if {fn}, {gn} are sequences with the property that f~,g~ E L(JL), n = 1,2, ... , and if lim [Un - f? dJL = lim [(gn - g)2 dJL = 0, n~oolx

n~oolx

then

3.2 In the spirit of 3.1, show that if f E L(JL) and {fn} is a sequence of integrable functions so that limn~oo dUn, f) = 0, and if {gn} is a sequence of measurable functions such that Ignl ~ M JL-a.e.,

limgn = 9 JL-a.e.,

then lim [ fngn dJL = [ fg dJL. n~oolx

lx

= 1,2, ... , and limn~oo dUn' f) = 0. Show that there exist an integrable function F and a sequence {nk} such that

3.3 Suppose that f, fn E L(JL), n

Ifnk 1 :$ F JL-a.e.

and

lim fnk

k~oo

=f

f E L(R) with the followRand M > 0,

3.4 Construct a Lebesgue integrable function

ing property: For any interval I

~

JL-a.e.

I{x E I: If(x)1 > M}I > O.

VIII.

144

3.5 Suppose

I is a Lebesgue integrable function on R+ and put

f

g(x) =

x>

I(t) dt,

1[0,00)

x

+t

Is 9 continuous? Does 9 have a limit as x 3.6 Suppose

More About L1

I

E

--+-

o.

oo? Is 9 differentiable?

L([a,b]) and put F( x)

= f I dy , 1[a,x]

a

~ x ~ b.

Show that F is a continuous BV function on [a,b] that satisfies the following property: Given e > 0, there exists 8 > 0, such that for any finite collection {[ai,bi]} of nonoverlap ping subintervals of [a,b], we have

Functions which satisfy (3.2) are said to be "absolutely continuous" in the sense of Vitali; we will have more to say about absolutely continuous functions in Chapter X. 3.7 Show that the integral of an arbitrary I E L(J.') is "absolutely continuous" in the following sense: Given e > 0, there exists 8 > 0, such that

L

III dJ.' < e,

whenever J.'(E) < 8.

We will have more to say about this notion of absolute continuity in Chapter XI. 3.8 Let (X,M,J.') be a finite measure space, and

I

a measurable extended real-valued function defined on X. Show that I E L(J.') iff

Lk:l J.'( {III 2:: k}) < 00. 3.9 Show that if I E

L(J.'), then lim

f

>'-+00 1{IJ1>>'}

III dJ.' =

(3.3)

0.

3.10 A sequence {In} of integrable functions is said to be "uniformly integrable" if (3.3) holds uniformly in n. More precisely, {In} is

uniformly integrable iff

lim sup

>'-+00 n

f

J{l/nl>>'}

I/nl dJ.' =

O.

3.

Problems and Questions

145

Show that if Jl(X) < 00 and {In} is uniformly integrable and limn->oo In = I Jl-a.e., then IE L(Jl) and

r In dJl = Jxr I dJl .

lim

n->oo Jx 3.11 If

sup

Ix

I/nlI+7J dJl

< 00 , 17 > 0,

show that {In} is uniformly integrable. The same conclusion is true if there exists 9 E L(Jl) so that I/nl ::; 9 Jl-a.e.j on the other hand the sequence

In

= (n/lnn)X(O,l/n)'

n

= 2,3, ...

is uniformly integrable with respect to the Lebesgue measure of 1= [0,1]. In fact, II In dx - 0, and yet the In's are not dominated by any integrable function. 3.12 Suppose Jl(X)

<

and show that uniformly integrable sequences are precisely those sequences with uniformly bounded integrals that are uniformly absolutely continuous. More precisely, {In} is uniformly integrable iff (i) There is a constant M such that I/nl dJl ::; M for all n, and, (ii) Given c > 0, there exists 6 > 0 such that Jl(E) < 6 implies I/nldJl < c for all n. 00

Ix

IE

< 00 and In ~ 0 Jl-a.e. for all n. Show that if the sequence {In} is uniformly integrable, then

3.13 Suppose Jl(X)

lim sup

Ix

In dJl ::;

Ix

lim sup In dJl .

= 1,2, ... , and limn->ood(fn,f) = o. Show that if Ihnl - 0, then limn->oo I/n(Y + hn) - l(y)1 dy = O.

3.14 Suppose 1,ln E L(Rn),n 3.15 Suppose that

IRn

IE L(Rn) and compute

r

lim I/(y Ihl->oo Jnn

+ h) + l(y)1 dy.

3.16 Prove that if A is a Lebesgue measurable subset of R n of positi ve measure, then the difference set A - A = {x E R n : x = Yl - Y2, Yl, Y2 E A} contains a neighbourhood of the origin.

VIII.

146

More About L1

State and prove a similar statement for A + A, and for A ± B, where B is another Lebesgue measurable set of positive measure. 3.17 Suppose I is a measurable function defined on R which assumes finite values on a set of positive measure and such that I(x + y) = I(x) + I(y) for all real x, y. Show that I is of the form I(x) = ex. 3.18 Prove that CJ(R n ) = {g: 9 is real-valued, it vanishes off a compact set, and its first order partial derivatives are continuous} is dense in L(Rn). Also, Ci(R n ) = {g:g and allits partial derivatives of order ~ (k-1) belong to CMR n )}, k = 2,3, ... , and Co(Rn) = n~l Ci(R n ), are dense in L(Rn). 3.19 If IE L(Rn), prove that there exists a sequence {Id of continuous functions such that limk.... oo Ik = I a.e. Further show that we may also require that each fk vanish off a compact set, and that it belong to CJ(Rn), or even to co(Rn). 3.20 Given

IE L(Rn), F(x,r)

put

= II( x,1 r )1 JI(x,r) f I(y)dy,

x E R n ,r > O.

Is F continuous as a function of x, for each r fixed? Is F continuous as a function of r, for each x fixed? 3.21 In the notation of 3.17, show that

limsupF(x,r) r .... O

and

liminf F(x,r) r .... O

are Lebesgue measurable. 3.22 Prove that Theorem 2.1 is true if we replace intervals by balls, i.e., (2.10) holds with M I there replaced by

Md(x)

= sup IB/ )1 f i l l dy, r>O x, r JB(x,r)

where B(x, r) = {y E Rn: Ix - yl < r} denotes the ball of radius r centered at x. 3.23 Prove the general version of the Hardy-Littlewood maximal the-

orem, i.e., remove the assumption that the integrable function vanishes off a bounded set.

I

3.

147

Problems and Questions

3.24 A point x at which lim II( 1 )1 x, r

r--+O

f

JI(x,r)

If(y) - f(x)1 dy

=0

is called a Lebesgue point of f, and the collection of all such points is called the Lebesgue set of f. Prove that if f is integrable, then almost every point is in the Lebesgue set of f. This notion is extremely important in the convergence of Fourier series, cf. Theorem 3.1 in Chapter XVII. 3.25 A family R = {R} of subsets of Rn is said to be regular at x provided that: (i) The diameters of the sets R tend to 0, and, (ii) There is a constant c such that if I( x, r) denotes the smallest interval centered at x containing R, then II(x,r)1 ~ clRI; the sets R need not contain

x. Show that if f is integrable, R is regular at x, and x is in the Lebesgue set of f, then . lim

diam(R)--+O

IRll JfR If(y) -

f(x)1 dy

= o.

3.26 Let E be a measurable subset of Rn; a point x E R n for which lim IE n I(x, r)1 II( x, r)1

=1

r--+O

is called a point of density of E. If the above limit equals 0, x is called a point of dispersion of E. Prove that almost every point of E is a point of density of E and a point of dispersion of Rn \ E. 3.27 Suppose that

hf(Y)dY = 0

(3.4)

for every subinterval I of R, and show that f = 0 a.e. In fact, the same conclusion is true if (3.4) holds for every I with III = c > 0, c a fixed constant. 3.28 Suppose that and let

f E L(R) vanishes off a bounded interval I F(x)=

r

J[a,x]

fdy,

= [a,b],

VIII.

148

More About L1

Is is true that

f

llim F(x + h) -

J[a,b] h-O

F(x) _

I(X)I

dx

= 07

h

Calderon and Zygmund, while considering problems related to the "norm" convergence of Fourier series of functions of several variables, introduced a decomposition of an integrable function into an essentially bounded "good" part and a "bad" part. The decomposition is described in the next four problems. 3.29 Let I be a finite interval in R and suppose I is a nonnegative integrable function which vanishes off I. Show that for any ,X satisfying

there is a sequence {Ik} of nonoverlapping intervals contained in I such that (i) ,X < Ilk Idy ~ 2,X, k = 1,2 ... (ii) I ~,X a.e. on 1\ Uk1k . (iii) \ Uk1k\ ~ ~ IUlk Idy ~ ~ Illdy.

Ttl

3.30 Referring to 3.26, consider the "good" function

and the "bad" function

b=l-g· Show that these functions enjoy the following properties (i) 0 ~ 9 ~,X a.e. on 1\ U Ik. (ii) 0 ~ 9 ~ 2'x on U Ik. (iii) b = 0 in 1\ Ulk ; Ilk b = 0 for all k. (iv) \b\~/+2'x. 3.31 Show that 3.29 and 3.30 are valid for I ,X>O

= R, I

E L(R), and any

3.32 The reader is invited to prove the n-dimensional version of 3.29, 3.30 and 3.31.

CHAPTER

IX

Borel Measures

In this chapter we study Borel measures on Euclidean space, their regularity properties, and the distribution functions associated with them. 1.

REGULAR BOREL MEASURES

A measure J.t on (R n , Bn) is called a Borel measure; the restriction of the Lebesgue measure to Bn is a familiar example of a Borel measure. In working with measures it is apparent that "regularity" plays an essential role. Now, in the case of the Lebesgue measure, regularity is built into its definition. We begin by showing that the same is true for those Borel measures which are finite on bounded sets; first the precise definition of regularity. A Borel measure J.t is said to be regular if for any E E Bn, J.t(E) may be computed by either of the expressions

J.t(E) = sup{J.t(K): K is compact, and K

~

E} ,

(1.1)

J.t(E) = inf{J.t(O):O is open, and 0 ~ E}. (1.2) These conditions roughly state that J.t is determined by the compact, or open, sets in Rn. A convenient way to verify these conditions is to consider the following equivalent formulations. For (1.1) we have: If J.t(E) < 00, given c > 0, there exists a compact set K ~ E such that

J.t(E \ K) = J.t(E) - J.t(K) and if J.t(E) = that

00,

~

given M > 0, we can find a compact set K

J.t(K)

~

M.

(1.3)

c, ~

E such (1.4)

14&

IX.

150

As for (1.2), if J.L(E) such that

Borel Measures

< 00, given £ > 0, there exists an open set 0 :2 E

J.L(O \ E)

= J.L(O) -

J.L(E):S

£,

(1.5)

and if J.L(E) = 00, by monotonicity any open set containing E also has infinite measure. We consider the regularity of finite Borel measures first; although the idea for proving this assertion is clear, it takes some time to carry out the details.

Theorem 1.1. Proof.

Suppose J.L is a finite Borel measure, then J.L is regular.

Let

A = {E E Bn : (1.1) and (1.2) are true for E}. The idea of the proof is to show that A is a u-algebra that contains the closed intervals, and which therefore coincides with Bn . It is not hard to check that A contains the closed intervals of Rn: Let I be a closed interval, then I is also compact and (1.1) holds. As for (1.2), let {Ik} be a decreasing sequence of open intervals that converges to I. By (iv) in Proposition 3.1 in Chapter IV it follows that

and (1.2) holds as well. Next, if {Em} ~ A and E = Um Em, then we have E E A. Indeed, suppose that £ > 0 has been given, and invoke (1.3) to find a sequence {Km} of compact sets such that

Km ~ Em, Furthermore, since E \

J.L(Em \ Km)

Um Km

~

:s £/2 m+1 ,

m = 1,2, ...

Um(Em \ K m), we have (1.6)

Now, since J.L (Um Km) <

00,

get at once that limM..... oo J.L

by (iii) in Proposition 3.1 in Chapter IV we

(U~=l Km) = J.L (U m Km). Whence, for M

sufficiently large, the compact set K = U~=l Km ~ E satisfies (1.7)

1.

Regular Borel Measures

151

Moreover, since

E\ K

= (E \ UmKm) U (UmKm \ K)

,

by (1.6) and (1.7) it follows that J.L(E) ::; J.L(K) + c. As noted in (1.3) above, this estimate gives that (1.1) is true for E. Along similar lines, let {Om} be a sequence of open sets such that

Em

~

Om,

J.L(Om \ Em) ::; €/2 m ,

m = 1,2, ...

Since 0 = UmOm is an open set containing E and O\E ~ we get that

J.L(O\E)::; LmJ.L(Om \Em)::;

Um (Om \

Em),

€,

and consequently, (1.2) is also true for E. Thus E E .A and .A is closed under countable unions. Finally we check that .A is closed under complementation. Suppose E E .A and € > 0 is given. Since J.L is finite, there exist an open set 0 ~ E and a compact set K ~ E such that

/J,(O \ E) ::; €/2,

and

J.L(E \ K) ::; €.

Moreover, since E \ K = (Rn \ K) \ (Rn \ E) and R n \ K = 0' is open, we also have

J.L( 0' \ (Rn \ E)) ::;

€ ,

0'

~

(Rn \ E) ,

and (1.2) holds for Rn \ E. A similar argument gives that

but we can only assert that R n \ 0 is closed. This is not a major difficulty: Since say, the sequence of compact sets {(Rn \ 0) n Bm} converges to R n \ O. Whence, by Proposition 3.1 (iii) in Chapter IV, we get that

and consequently, for m sufficiently large it follows that

152

IX.

Let J('

Borel Measures

= (Rn \ 0) n Bm; J(' is a compact subset of Rn \ E

and since

(Rn \ E) \ J(' = ((Rn \ E) \ (Rn \ 0)) U ((Rn \ 0) \ J('), by the above estimates it follows that 1-'( Rn \ E) - 1-'( J(') ~ c. Thus, (1.1) holds for R n \ E, and A is closed under the taking of complements. Whence A is a a-algebra that coincides with Bn , and we have finished .

•

Since R n is a-compact it is possible to extend Theorem 1.1 to more general Borel measures. More precisely, we have Theorem 1.2. Suppose I-' is a Borel measure which is finite on bounded subsets of Rn. Then I-' is regular. Proof.

Since Rn = Um{x:

Ixi

~

m} =

Um B m, say, for

each E in

Bn we have E

= Um (E n Bm) ,

with

En Bm E Bn

,

m

= 1,2 ...

Thus, from Proposition 3.1 (iii) in Chapter IV we get

I-'(E)

= m-+oo lim I-'(E n Bm).

(1.8)

The idea is to approximate the measure, of the sets that appear on the right-hand side of (1.8), and this will be achieved by "restricting" I-' to Bm. More precisely, consider the sequence of Borel measures given by

I-'m(E) = I-'(E n Em),

m = 1,2, ...

(1.9)

Since I-' is finite on bounded sets, the I-'m'S are finite Borel measures, and, by Theorem 1.1, they are regular. Fix now E E Bn , let £ > 0 be given, and put Em = E n B m , m = 1,2, ... By regularity, there exist compact sets Pm and open sets G m such that

(1.10) and

(1.11) We rewrite (1.10) and (1.11) in terms of 1-'. Since subset of Em, (1.10) reads

J(m

=

Pm

is a compact

(1.12)

2.

Distribution Functions

153

As for (1.11), let {Im,k} be a decreasing sequence of bounded open balls which converges to Em. Since f.l(im,d < 00, by (iv) in Proposition 3.1 in Chapter IV, we have

and consequently we can find a sequence of km's such that

Now, Om becomes

= Gm n Im,km

is an open set that contains Em, and (1.11) (1.13)

We are now ready to show that f.l(E) may be computed by both (1.1) and (1.2); we do (1.1) first. Combining (1.8) and (1.12) it readily follows that we may find a sequence of compact sets {Km} with the property that this gives (1.1) whether f.l(E) is finite or not. As for (1.2), we must only do the case f.l(E) < 00. If {Om} is the sequence of open sets introduced in (1.13), put 0 = Um Om and observe that 0 is an open set which contains E. Furthermore, since 0 \ E ~ Um(Om \ Em), by (1.13) it follows that

f.l(O) - f.l(E) = f.l(O \ E) $ Lm f.l(Om \ Em) $ Thus (1.2) is also true, and f.l is regular.

E:.

•

2. DISTRIBUTION FUNCTIONS Borel measures on the line which are finite on bounded sets are important in applications and there is a useful way to describe them. Let BB denote the collection of those Borel measures which are finite on bounded sets, assume that f.l E BB is finite and, referring to 4.28 in Chapter IV, let Fy be a distribution function induced by J-L. Since J-L is finite a way to normalize the Fy's is to consider not the expression given there but rather the distribution function F corresponding to y = -00, namely

F(x) = Jl«-oo,x]).

(2.1)

IX.

154

Borel Measures

F is called the distribution function of JL, it is nondecreasing and rightcontinuous, and it satisfies lim F(x)

x--+-oo

= 0,

lim F(x) <

x--+oo

Also, as a consequence of (2.1), it follows that for we have

JL«x, y]) = F(y) - F(x) , JL([x, y)) = F(y-) - F(x-),

(2.2)

00.

-00

<x
00

JL([x, y]) = F(y) - F(x-) JL«x,y)) = F(y-) - F(x).

Furthermore, if D is a dense subset of R, then the relation (2.1) is already determined by XED, or by any of the four above relations when x, y ED. The remarkable fact is that, conversely, any non decreasing rightcontinuous function F that satisfies (2.2) determines a unique finite Borel measure JL such that (2.1) is true. Rather than proving this result we discuss a more general one that also includes F(x) = x, which intuitively corresponds to the Lebesgue measure on the line. The precise statement IS

Theorem 2.1. Let BB = {JL: JL is a Borel measure on the line which is finite on bounded sets}, and 1) = {F: F is non decreasing and rightcontinuous}; we identify those functions in 1) that differ by a constant. Then, there is an injective mapping T from BB onto 1) which satisfies the following property: If TJL = F and c is an arbitrary constant, we have

c + JL«O,xD

F(x) = { c

c - JL«x,OD

°°

if x> if x = if x < 0.

(2.3)

Clearly (2.3) is equivalent to

F(y) - F(x)

= JL«x,y]) ,

all real x < y.

(2.4)

Proof. It is not hard to check that if T is given by (2.4), then T is one-to-one. Indeed, if TJL is constant, from (2.4) it follows that JL vanishes on all half-open intervals (x,yj. Since any open subset of R is a countable union of such intervals, JL also vanishes on all open sets. Furthermore, since JL is regular, by (1.2) we get JL(E) = for all E E B1 • Thus JL is the measure, and T is injective. To show that T is onto requires some work. Suppose F E 1) and observe that since F is nondecreasing the (bad) set

°

°

B = {r:F-l({r}) consists of more than one value x E R}

2.

Distribution Functions

155

is at most countablej B consists of those points in the range of F which correspond to the intervals of constancy of F. Now, let ~ be the interval-valued function defined as follows: Since FE V,F(x-) and F(x+) = F(x) exist for each x E Rj then put ~(x)

= [F(x-),F(x)]

,

x

E

R.

(2.5)

Thus, for each real x, ~(x) is a closed interval, degenerate if F is continuous at x, and closed and bounded otherwise. For each subset E of R let (2.6) ~(E) = ~(x),

U

xEE

and put J =

~(R)j

clearly J is also an interval. Now set

A = {E E 8 1 : ~(E) E £} j we claim that A is a u-algebra which contains all intervals, and which consequently coincides with 8 1 . First observe that if E is an interval, then so is ~(E); thus A contains all intervals. Next we verify that A is closed under complementation; this is clear if F is strictly increasing for then ~(R

\ E)

=J

\

~(E)

E £,

whenever E EA.

(2.7)

In the general case, i.e., when F is only monotone non decreasing, a slight complication occurs when E includes an interval of constancy of F. For instance, if F(x) = r for x E I = [a,b] and E = [a,(a + b)/2], then ~(R\E) = J, but J\ ~(E) = J\ {r}, and equality does not hold in (2.7). Nevertheless, equality will hold there if we add a subset of B, namely {r}, to J \ ~(E). The same reasoning applies to the general case. More precisely, for any E E A there exists an (at most countable) subset Bl of B such that (2.8) ~(R \ E) = (J \ E) UBI. From (2.8) it is clear that ~(R \ E) E £, which in turn implies that R\E E A, and consequently, A is closed under the taking of complements. Finally we check that if {En} ~ A and E = Un En, then E also belongs to A. This is a simple consequence of the readily verified identity

Thus, A is the u-algebra 8 1 •

IX.

156

Borel Measures

Let now p. be the set function given by

p.(E)

= 1cT>(E)I,

E E B1 •

(2.9)

Since for each Borel set E, cT>(E) E C, p. is well-defined. We claim that p. is a Borel measure which is finite on the bounded sets of R. That p.(0) = 0 is obvious. Next let {En} be a sequence of pairwise disjoint Borel subsets of the line and let E denote their union. In general the cT>(En)'s are not pairwise disjoint, as there may be overlaps created by the intervals of constancy of Fi this is not a serious inconvenience. First recall that since B is at most countable, for any A E C we have IA \ BI = IAI, cf. 3.2 in Chapter V. Next observe that the sets cT>(En) \ B, n = 1,2, ... , are pairwise disjoint and Lebesgue measurable. Thus

p.(E)

= 1cT>(E)1 = IUncT>(En) I = I(UncT>(En)) \ BI = IUn(cT>(En) \ B)I = Ln 1cT>(En) \ BI = Ln 1cT>(En)1 = Ln p.(En).

Whence, p. is a Borel measure on the line. It only remains to show that p. is finite on bounded sets, and that Tp. = F, i.e., that (2.3) holds. Let E ~ [-n,n] be a bounded Borel set. Since cT>(E) ~ cT>([-n,nD we obtain

p.(E)

= 1cT>(E)1 ~ 1cT>([-n,nDI < 00,

and p. is finite on bounded sets. Finally, let 1= (x,y]. Since lim F(z-)

z-+x+

= F(x+) = F(x)

by (2.5) it is apparent that either

cT>(I) = (F(x),F(y)] ,

or

cT>(I) = [F(x),F(y)].

In either case we have

p.(I) = IcT>(I) I = F(y) - F(x) , and (2.4) holds. It is now easy to see that (2.3) holds as well, with c there. •

= F(O)

To emphasize the fact that Tp. = F, and that, consequently, p. and F are related by (2.3), or (2.4), we denote p. by P.Fi P.F is called the Lebesgue-Stieltjes measure induced by F, and

In f

dp.F

,

f Borel measurable ,

2.

Distribution Functions

157

1

3

i

!

2

1

i

.1

o

a0

.1 3

Figure 4 is called the Lebesgue-Stieltjes integral of lover R with respect to dJ.LF' The following is a natural question to ponder: If F E 1) has a measurable derivative F' a.e., under what circumstances is dJ.LF(x) = F'(x) dx? That equality does not always hold follows by considering the Cantor Lebesgue function, which we construct next. Referring to the construction of the Cantor set and to the constructIOn in Section 1 of Chapter VI, let I! be the 2n -1 open intervals whose union is D n , and for n = 1,2, ... , let In be the continuous function defined on I = [0,1] which satisfies In(O) = 0, In(1) = 1,ln(x) = k2- n for x E I!, and which is linear on each interval of Cn. It is clear that each In is monotone non decreasing, and since the values only change on the Cn's, that In = In+! on I!, k = 1, ... , 2n - 1, and

I/n(x) - In+!(x)1

~

2- n ,

all x in I.

This estimate allows us to show that the In's converge uniformly on I. Indeed, for any k < m we have m-l

I/m(x) - Ik(X)1 = L(Jn(X) - In+l(X)) n=k

IX.

158 00

~

Borel Measures

00

L I/n(x) - In+1(x)1 ~ L2-n = 2- +1. k

n=k

n=k

Thus, given c > 0, we may choose N so large that I/m(x)-!k(x)l~c,

m,k~N,allxEI,

and the sequence {In} is uniformly Cauchy on I. Whence lim n -+ oo In(x) = I( x) exists on I, and since I is compact and the In's continuous and nondecreasing, I is also continuous and non decreasing. Also note that 1(0) = 0, 1(1) = 1, and that I is constant on every interval removed in the construction of Cj I is called the Cantor-Lebesgue function. Let now 0 if x < 0 F( x) = { I( x ) if 0 ~ x ~ 1 1 if x > 1. Clearly F E 1), and since F is not constant, fLF is not the 0 measure. On the other hand it is apparent that F'( x) = 0 for x ~ C, that is a.e. Whence it readily follows that

L

dfLF = fLF([O, 1]) = F(I) - F(O)

= 1:1

L

F' dx

=0 .

It is also clear that for the Cantor-Lebesgue function is true: [ !' dx = 0 :I 1(1) - 1(0) = 1.

I

the following

l[o.l]

This puzzling fact will be discussed in full detail in Chapter X, where we characterize those functions which coincide with the integral of their derivatives. We close this section with two interesting remarksj since the proofs follow along familiar lines we leave it to the reader to carry them out, cf. Section 3 in Chapter VII. Theorem 2.2.

Let 9 be a bounded real-valued function defined on

I = [a,b], I a nondecreasing right-continuous function defined on I, and put I(a) F(t) = { I(t)

I(b)

if t ~ a if a < t < b if t ~ b.

3.

159

Problems and Questions

Then, 9 E 1l(I) iff p F( {x E I: 9 is not continuous at x}) =

Furthermore, if

J:

°.

9 df exists, then 9 E L(p F), and

3. PROBLEMS AND QUESTIONS 3.1 What is the cardinal number of Bn? 3.2 Suppose p is a Borel measure on the line such that p([O,l]) = 1, and for each real x,

Show that p coincides with the restriction of the Lebesgue measure to B 1 • 3.3 Suppose that p is a probability Borel measure on [0,1], and that for each Borel set E C [0,1] with lEI = 1/2 we also have p(E) = 1/2. Does it follow that p coincides with the restriction of the Lebesgue measure to B1? 3.4 Suppose that p is a finite Borel measure on Rn, and that A C Rn is closed. Show that (x) = p(x + A) is upper semicontinuous, and consequently, measurable. 3.5 Let p be a finite Borel measure on a bounded interval of the line such that for each real x, p( {x}) = O. Show that given £ > 0, there exists 0 = 0(£) > 0 with the property that if E E B1 and diam (E) < 0, then p(E) < £. 3.6 Suppose p is a nonzero Borel measure on the line which is finite on bounded sets. Show that if

then I' is a Dirac measure.

IX.

160

Borel Measures

3.7 Let 4> be a nonnegative additive set function defined on Bn , and suppose that for E in Bn we have

4>(E) = sup{4>(K):Kcompact,K

~

E}.

Show that 4> is O'-additive, and hence a Borel measure. 3.8 Suppose J.L is a regular Borel measure on R n , and let E E Bn. Show that there exist a G s set U and an Fq set V such that

V ~ E ~ U,

J.L(U \ V)

= O.

3.9 Suppose J.L is a regular Borel measure on Rn, and let f be a nonnegative integrable function. Show that the set function

is also a regular Borel measure on Rn. 3.10 Suppose J.L, A, are Borel measures on Rn and let

(J.L

V

A)(E) == sup{J.L(A) + A(E \ A): A

~

E, A

E

Bn} ,

(J.L

A

A)(E)

= inf{J.L(A) + A(E \

~

E,A

E

Bn}.

and

A): A

Show that J.L V A and J.L A A are Borel measures and that

(J.L V A)(E) + (J.L A A)(E)

= J.L(E) + A(E) ,

all E E Bn.

If J.L and A are regular, are J.L V A and J.L A A also regular?

3.11 An atom of a Borel measure J.L is a singleton {x} such that J.L( {x}) > O. Show that the number of atoms of a O'-finite Borel measure J.L is at most countable. 3.12 Suppose J.LF is the Borel measure on the line induced by F E V. Show that

J.L F( {x}) = 0 iff F is continuous at x • Moreover, if {x} is an atom for J.LF' then we have

3.

Problems and Questions

161

3.13 Show that a regular Borel measure J.L on the line is a probability measure iff there exists FE V such that J.L = JLF and lim F(x)

x ..... -oo

= 0,

and

lim F(x) = 1.

x ..... oo

3.14 We say that a Borel measure JLF is atomic, or discrete, if J.LF(E) = 0 whenever E E B1 does not contain any atom of JLF; in this case the associated distribution function F is said to be discrete. On the other hand, if F is continuous, or equivalently when JLF has no atoms, we say that the Borel measure is continuous. Show that if J.L is a regular Borel measure on the line, then there exist a discrete measure JLd and a continuous measure J.Lc such that J.L = JLd + J.Lc. Is this decomposition unique? 3.15 Suppose J.L is a regular Borel measure on the line with no atoms, and let 0 < 11 < J.L(R). Show that there exists E E B1 such that JL(E) = 11· 3.16 Let J.L be a regular Borel measure on Rn. Show that there exists a unique closed subset C of R n with the following two properties: (a) J.L(R n \ C) = 0, and (b) If 0 is an open set such that C n 0 "I 0, then JL( C n 0) "I O. This closed set C is called the support of J.L, and we also say that J.L is supported in C, and denote this relation by supp J.L = C. What is the support of the Dirac 6x measure? Given a compact subset ]( of R, construct a measure J.L such that supp J.L = ](. 3.17 Show that supp (J.L V ,x)

= supp J.L U supp,x,

and that supp (J.L " ,x) ~ supp J.L n supp,x .

(3.1)

By means of an example show that the inclusion in (3.1) may be proper. 3.18 Suppose J.L is a regular Borel measure on the plane such that for any horizontal or vertical line L we have J.L(L) = o. Show that the function ¢>(x) = J.L(I(x,l)) , x E R2, is continuous. Can you think of n-dimensional extensions?

IX.

162

Borel Measures

3.19 Let J.L be a finite Borel measure on the plane, and suppose that for any line L we have J.L(L) = o. Show that if E E B2 and 0 < "I < J.L(E), there is a Borel set ACE such that J.L(A) = "I. 3.20 Suppose J.L P is the Borel measure induced by the distribution function F E V given by

F(x) =

o { x2

ifx
<

ifx~1.

2

Find the measures J.LP,c' J.LP,d associated to J.Lp as in 3.14 above, and compute

lXdJ.Lp(x). 3.21 Suppose F E V satisfies: F(x) x ~ 1. Compare

[

= 0 for

x

and

if

(1- F(x))dx

i[O,l]

~

0, and F(x)

= 1 for

[RXdJ.LP(x).

= 0 and that J~oo Ixl dJ.Lp(x) < 00 iff the integrals F(x)dx and ito,oo)(l- F(x)) dx are finite.

3.22 Given a distribution function F such that lim x -+_ oo F( x)

lim x-+ oo F(x)

J<-oo,O]

= 1, show

3.23 Show that if FE V is odd, then for any 9 E Co(R) we have

19(X)dJ.Lp(X)

= 19(-X)dJ.Lp(x).

3.24 (Change of Variable). Let A be a finite Borel measure on Bn , continuous real-valued function defined on R n , and set

F(x) = A(T-l(-oo,xD,

T

a

x E R.

Show that F E V, and that for any 9 E Co( R) we have [ gOT dA = [ 9 dJ.L P . iRn iR

In what follows we assume that the distribution functions F E V we work with are normalized so that lim F(x)

x-+-oo

= 0,

and

lim F(x)

x-+oo

= 1.

3.

Problems and Questions

163

3.25 Given F, Fn E V, n = 1,2, ... we say that Fn converges weakly to F if at each point x at which F is continuous we have lim Fn(x)

n-+oo

= F(x).

Prove that Fn converges weakly to F iff

lim JLF.n (( -oo,x])

n--+oo

= JLF(( -oo,x]) ,

at every point x for which JL( {x} ) = O. 3.26 Among other reasons, the notion of weak convergence is important in Probability because of the following approximation property: For every F E V there exists a sequence {Fn} ~ V such that (i) Fn converges weakly to F. (ii) Fn is continuous everywhere on R. (iii) Fn is constant on each interval ofthe form ((k-1)jn,kjn] , k = O,±1,±2, ... 3.27 The notion of weak convergence corresponds to a "metric" convergence. The Levy distance d (F, G) between F, G E V is defined as the infimum of those c > 0 for which

G(x - c) - c

~

F(x)

~

G(x + c) + c,

all x E R.

Prove that a necessary and sufficient condition for Fn to converge weakly to F is that d (Fn, F) --t O. 3.28 Show that if F E V is everywhere continuous, then F is uniformly continuous. 3.29 Given a real-valued function G defined on R, we define its modulus of continuity w(G,c) by the expression

w(G,c) = sup{G(x) - G(y): Ix -

yl

~

c}.

Show that if F, G E V, and d( F, G) < c, then we have sup IF(x) - G(x)1 ~ c + w(F,c). As a consequence of this prove that if Fn converges weakly to an everywhere continuous distribution F, then Fn converges uniformly. to F. Is this statement true if F is only continuous on a bounded closed interval?

IX.

164

Borel Measures

3.30 Suppose J.L,J.Ln EBB, J.L(R) = J.Ln(R) = 1 for all n. We say that J.Ln converges weakly to J.L if lim J.Ln« -oo,x]) = J.L« -oo,x])

n-+oo

at every point x at which J.L( {x}) = O. For instance, the statements Fn converges weakly to F and J.L Fn converges weakly to J.L F are only different expressions of the same fact. Prove that if J.Ln is the Dirac measure at x n , and J.L the Dirac measure at x, then J.Ln converges weakly to J.L iff Xn --+ x. 3.31 Let J.L, J.Ln be Borel measures on the line, n following conditions are equivalent: (i) J.Ln converges weakly to J.L. (ii) For every I E Co(R) we have lim n-+oo

=

1,2, ... Then the

JRf I dJ.Ln = JRf I dJ.L •

(iii) For every Borel set E with the property that its boundary BE = En (R \ E) is J.L-null, we have lim J.Ln(E)

n-+oo

= J.L(E).

3.32 If I denotes the Cantor-Lebesgue function in I = [0,1] and x E C is of the form x = 2:~=1 2an /3 n , where each an = 0 or 1, show that I(x) = 2:~=1 an/2n. Also compute

i

xnl(x)dx,

n

= 0,1,2.

CHAPTER

x

Absolute Continuity

In this chapter we discuss the class of absolutely continuous functions, namely, those functions which may be recovered by integrating their derivatives. 1.

VITALI'S COVERING LEMMA

In dealing with the question of whether the indefinite integral of I differentiates to I(x) it was essential to handle families of intervals. The same is true in general problems of differentiation, or any other area of Analysis where intervals are sorted out. For the problem at hand this process is carried out by means of Vitali's covering lemma; first a definition. A family V of closed intervals of R is said to be a covering of E in the sense of Vitali if for any x E E and c > 0 there is an interval I in V which contains x and so that III ~ c. In other words, every point of E belongs to arbitrarily small intervals of V. Such coverings satisfy the following remarkable property. Theorem 1.1 (Vitali's Covering Lemma). Suppose E is a subset of the line with IEle < 00 and V is a covering of E in the sense of Vitali. Then there exists an at most countable family {Ik} of pairwise disjoint intervals of V such that (1.1) Proof. Let 0 be an open set of finite measure which contains E, and discard from V those intervals that are not totally contained in o. It is clear that this new family, which we call V again for simplicity, is also a Vitali covering of E.

165

X.

166

Absolute Continuity

Having done this, pick an interval h, say, of V. If IE \ hi = 0 we are done, otherwise we choose recursively a family of intervals of V according to the following rule: Suppose that the pairwise disjoint intervals h, ... ,In of V have been chosen and that IE\U k=1 Ikle > o. Consider then the open set

Gn

=0

\ Uk=IIk

f: 0,

and the class of those intervals of V totally contained in G n ; the idea is to select In+1 as a largest interval in this class. Thus, if kn

= sup{III:I E V

and I C G n } > 0,

let In+1 be any subinterval of V contained in G n such that

IIn+11 > k n/2; by construction it is clear that In+1 n (II U ... U In) = 0. Either the selection process stops after a finite number of steps, and if this is the case we have finished, or else there exists a pairwise disjoint sequence {Id of intervals of V such that

Uklk ~ 0,

Lk IIkl ::S

and

101 < 00.

In this case, given TJ > 0, we may find N such that l:~N+1IIkl < TJ. We consider RN = E \ U~=1 h and estimate its Lebesgue outer measure in terms of TJ. Since each x in RN belongs to the open set GN, by assumption there is an interval I E V containing x such that I n (II U ... U IN) = 0. We claim that there is an index n > N so that I n In f: 0. Indeed, if I E V and for all m we have I n 1m = 0, it follows that

III ::S k m < 21Im+11 ~ 0,

as m ~

00 ,

which is impossible. Let n be smallest index so that I n In f: 0; clearly n > N. Furthermore, since by the way the In's were selected we have III ::S k n < 2IIn+1l, by simple geometric considerations we obtain

+ IIn+1I/2 < 21In+11 + IIn+1I/2 = 5IIn+1I/2.

d(x,midpoint of In+J)::S III

Let I n+1 denote the interval concentric with In+1 with sidelength 5 times that of I n +1 • By the above estimate it follows that x E I n +1 and consequently, Rn ~ U~N+1 Jk. Thus

IRNle < ,",00 IJkl - L...Jk=N+l

= 5,",00 IIkl = 5TJ. L...Jk=N+1

Whence by (1.2) we have

IE \ U~lhle ::S

IE \Ui"=IIkle < 5TJ,

and, since TJ is arbitrary, (1.1) holds.

•

(1.2)

2.

Differentiability of Monotone Functions

167

Corollary 1.2. Under the assumptions of Theorem 1.1, given c > 0, there exists a finite family II, ... , IN of pairwise disjoint intervals of V such that (1.3) Proof. Pick TJ = c/5 in the proof of Theorem 1.1; then (1.2) gives the desired conclusion. • Note that whereas the validity of Corollary 1.2 requires that IEle < 00, the conclusion of the Vitali covering lemma is true for an arbitrary subset Eof R.

2. DIFFERENTIABILITY OF MONOTONE FUNCTIONS Suppose I is a real-valued function defined on 1= (a,b), and for x E I and h -:f:. 0 with x + h E I put DI(x,h) = I(x

+ h) -

I(x) .

h Whether I is differentiable at x E I, or has a one-sided derivative at x, or not, the following four quantities, called the Dini numbers of I at x, are well-defined: D+ I(x) = lim sup D I(x, h),

D+/(x) = liminf D I(x, h),

h-+O+

h-+O+

D- I(x) = lim sup DI(x, h) ,

D_/(x) = liminf DI(x,h).

h-+O-

h-+O-

Clearly D+/(x) ~ D+ I(x) and D_/(x) ~ D- I(x) and I'(x) exists iff all four Dini numbers of I at x are equal. The stage is now set for

Theorem 2.1 (Lebesgue). Let I be an open subinterval of the line and suppose I is a monotone real-valued function defined on I. Then I' exists a.e. on I. Proof. We may assume that I is nondecreasing, and consider first the case when I is bounded. We will be done once we show that D_I ~ D- I ~ D+I ~ D+ I ~ D_I,

a.e. on I,

(2.1)

x.

168

Absolute Continuity

for then all the Dini numbers of f are equal at those x's where (2.1) holds, and f' exists a.e. on I. As noted above, the first and third inequalities in (2.1) are always true, so we only need to establish the second and fourth inequalities there. Now, this amounts to showing that the (bad) sets

B

= {x E I:D+ f(x) > D-f(x)} and B' = {x E I:D- f(x)

> D+f(x)}

are null. Since the proof for both sets follows along similar lines we only consider B. First observe that all Dini numbers are nonnegative, and if for rational numbers u > v > 0 we put

= {x E I:D+ f(x) > u > v > D-f(x)} , then we have B = Uu,v Bu,v. Thus the desired conclusion will follow once we show that each of the Bu,v's is null. So we suppose that IBu,vle = T/, Bu,v

and show that T/ = O. The idea of the proof is to approximate Bu,v by a simpler set consisting of pairwise disjoint intervals (here we use the fact that D-f < v and the Vitali covering lemma) and then to further approximate the part of Bu,v which lies within those intervals by another family of intervals (here we use the fact that D+ f > u and Vitali's covering lemma again). First observe that since Bu,v ~ I we have T/ < 00, and, by (1.8) in Chapter V, given € > 0, there exists an open set 0 ;2 Bu,v such that

Moreover, since for each x in Bu,v we have D-f(x) < v, there exists a sequence hx,n > 0 approaching 0 such that the intervals [x - hx,n,x] C 0 and (2.2) f(x) - f(x - hx,n) < vhx,n all n. Clearly V = {[x - hx,n,x]} is a covering of Bu,v in the sense of Vitali and consequently, by Theorem 1.1 there is a finite collection II = [XI ,XI - hI], ... ,In = [xn,x n - h n], say, of pairwise disjoint intervals of V such that

(2.3) Next let jj denote the interior of Ij, 1 ::; j ::; n, and observe that since for each x in

B~,v

= Bu,v n

(Ui=1 jj)

we have D+ f( x) > u, there is

2.

Differentiability of Monotone Functions

a sequence kx,m 1 :$ j :$ n, and

>

169

0 tending to 0 such that [x,x

I(x

+ kx,m) -

I(x) > ukx,m,

+ kx,m]

~

Ij for some (2.4)

all m.

Since V1 = {[x ,x + kx,m]} is a covering of B~,v in the sense of Vitali, we can find a finite collection J 1 = [x~ ,x~ + k1], ... , J m = [x~,x~ + km], say, of pairwise disjoint intervals of V1 with the property that

(2.5) Let now

m

0:$ D..J = 2:(f(xi + ki) - I(xi» i=l denote the increase of I along the Ji'S, and, similarly, let D..] denote the increase of I along the I/s. It is not hard to check that (2.6) Indeed, suppose that J1 , ••• , Jml are ordered from left to right and are contained in /1 j that Jml +1, ... ,Jm2 are ordered from left to right and are contained in 12 , and so on. Since {Ji}~l is a pairwise disjoint collection of intervals contained in /1 it readily follows that

2::1(f(xi

+ ki) -

I(xi»

= I (X~l

:$

I(X~l

+ kml ) + km1 )

-

••• -

(J(x~) - I(x~

+ k1» -

I(xt)

l(xD :$ I(xt) - I(X1 - ht).

Whence, by adding up the increase of I along these blocks of Ji'S it follows that (2.6) holds. Now, by (2.2), and since the I/s are all contained in 0, it is clear that

D..] <

2:;=1 vlIjl

= viOl :$ v(1] + oS).

(2.7)

On the other hand, by (2.4) we get

D..J >

U

2::1IJil,

(2.8)

and consequently, we need a lower bound for the right-hand side of (2.8). Since Bv.,v = (Bv.,v \ Ui=l Ij) U B~,v UN, where N Ij 's} is a finite set, it is clear that

Bu,'IJ

~

(Bu,'IJ \ Ui=l I

= {endpoints of the

j) U (B~,'IJ \ U~lJi) U (U~lJd UN.

X.

170

Absolute Continuity

Thus, on account of (2.3) and (2.5) it readily follows that

1] =

IBu,vle ~ c + c + I:: 1IJil,

or

I::1IJil 2:: 1] -

2c.

Substituting this estimate in (2.8), and combining it with (2.6) and (2.7), we see that

(1]-2c)u <

/}..J ~ /}..[

< V(1]+c),

c > O.

(2.9)

Since c in (2.9) is arbitrary, we also have 1]U ~ 1]V, and since 0 < v < u, this can only hold if, as asserted, 1] = O. This completes the proof when I is bounded. On the other hand, if the interval I is unbounded, observe that

say, where h is a bounded open interval, k on each Ik, and consequently also a.e. on I.

= 1,2, ... Thus I' exists a.e. •

Not only does the derivative of a monotone function exist a.e., but it also is integrable. More precisely, we have Lebesgue's theorem, Theorem 2.2. Suppose I is a nondecreasing real-valued function defined on I = (a,b). Then I' E L(I) and

(2.10) Proof. Extend I to R by setting I( x) = I( a+) if x ~ a, and I( x) I(b-) if x ~ b. Put now

In(x) = n(f(x

+ lin) -

and observe that since by Theorem 2.1 lim In = I'

n-+oo

x E R, n

I(x)) ,

I'

= 1,2, ...

=

,

exists a.e., we have

a.e. on I.

Thus, by Fatou's Lemma, it follows that

i /'

dx

~ lim inf

i

In dx .

(2.11)

3.

Absolutely Continuous Functions

171

It is rather straightforward to compute the integral on the right-hand side of (2.11). Indeed, by 4.6 in Chapter VII, and with n sufficiently large, we have

h Jn dx = n h J( x + 1/n) dx - n

=n f

J dx - n

J[b,b+l/n)

i

J dx

f

J dx = A + B ,

J(a,a+1/n]

say. Clearly A = J(b-), and B ~ J(a+). Whence, for all sufficiently large n the integral in question is dominated by J(b-) - J(a+), and (2.10) holds. •

Corollary 2.3. Then l' E L(I) and

Suppose

J is

hlJ/ldx

BV on a bounded interval I = [a,b].

~ V(J;a,b).

(2.12)

Our aim now is to discover when equality holds in (2.10); for this we need the concept of absolutely continuous functions.

3. ABSOLUTELY CONTINUOUS FUNCTIONS Absolutely continuous, or AC, functions were introduced in 3.6 in Chapter VIII. They are continuous functions whose increment along any collection of pairwise disjoint intervals of sufficiently small total length is arbitrarily small. This concept excludes the Cantor-Lebesgue function J which, although being locally constant on a subset of [0,1] offull measure, it nevertheless increases from 0 to 1 there. To see this we cover the Cantor set C by a union Un( an, bn ), say, of pairwise disjoint open intervals with L:n(bn - an) is arbitrarily small. Extend J so that J(x) = 0 for x < 0 and J( x) = 1 for x > 1. Then it is not hard to verify that L:n(J(bn ) - J( an)) = 1, and consequently, L:~=l(J(bn) - J(a n )) > 1/2 for sufficiently large N, while at the same time, L:~=1 (bn - an) is arbitrarily small. So, which among the continuous functions are AC, and what properties do AC functions satisfy? Proposition 3.1.

Let I = [a,b] and suppose I is AC on I. Then

I is BV on I, and consequently, by Corollary 2.3, I' exists a.e. and it is integrable there.

x.

172

Absolute Continuity

Proof. Let 8 be the real number that corresponds to the choice c = 1 in the AC definition of I, and let the integer N > (b - a)/8. Note that, in particular, we have

v (f; x, x + "7) ~

1,

any x E I , 0

< "7 ~

(3.1 )

8.

The idea now is to use (3.1) to put together the estimates along the partitions of I. So, let P = {a = Xo < ... < Xn = b} be a partition of I and let P' be the partition of I obtained by adjoining the points a + (b - a)/N, a + 2(b - a)/N, ... ,b to P. Since P' is finer than P it readily follows that

2:

2:

I~k/l ~

over l'

I~k/l·

(3.2)

over 1"

It is not hard to estimate the right-hand side of (3.2); indeed, by (3.1) it does not exceed

V(f;a,a + (b - a)/N) + ... + V(f;a + (N -1)(b - a)/N,b) Since P is arbitrary, by (3.2) it follows that V(f; a, b) on I . •

~

~

N.

N and I is BV

So, AC functions are continuous and BV, but is the converse to this statement true? It is partially true, and to discuss it we need some preliminary results.

Lemma 3.2. Suppose A function defined on I such that

~

II' (x) I ~

I = [a,b], and let M,

I

be a real-valued

x EA.

Then

I/(A)le ~ MIAle. Proof.

(3.3)

Given c > 0, let 0 be an open set such that

101

~

IAle + c ,

0

2 A.

(3.4)

Break up A into two disjoint parts, Al = {x E A: I is constant in a neighbourhood of x}, and A2 = {x E A: I is not constant in any neighbourhood of x}, say. We construct a covering V of I(A) in the sense of Vitali as follows: If I(x) E I(A) and x E At, then there is an interval J ~ 0 such

3.

Absolutely Continuous Functions

173

that f(x) = fey) for all y E J. Now, if I( w, w') denotes the closed interval with endpoints w and w' (note that w' < w is possible), then we can find h so that I(x,x+Mh) C 0 and f is constant on I(x,x+Mh). Such values f(x) are then assigned the intervals I(f(x),f(x) + Mb), where b satisfies

I(x,x

+ Mb) ~ I(x,x + Mh).

On the other hand, if f(x) E f(A) and x E A 2 , then there is a sequence hx,n '=I 0, n = 1,2, ... , converging to 0, such that I(x,x + hx,n) C 0 for all n, and

°< If(x + hx,n) - f(x)1

~ (M

+ c)lhx,nl,

n = 1,2, ...

To these values f(x) we assign the intervals I(f(x),f(x + hx,n)), n = 1,2, ... Now, the collection V of all the intervals introduced above is a covering of f(A) in the sense of Vitali and consequently, there is an at most countable family consisting pairwise disjoint intervals II' ... ' Ik ... , say, such that Whence, we also have (3.5) To estimate the right-hand side of (3.5) we separate the Ik's into two families: Those that correspond to f(x) with x E AI, call them Il's, and those corresponding to f(x) with x E A 2 , call them ~'s. Furthermore, if Il = I(f(xk),/(xk) + Mb k ), let Jl = I(xk,xk + bk ), and if I~ = I(f(xk),f(xk + h k )), let J~ = I(xk,xk + h k ). Since the Ik's are pairwise disjoint, so are the Jk'S, and they are also contained in O. Consequently, by (3.4) we have

2: IIkl

2: jIll + 2: II~I ~ 2: M lbkl + 2:(M + c)lhkl = M 2:IJll + (M + c) 2:IJ~1 ~ (M + c) 2:IJkl ~ (M + c)IOI ~ (M + c)(IAle + c), =

which substituted into (3.5) gives If(A)le ~ (M + c)(lAle + c). Moreover, since c is arbitrary, the above inequality is also true with c = 0, and (3.3) holds. • An interesting consequence of Lemma 3.2 is

x.

174

Absolute Continuity

Lemma 3.3. Suppose I is a real-valued function defined on I = [a,b], and let A be a measurable subset of I so that I'(x) exists everywhere on A and is measurable there. Then

I/(A)le ~

i

11'1 dx.

(3.6)

Proof. First suppose that for some integer M we have on A and consider the level sets

II'(x)1 < k/2n},k =

Ak,n={X E A:(k-1)/2n ~

1/'(x)1 ~ M

1, ... ,M2n,n

= 1,2, ...

Since for each n we have A = Uk Ak,n, by Lemma 3.2 and Chebychev's inequality it readily follows that

I/(A)le ~

I/(Ak,n)le ~

L k

L(k/2 n )I A k,nl k

= L((k - 1)/2 n )I A k,nl k

1

~ Lk

Ak,n

1

+ 2n L

IAk,nl

k

1/'ldx + 2~IAI

~ JAf Il'ldx + 2~IAI.

Since this estimate holds for every n, (3.6) is true in this case. As for the general case, note that 00

U{x E A: k -

A=

00

1 ~ 11'1 < k}

k=l

=

U Ak , k=l

say, where the Ak's are pairwise disjoint and 1/'1 exists and is bounded and measurable on each A k • Then, by the first part of the proof we have

I/(A)le ~ and (3.6) holds.

L

I/(Ak)le

k

~ L f 1/'1 dx = f 1/'1 dx, k

~k

~

•

We are now ready for

Theorem 3.4 (Banach-Zarecki). Suppose I is a continuous, BV, real-valued function defined on I = [a,b]. Then I is AC on I iff I maps null sets into null sets, i.e.,

IAI =

0

implies

I/(A)I = O.

3.

Absolutely Continuous Functions

175

Proof. We show the necessity first: It is enough to prove that given a null set A ~ (a,b) and c > 0, we have

I/(A)le

~

(3.7)

c.

We invoke (iii) in 4.14 below: From the hypothesis of AC there exists 8 > 0 such that no matter what finite pairwise disjoint family {Ik = (ak,bk)} of subintervals of (a,b) we take, with the notation w(j, J) = sUPJ I - infJ I, we have 2:(bk - ak) < 8 implies 2:w(j,Ik) < c. (3.8) Also observe that since I/(Ik)le ~ w(j,Ik) for each k, by (3.8) it readily follows that

Choose now an open set 0 with 101 < 8 so that A C 0 = U~l(ak,bk) ~ (a,b), where the (ak,bk),s are pairwise disjoint, and note that the above estimate implies that

(3.7) follows at once from this. As for the sufficiency, suppose that c > 0 is given, and let {(ak,b k )} be a finite pairwise disjoint family of subintervals of I. Then, if Ak = {x E [ak,b k]: I'(x) exists }, we have I[ak,bk] \ Akl = 0 for all k. Furthermore, since I is continuous we also have

Whence, combining these remarks, and by our assumption and Lemma 3.3, we obtain

2: 1/([ak,bkDle ~ 2: 1/([ak,bk] \ Ak)le + 2: I/(Akle

2: I/(bk) - l(ak)1 ~ k

k

k

~

2: lAic f 1f'1 dx = k

k

f

lUAIc

1/'1 dx .

(3.9)

Now, since / is BV on I, by Corollary 2.3 /' is integrable on I, and we are in a position to invoke 3.7 in Chapter VIII: Choose 8 > 0 so that the

x.

176

Absolute Continuity

conclusion there holds for the c we fixed at the beginning of the argument, and observe that since Uk Ak ~ Uk[ak,bk], we also have

IUkAkl ~ 6 whenever

2:(bk - ak) ~ 6. k

Therefore, by (3.9) it follows at once that

2: I/(bk) -

l(ak)1 ~ c whenever

k

and

I is AC on I.

•

That the assumption that I is BV is necessary for the validity of Theorem 3.4 follows from a construction which is reminiscent of the discussion preceding (1.5) in Chapter III. Consider 1= [0,1] and a Cantor-like subset J( of Ij the measure of J( may be positive or not. Write the set 1\ J( = Un(an,bn ) as the at most countable pairwise disjoint union of open intervals, and let Cn denote the midpoint of (an, bn ). If dn is a sequence of positive numbers with limit 0, define Ion I as follows: I( x) = 0 for x E J(, I(c n ) = dn for all n, and I is linear in [an, cn] and [cn, bn]. Then I is continuous, and V(fj 0, 1) = 2 L~=l dn • To see that I maps null sets into null sets, consider a null subset A of I. By 5.8 in Chapter I, we have I(A) = I(An J()U Un I (A n (an, bn )). Since I is linear in [an, cn] and in [cn,b n], it readily follows that I/(A n (an,bn)1 = 0 for all n, and so I/(A) ~ I{O}I + Ln I/(A n (an, bn)1 = o. If Ln dn = 00, then I fails to be AC on I since it is not BV there. In order to establish further properties of AC functions we introduce the following definition: Suppose I is a real-valued a.e. differentiable function on an interval I. We then say that I is singular if I' = 0 a.e. on I. How do AC singular functions look?

Proposition 3.5. Suppose I is an AC singular function defined on an interval I. Then I is constant. Proof. Let A be a subset of I of full measure so that I'(x) x E A. By Lemma 3.3 we have

I/(A)le

~

L1/'1

dx = O.

= 0 for (3.10)

Further, since II \ AI = 0, by the necessity of Theorem 3.4 it follows that

If(I\A)1 =

o.

(3.11)

3.

Absolutely Continuous Functions

177

Whence combining (3.10) and (3.11) we get

I/(I)le ~ I/(A)le + 1/(1 \ A)le =

o.

(3.12)

Now, since I is continuous, unless I is constant, 1(1) contains an interval and (3.12) does not hold. Therefore I must be constant. • We are now ready to characterize, following Lebesgue, the class of functions which may be reconstructed by integrating their derivatives. Theorem 3.6. Suppose I is a real-valued function defined on I [a,b]. Then, I' exists a.e. in (a,b), it is integrable there, and I(x)-/(a)=

f

I'(t)dt,

J[a,x]

a~x~b,

=

(3.13)

iff I is AC on I. Proof. We do the sufficiency first: Since I is AC on I, I' exists a.e. on I and it is integrable, therefore we only need to show that (3.13) holds. For this purpose put F( x) = i[a,x] f' (t) dt, a ~ x ~ b, and observe that by 3.6 in Chapter VIII also F is AC on I, and, by the Lebesgue differentiation theorem, F' = I a.e. Let 9 = F - Ii 9 is AC and singular on I, and, by Proposition 2.6, 9 is constant there. More precisely, F( x) - I( x) = F(a) - I(a), a ~ x ~ b, and since F(a) = 0, it readily follows that F(x)=j

1'(t)dt=/(x)-/(a),

[a ,x]

a~x~b,

as we wanted to show. Conversely, since I(x) = I(a)

+f

f'(t)dt,

J[a,x]

a~x~b,

the sufficiency follows from Theorem 2.2 in Chapter VIII.

•

Implicit in the proof of Theorem 3.6 is the following important result concerning BV functions. Theorem 3.7 (Lebesgue). Suppose I is BV on 1= [a,b]. Then there exist an AC function 9 and a singular function h such that I(x)

= g(x) + h(x),

x E I.

Up to constants, the decomposition in (3.14) is unique.

(3.14)

x.

178

Absolute Continuity

Proof. Since I is BV on I, I' exists a.e. there, and it is integrable. Let g(x) = Ira,x] I'dt, a ~ x ~ b, and set

h(x)=/(x)-g(x),

a~x~b.

Then 9 is AC on I and by the Lebesgue differentiation theorem, h' = I' - g' = 0 a.e. on I. Thus 1= 9 + h is a desired decomposition. As for the uniqueness (modulo constants), suppose that also I = gl + hI, where gl is AC on I and hI is singular. We then have

(3.15)

9 - gl = hI - h,

where the expression on the left-hand side of (3.15) is AC and that on the right-hand side is singular. By Proposition 3.5 it readily follows that this function is constant, c say. Thus 9 = gl + c, and h = hI - c. •

4. PROBLEMS AND QUESTIONS 4.1 Show that in Vitali's covering lemma we may also demand that given E > 0, L:k IIkl ~ (1 + E)IEle. 4.2 State and prove Vitali's covering lemma, including the conclusion of 4.1, for subsets E of Rn with IEle < 00, and covered in the sense

of Vitali by closed n-dimensional intervals. 4.3 Let E be a subset of Rn that is the union of sets, each being an open

interval together with any of its edges. Prove that E is Lebesgue measurable. 4.4 A measure J.L on (Rn,£.) is said to be doubling provided there exists an absolute constant c such that

J.L(I(x,2r»

~

cJ.L(I(x,r)) ,

all x E R n ,r

> o.

Given a doubling measure J.L, a family V of closed intervals of R n is said to be a covering of E E £. in the sense of Vitali if for any x E E and E > 0, there is an interval I E V which contains x and so that J.L(I) ~ E. Show that if V is a covering of E in the sense of Vitali, and J.L(E) < 00, then there exists an at most countable family {Ik} of pairwise disjoint intervals of V such that J.L (E \ Uklk) = O. 4.5 Suppose I is a real-valued function defined on an interval I, and that all the Dini numbers of I for x in I lie between -k and k,

4.

Problems and Questions

179

where k is some positive constant. Must I be Lipschitz on I? If so, what is the relation between the Lipschitz constant of I and k? 4.6 Let I be a real-valued function defined on I there exist real constants u, v such that

u~

n+ I( x)

Is it then true that for all a

~ v,

~ x

<

x

= [a,b]

and suppose

all x E I.

+ h ~ b we have

uh ~ I( x + h) - I( x) ~ vh ? 4.7 Suppose I is a continuous real-valued function defined on 1= [a,b], A c I is at most countable, and n+ I( x) ~ 0 on I \ A. Prove that I is nondecreasing on I. 4.8 Compute the Dini numbers of the Cantor-Lebesgue function at each point x of [0,1]. 4.9 (Fubini's Lemma) Let {Ik} be a sequence of non decreasing functions defined on an interval I of the line. Show that if the series I(x) = Lk Ik converges to a finite limit on I, then f' = Lk Ik a.e. on I.

4.10 Let (an) be a sequence of distinct points in an interval I of the line, and suppose (un), (vn ) are sequences ofreal numbers such that L lunl, L Ivnl < 00. Put

and show that 8 ' = 0 a.e.

8

=

L In

has a finite derivative a.e. on I and that

4.11 Let E be any subset of R. Show that for almost all x in E we have

1.

1m

h-+O

IE n [x -

h,x 2h

+ h]le

_ 1 - .

4.12 Does there exist a strictly increasing function terval I so that I' = 0 a.e. on I?

I

defined on an in-

4.13 Suppose I is a real-valued function defined on I. Show that if not constant and I' = 0 a.e., then I cannot be Lipschitz on I.

I

is

4.14 Let I be a real-valued defined on I = [a,b]. Prove that the following statements are equivalent:

x.

180

Absolute Continuity

(i) I is AC on I. (ii) Given E > 0, there exists 6 > 0 such that for any finite collection {[ai,bi]} of nonoverlap ping subintervals of [a,b] we have IEi(j(bi) - l(ai»1 < E, whenever Ei(bi - ail < 6. (iii) If w(j, J) = sup I - inf I , J

J

then for each E > 0, there is 6 > 0 such that for any finite collection {Ik = [ak,b k]} of nonoverlapping subintervals of I we have Ekw(j,Ik) ~ E, whenever Ek(b k - ak) < 6. 4.15 Let I be a real-valued continuous function defined on I = [a,b], and suppose that I is AC on [a,ci], for any d < b. Show, then, that I is AC on I. Is this result true if the assumption that I is continuous on I is dropped? 4.16 If the functions 1,9 are AC on an interval I, show that their difference, sum and product are also AC on I. If 9 vanishes nowhere on I, then the quotient 1/9 is also AC on I. 4.17 Show that the composition of AC functions need not be AC, or even BV for that matter. 4.18 Let I be AC on I, and 1(1) ~ J. If 4>: J that 4> 0 I if AC on I.

-+

R is Lipschitz, show

4.19 Suppose I is a non decreasing, AC function on I, and 1(1) ~ J. Show that if 4> is AC on J, then 4> 0 I is AC on I.

4.20 Suppose I is a monotone function defined on I = [0,1], and let E = {x E I: n+ I(x) = oo}. Prove that I is AC on I iff I/(E)I = 0. 4.21 Suppose I is a real-valued continuous function defined on I, and let A be an Fq subset of I. Prove that I(A) is Fq too, and show that as a consequence of this, if I is AC on I, then it maps Lebesgue measurable sets into Lebesgue measurable sets.

4.22 Let N be a Lebesgue null subset of I = [0,1]. Show that there is a real-valued function I defined on I which is AC there, and such that f'(x) = 00 for each x E N. 4.23 Show that there is an AC function tone on no subinterval of I.

I defined on [0,1] which is mono-

4.24 Let I be a real-valued Lebesgue measurable function defined on I = [a,b], and suppose e, fJ > 0 are given. Show that there exist a

4.

Problems and Questions

181

Lebesgue measurable set B C I, and an AC function F defined on I such that II - FI < c and IBI < "I.

II\B

4.25 Let E be a bounded Lebesgue measurable set in the line and let

In(x)=n [

XEdy,

n=1,2, .•.

i[:c,X+l/n]

Show that each In is AC on every bounded interval of the line, that ~ 1, that lim n -+ oo In = XE a.e., and that lim n -+ oo d (In,XE)

o ~ In = o.

Is this result sufficient to prove that AC functions are dense in the metric of L(R)? 4.26 Let 9 be a continuous function defined on 1= [a,b] and suppose is AC there. Prove that I: 9 dl = II gl' dy.

I

4.27 Let 1= [a,b], and suppose I is BV on I. Prove that for each Borel set E ~ I we have 1/'1 dy ~ 1V(J)(E)le, and that there is equality here provided that I is AC on I. A related result is the following: Suppose I is a strictly monotone AC function defined on I, and let I(I) = J. Show that for every Borel subset E of J we have 1,-1 (E) I'(y) dy = lEI.

IE

4.28 (Change of Variable) Let I = [a,b], and g: I -+ R, g(I) C J C R, be continuous there. Furthermore, if J = [c,d) and I: J -+ R is integrable, put F(x) = irc,x] I dy, c ~ x ~ d. Now, suppose that 9 and Fog are a.e. differentiable on their domains of definition, and prove that the relation (F 0 g)' = (J 0 g) g' holds a.e. on I. Finally, show that Fog is AC on I iff (i) (J 0 g) g' E L([a, b)). (ii) For each subinterval I' = [a',b'] of I we have

1

9 (b')

g(a')

I dy =

1

(J 0 g) g' dy .

l'

4.29 (Integration by Parts) Let I = [a, b] be a bounded interval, and suppose F, 9 E L(I). Show that if

F(x)= [

Idy,

J[a,x]

G(x)= [ J[a,x]

gdy,

a~x~b,

then IG,gF E L(I) and

i

lGdY +

i

Fgdy = F(b)G(b) - F(a)G(a).

X.

182

Absolute Continuity

Also, if I,g are AC on [, we have

iI9'dY+ il'9dY=/(b)9(b)-/(a)g(a). 4.30 Suppose I, I' E L(R). Prove that

JR I' dy = O.

4.31 Let [ = [a, b], and suppose I is a continuous real-valued function defined 'on [. The estimate (1.5) in Chapter III implies that if V(x) is AC on [, then also I is AC on [. Discuss whether the converse is true, to wit, does the assumption that I is AC on [ imply that V(x) is AC on I? 4.32 Let {In} be a sequence of AC functions defined on [ = [0,1] such that In(O) = 0 for all n. Assume that the sequence of derivatives {/~} is Cauchy in L(I), i.e., limn,m-+ooJII/~(x) - 1~(x)ldx = o. Prove that {In} converges uniformly to a function I, and that I is AC in [. 4.33 Prove that a real-valued function I(x)

=j

(-oo,x]

dy,

I defined on R is of the form where E L(R),

iff (a) I is AC on [-n,n] for all n, (b) V(J; -n,n) n, and, (c) limlxl-+oo I(x) = O. Prove it.

~ k

<

00

for all

4.34 Let [ = [a,b], and suppose I is a continuous function defined on [. Show that I is AC on [iff there exists a sequence {In} of Lipschitz functions defined on [ such that lim n-+ oo V(J - In; a, b) = O. 4.35 Suppose I, In are BV on [

V(Jn - I; x)

= [a, b], n = 1,2, ... , and -t

0,

for some a < x < b.

Prove that there exists a subsequence

nk

-t

00

such that

CHAPTER

XI

Signed Measures

In this chapter we consider u-additive set functions of arbitrary sign, or signed measures, establish their basic properties and describe, in the Theorems of Lebesgue and Radon-Nikodym, their basic form. I learned the proof of these theorems from R. Bradley.

1.

ABSOLUTE CONTINUITY

In Chapter IV we dealt briefly with additive set functions of arbitrary sign; these functions appear quite naturally in applications. We consider here extended real-valued u-additive set functions; we motivate our interest in them with two simple examples. Let 1'1' 1'2 be measures on (X, M), and let v be the set function defined on M by veE)

= 1'1 (E) -

J.t2(E) ,

E EM.

(1.1)

Although v is not necessarily nonnegative, it satisfies many of the properties of a measure provided, of course, that the right-hand side of (1.1) is defined. Similar considerations apply to the set function v( E) =

L

f dJ.t1 , E EM,

(1.2)

where f is an extended real-valued measurable function defined on X for which the integral in (1.2) exists. To consider the general setting we introduce the following definition: Given a set X and au-algebra M of subsets of X, we say that a set

183

XI.

184

Signed Measures

function v defined on M is a signed measure provided the following three properties hold: (i) v:M --+ [-00,00], and v assumes, at most, one of the values-oo or 00. (ii) v(0) = O. (iii) If {EkH~l !;; M is a sequence of pairwise disjoint sets, then (1.3) The equality in (1.3) means, in particular, that if Iv(Uk Ek)1 < 00, then the series on the right-hand side of (1.3) converges absolutely and unconditionally, and that it diverges properly to ±oo otherwise. Also note that property (ii) rules out the possibility that v is identically -00 or identically 00. The usual properties of a measure are true in this more general setting as well. For instance, properly interpreted, the results concerning limits discussed in Proposition 3.1 of Chapter IV hold; it is incumbent upon the reader to verify that this is the case. How do signed measures look, and how can they be represented? Referring to 4.8-4.12 in Chapter IV, given a signed measure v on (X,M), let v+, v_ and Ivl denote the positive variation, the negative variation, and the total variation of v, respectively. By 4.14 in Chapter IV, all the variations are actually measures on (X,M) and, by 4.11 in that Chapter, the Jordan decomposition

v(E) = v+(E) - v_(E) ,

E EM,

obtains. Thus, a general representation in the spirit of (1.1) holds for arbitrary signed measures. We note in passing that the Jordan decomposition is not unique: If J.L is any finite measure on (X,M) and vI = v+ + J.L and v2 = v_ + J.L, then we also have

v(E)

= vI(E) -

v2 (E) ,

E EM.

However, the Jordan decomposition satisfies a "minimality" condition, cf. 3.8 below. As for (1.2), it also leads to an interesting theory; the observations in 3.7 in Chapter VIII are relevant here. In fact, motivated by those considerations we introduce the following definition: Suppose J.L is a measure and v is a signed measure on (X,M)i we say that v is absolutely continuous with respect to J.L, and denote this by v <: J.L, if v(A) = 0 for any A E M with J.L(A) = O. Informally, each J.L-null set has v measure O. Our first result explains how this nomenclature is derived from our usual understanding of absolute continuity.

1.

Absolute Continuity

185

Proposition 1.1. Let J-LF be the Borel measure induced by the distribution function FE V. Then J-LF is absolutely continuous with respect to the Lebesgue measure iff F is AC on every bounded interval of R. Proof. We show the necessity first: IfF is not AC on every bounded interval of R, then there exist an interval 1= [a,b] and € > 0, such that for every 0 > there is a family {[ak,b k]} of nonoverlapping subintervals of I so that

°

L IF(bk) -

(1.4)

F(ak)1 ;?: e,

k

Now, since as is readily seen, cf. Theorem 2.1 in Chapter IX,

(1.4) implies that the set B = Uk[ak,b k] verifies (1.5) We now invoke (1.5) with 0 = 1/2 n , n of (bad) sets {Bn} so that

J-LF(Bn) > e,

IBnl ~

= 1,2, ... ,and construct a sequence

1/2 n ,

Bn ~ I,

all n.

Thus, on the one hand by the Borel-Cantelli Lemma it follows that Ilim sup Bn I = 0, and, on the other hand, by 4.25 in Chapter IV, we have J-LF(limsup Bn) > 0, contradicting the fact that J-LF is absolutely continuous with respect to the Lebesgue measure. As for the sufficiency, suppose that IAI = 0, and given e > 0, let 0 > be the number that corresponds to the choice of € in the AC definition of F on [-2n,2n], n > 0. Observe that since IA n [-n,n]1 = 0, there exists an open set 0 = Uk(ak,bk) ~ [-2n,2n] such that

°

An [-n,n] C 0,

101 < O.

Also note that by (2.4) in Theorem 2.1 in Chapter IX,

EZ"=l (bk - ak) ~ 101 ~ 0, we get J-LF (U:=l (ak,b k ») ::; :E:=1 (F(bk) - F(ak» ::; e,

and consequently, since

all N.

XI.

186

Signed Measures

Since this inequality holds for all N, it follows that J.LF(O) c is arbitrary and J.LF is regular, we have

J.LF(A n [-n,n]) = 0,

~

c, and since

all n.

But this can only be true if J.LF(A) = 0, and consequently, J.LF is absolutely continuous with respect to the Lebesgue measure. • Also corresponding to the following result.

Suppose J.L is a measure and v is a signed measure

Proposition 1.2. on (X, M) so that

Iv(E)1 < Then, v

~

c-o definition of AC functions, there is the

00

whenever

J.L(E) <

00.

(1.6)

J.L iff given c > 0, there exists 0 > 0, such that Iv(E)1 < c

whenever

J.L(E) <

o.

(1.7)

To show that the condition is sufficient observe that if J.L(E) < 0 for all 0 > 0, and (1.7) gives that v(E) = 0. As for the necessity, suppose that v ~ J.L and that (1.7) is false. Then there exist a sequence {Bn} ~ M and c > 0, such that Proof.

= 0, then J.L(E)

Iv(Bn)1 > c

J.L(Bn) ~ 1/2n ,

and

n

= 1,2, ...

Pick now a subsequence {B nk }, say, so that all the v(Bnk)'s are of the same sign, and observe that since J.L(Uk B nk ) < 00, by (1.6) it also follows that Iv(Uk Bnk)1 < 00. The proof may now be finished in a stroke: By 4.25 in Chapter IV, Iv(lim sup Bnk)1 > 0, and by the Borel-Cantelli Lemma, J.L(lim sup B nk ) = 0; this contradicts the fact that v ~ J.L. • Observe that even if v is a measure, (1.6) is necessary for Proposition 1.2 to hold. Consider, for example,

v(E) =

°

l

x 2 dx,

E E C.

Then lEI = implies v(E) = 0, but since for Ixllarge the set E = (x,x+7]) has lEI = 7] and v(E) is large, (1.7) fails. Next suppose that p, is a probability measure and v is a signed measure defined on (X, M), and that

Iv(E)1

~

p,(E) ,

E EM.

(1.8)

1.

Absolute Continuity

187

Clearly v <:: pj the question is, then, how to go about constructing a measurable function I so that

veE) =

Lld

P,

E EM.

(1.9)

Now, for A EM, the function

(1.10) = (v(A)/ peA)) XA + (v(X \ A)/p(X \ A» XX\A , satisfies (1.9) for E = A,X\A. I is a first and crude approximation to the

I

solution to our problem, and it is well-defined if the convention % = 0 is in forcej in fact, by (1.8), it is natural to adopt this convention. Observe that I is measurable and III ~ 1. We may think of I as Ip, namely, as a function associated to the measurable partition P = {A, X \ A} of X. Intuitively, the more refined the partition P is, the more spread out the function Ip associated to P by a formula similar to (1.10) will be. We remind the reader that we only consider measurable partitions of X, and that, given partitions P and pI of X, we say that pI is finer than P if given A' E pI, there exists A E P such that A' ~ A. Now, on a probability measure space a natural measure ofthe amount of spread of a measurable function is its "variance"j in the process described in (1.10) above the finer the partition P becomes, the greater the variance of the associated function Ip is. We expect, then, that the function I that verifies (1.9) will emerge as the limit function of the Ip's as the partitions P get finer, or as the variance of the Ip's approaches a supremum. The "expected value" of the Ip's is the same, and, as in the case of the function defined by (1.10), it equals I dp = veX). Thus by the well-known formula

Ix

variance = (second moment) - (expected value)2 , it is clear that maximizing the variance of the h's is equivalent to finding the maximum of the second moments of these functions. The proof of (1.9) we present below follows along these lines. More precisely, we construct the function I by maximizing the second moments of an appropriate family of functions. We begin by formalizing the relation between the partitions P of X and the functions Ip associated with them. Specifically, given a partition P = {At, ... ,An } of X, let Ip:X --+- [-1,1] be defined by the expression n

Ip

=

L (V(Ak)/ peAk)) XA,. , k=l

(1.11)

XI.

188

where the convention % = 0 is in force. Clearly, if A E M is any set of the form A k m ::; n, we have

Signed Measures

= Uj=l Ak"

1 ::;

kl < ... <

and, in particular,

v(X)

=

Ix 11'

dp.,

all partitions P.

The following properties of partitions are essential to carry out the verification of (1.9). Lemma 1.3. Let (X,M,p.) be a probability measure space, and suppose P and pi are partitions of X, with pi finer than P. If v is a signed measure on (X,M) which satisfies (1.8), A E P, and 11' is given by (1.11), the following five properties hold:

v(A)

=

Lh

L = Lli,

dp.

= 11"

L11' 11" L11' U1" - 11') = LIf" = LU1" - 11')2 + LIf, LIf, LIf" dp.

dp.

dp.

and

dp. ,

(1.13)

0,

(1.14)

dp.

dp. ::;

(1.12)

dp.,

dp. .

dp. ,

(1.15)

(1.16)

By the additivity of the integral it is clear that the above properties also hold for any set E E M of the form E = Uk Ak, where Ak E P for each k. In particular, they are true for E = X. Proof. Since pi is finer than P, each A E P may be written as a finite union A = Uk Ak, Ak E P'. Furthermore, since

11"(x) = V(Ak)/p.(Ak) , x E Ak, it readily follows that

f 11" lA

dp.

= I: I 11" Ie lAic

dp.

= I: (V(Ak)/ P.(Ak)) P.(Ak) = v(A) . Ie

1.

Absolute Continuity

189

Similarly, since fp(x) = II(A)/p(A) ,

we also have

x E A,

(1.17)

L

fp dp = (II(A)/ p(A)) p(A) = II(A) ,

and (1.12) holds. The verification of (1.13) is also simple: If A E P is partitioned into Ak's E pI as before, we have fp(x)fp,(x) = (II(A)/ p(A)) I)II(Ak)/ peAk)) XAk (x),

x E A,

k

and consequently, the left-hand side of (1.13) is equal to

f

JA

fp fpl dp = (II(A)/ p(A))

Lk (II(Ak)/ peAk)) JAf

dp k

= (II(A)/ p(A)) II(A). As for the right-hand side of (1.13), on account of (1.17) it equals

L

(II(A)/ p(A))2 dp = II(A)2 / p(A),

and (1.13) holds. (1.14) is a simple rewriting of (1.13). Next we consider (1.15). Since fpl = UPI - fp)+ fp, it readily follows that f~, = UPI - fp)2 + 2 UPI - fp) fp + f~ . (1.18) Whence, integrating the identity (1.18) over A, and invoking (1.14), we get at once that (1.15) holds. (1.16) follows from (1.15), and we have finished. • We are now ready to prove Theorem 1.4. Suppose (X,M,p) is a probability measure space and II is a signed measure defined on (X, M) such that III(E)I ~ p(E),

all E EM.

Then there exists a unique measurable function f: X

lIe E) =

kf

dp ,

all E in M .

(1.19) ---t

[-1,1] such that (1.20)

Uniqueness is understood in the following sense: If It is another measurable function defined on X for which (1.20) holds, then f = It p-a.e.

XI.

190

Signed Measures

Proof. Consider the collection of all the finite measurable partitions P of X, to each associate the function fp given by (1.11), and set

TI = sup { 'P Jx

I~ dJ.L .

Clearly 0 ~ TI ~ 1. Let now {Pn } be a sequence of partitions of X with the property that

IxI~ndJ.L;:::

TI-1/4 n ,

(1.21)

n=1,2, ...

It is more efficient, however, to work with the sequence {P~} consisting of the common refinement of the Pn's; specifically, let Pf = PI, P 2 be the common refinement of PI and P2, and so on. Denote I'P'n by In, and observe that since P~ refines P n for all n, by (1.16), (1.21), and the definition of TI, we have

(1.22) Next we construct a "maximal" element from the In's. Since for each n P~+! refines P~, by (1.15) and (1.22) it follows that

Ix (Jn+! - In)2 dJ.L = Ix

I~+! dJ.L -

Ix

I~ dJ.L

~TI-(TI-1/4n)=1/4n,

n=1,2, ...

Thus, by the Cauchy-Schwarz inequality, cf. (3.1) in Chapter VIII, we have

Ix lin+! - Inl dJ.L

~

(Ix (In+! - In)2 dJ.L) 1/2 (J.L(X))1/2

~

1/2n,

n = 1,2, ...

and consequently, since

we may invoke 4.7 in Chapter VII and obtain that L:n(Jn+l - In) converges absolutely, and pointwise, J.L-a.e. on X. Put now N

9 = lim L(Jn+l - In) = lim IN+l N-+oo n=l

N-+oo

It ;

1.

Absolute Continuity

191

9 is well-defined J.L-a.e. on X.

Next we show that the function I = 9 + It gets the job done; in other words, (1.20) is true for I and any E in M. Given E E M, let {Qn} be the sequence of partitions consisting of the common refinement of the partitions p~ and {E, X \ E}. Clearly Qn is finer than p~, n = 1,2, ... , and Qn+1 is finer than Qn, n = 1,2, ... Thus, if h n = hQn' n = 1,2, ... , by (1.22) and (1.16) it readily follows that

Ix I~ dJ.L ~ Ix h~ dJ.L ~ "', Ix (hn - In)2 dJ.L Ix h~ dJ.L - Ix I~ dJ.L '" - 1/4n

~

all n,

and, by (1.15) and (1.22), we also get =

~ '" - ('" - 1/4n ) = 1/4 n ,

all n. (1.23) Consequently, once again by the Cauchy-Schwarz inequality and (1.23), we have

Ik

(hn - In) dJ.L1

~ (k (hn ~

In? dJ.L) 1/2 J.L(E)I/2

(Ix (hn - In? dJ.Ly/2 ~

1/2 n ,

all n. (1.24)

We are now ready to compute veE). Since Qn is finer than {E,X \ E} for all n, by (1.12) we have

veE)

=k

hn dJ.L

=k

(hn - In) dJ.L + kIn dJ.L

= An + B n ,

say. By (1.24), lim n -+ oo An = O. As for the Bn's, first observe that lim In = I J.L-a.e., and I/nl ~ 1 all n. n-+oo

Whence, by LDCT it follows that lim Bn = n-+oo

JEf I dJ.L ,

and (1.20) holds. Further, if It is another measurable function for which (1.20) is true, then fEU - It) dJ.L = 0 for all E E M, and consequently, I - It = 0 J.L-a.e. • The applications of Theorem 1.4 are numerous and interesting; we begin by showing how a signed measure may be decomposed into "positive" and "negative" parts. Theorem 1.5, and the remarks that follow it, are known as Hahn's decomposition theorem. This decomposition is true in greater generality, cf. 3.6 below, but the result presented here is sufficient for the applications we have in mind.

XI.

192

Signed Measures

Theorem 1.5. Let v be a signed measure defined on (X, M), and suppose that its variation Ivl is a probability measure on (X, M). Then there exist two disjoint, measurable sets A and B, A U B = X, so that (i) veE n A) ~ 0, all E E M, (ii) veE n B) ~ 0, all E E M. Proof.

Since

Iv(E)1 ~ Ivl(E) ,

all E EM,

Theorem 1.4 applies with Il. = Ivl. In particular, there is a measurable function I, III ~ 1, that satisfies

veE)

= ~ I dlvl ,

all E EM.

We claim that III = 1 Il.-a.e. First note that since Ivl(X) = 1, by 4.12 in Chapter IV, given e > 0, there exists a partition P = {Al, ... ,Am } of X such that m

1- e~

L

Iv(Ak)1 ~ Ivl(X) = 1. k=l Thus, by (1.12), it readily follows that m

1-e ~ Llv(Ak)1 k=l =

t

Ivalue of

k=l

11'

on Akllvl(Ak) = 11/1'1 dlvl·

x

This estimate, together with the Cauchy-Schwarz inequality, and with the meaning of TJ introduced in Theorem 1.4, give

Since e is arbitrary, it follows at once that TJ = 1. Referring once again to the proof of Theorem 1.4, by LDCT we obtain 1

= TJ = n-oo}x lim { I~ dlvl = { 12 dlvl, }x

and consequently, since Ivl is a probability measure, we have a.e., and III = 1 Ivl-a.e., as claimed.

P

= 1 Ivl-

1.

Absolute Continuity

193

Let now A = {x EX: I(x) = 1}, and B = X \ Aj {A,B} is a disjoint partition of X. Since I = -1 Ivl-a.e. on B, for any E E M we have

v(E

n A) = f

lEnA

Idlvl = Ivl(E

n A) ~ 0,

and

v(EnB)=

f

lEnB

Idlvl=-lvl(EnB)$O._

Two remarks concerning this result: First, the decomposition X =

A u B is not unique, and second, the result is true in greater generality. For instance, if Ivl is O"-finite, then we have X = Uk Xk, where the Xk'S are pairwise disjoint and Ivl(Xk) < 00 for all k. By rescaling if necessary we may assume that Ivl(Xk) = 1 for all k, and apply Theorem 1.4 to each Xk, thus obtaining sequences {Ak}, {Bk} as in that theoremj then the sets A = Uk Ak and B = Uk Bk correspond to the "positive" and "negative" parts of v respectively. Another interesting consequence of Theorem 1.4 is Theorem 1.6. and that

Suppose that >',1' are O"-finite measures on (X,M)

>'(E) $ J.t(E) ,

all E EM.

Then there exists a unique (in the J.t-a.e. sense as explained above) nonnegative measurable function I: X - t [0,1] such that

>'(E) = lldJ.t,

all E EM.

(1.25)

Furthermore, if 9 is a measurable extended real-valued function defined on X, then (1.26) The equality in (1.26) is understood as follows: If the integral on either side of (1.26) exists, then the integral on the other side also exists and they are equal. Proof. Write X = Uk X k , where the union is pairwise disjoint and J.t(Xk) < 00 for all k. By rescaling if necessary, assume that J.t(Xk) = 1, and invoke Theorem 1.4 for the measures >'k,J.tk on (X,M) given by

XI.

194

Signed Measures

The function fin (1.25) is obtained as f = :Ek fkXXk' where the fk'S are the (unique) functions that satisfy (1.20) for the measures Ak and J1k. As for (1.26), suppose first that 9 ~ 0, and let {
1


X

j

=

j

Ix (L:

j

l

ajXEj) f dJ1

fdJ1

EJ

=

Ix


and, by MCT, it follows that (1.26) holds in this case. As for a general function g, note that 9 = g+ - g-, and apply the above result to g+ and g- separately. •

2. THE LEBESGUE AND RADON-NIKODYM THEOREMS Let J1 and v be signed measures on (X,M). We say that J1 and v are mutually singular, and denote this by J1 .1 v, if there exists a disjoint partition {A, B} of X such that IJ.LI(A) = Ivl(B) = O. In this case we also say that J.L is singular with respect to v, or symmetrically, that v is singular with respect to J1. For instance, in Theorem 1.5 the measures

v I (E) = veE n A),

v2 (E) = veE n B),

E EM,

are mutually singular. Another interesting example we have encountered is that of J1 = J1F' the Borel measure induced by the extension F to R of the Cantor-Lebesgue function, and v the Lebesgue measure on the line. To show that this last example is part of a general state of affairs, and to elucidate the notion of singularity, we present a preliminary result. Lemma 2.1. Suppose J.LF is a finite Borel measure induced by a distribution F E V. If A is a Borel set on the line such that F' exists on A and M, m ~ 0, then we have (i) If F'(x) ~ M for x E A, then J.LF(A) ~ MIAI. (ii) If F'(x) ~ m for x E A, then J1F(A) ~ miAI.

2.

The Lebesgue and Radon-Nikodym Theorems

195

Proof. Given c > 0, let In denote the collection of intervals of the form (u,v] that satisfy the following two properties: (a) u, v are rational, and 0 < v - u < lin. (b) F(v) - F(u) ~ (M + c)(v - u). Observe that the sets

An =

U

(An (u,vD,

n = 1,2 ...

(u,V]EIn

are Borel sets, and that under the hypothesis of (i), they increase to A. Let {In,k} be a sequence of nonoverlapping intervals, open on the left and closed on the right, such that

L IIn,kl ~ IAnl

+c .

k

By working with the In,k's it is possible to assume that each In,k meets An, has rational endpoints, and that IIn,kl < lin for all k. Then, (a) and (b) above apply to each In,k = (Un,k,Vn,k], and it readily follows that

JLF(An) ~ LJLF(In,k) = L(F(vn,k) - F(Un,k)) k

~ (M

k

+ c) L(Vn,k -

Un,k) = (M

+ c) L

k

~

(M + c)( IAnl

IIn,kl

k

+ c).

Since c is arbitrary the above inequality implies that

and since the An's increase to A, a similar inequality holds with A in place of An above, and (i) is true. The proof of (li) follows along similar lines, a Vitali covering argument also works in this case, and is therefore left to reader. • We are now ready to prove Proposition 2.2. Suppose JLF is a finite Borel measure induced by a distribution F E 'D. Then JLF is singular with respect to the Lebesgue measure on the line iff F is singular, Le., F' = 0 a.e.

XI.

196

Signed Measures

Proof. We first show that the condition is sufficient. Given "I by (i) in Lemma 2.1 we get

and consequently, by first letting "I -+ 0 and then n -+ follows that JLF( {F' = O}) = o. Put now A

= {F' = O} and B = R \ JLF(A)=IBI=O,

00,

> 0,

it readily

A. Since F is singular we have

AnB=0,

AUB=R,

(2.1)

and consequently, JLF is singular with respect to the Lebesgue measure. On the other hand, if (2.1) holds for Borel sets A and B, by the other half of Lemma 2.1, it readily follows that for "I > 0 we have

"I1{x: F' ~ "I} I = "I1{x E A: F' ~ "I}I + "I1{x E B: F' ~ "I} I = "I1{x E A: F' ~ "I}I ::; JLF(A) = O. So, for each "I > 0 we have I{F' ~ "I}I = 0, and by letting "I -+ 0 we get that F' = 0 except on a null Lebesgue set. • Propositions 1.1 and 2.2 suggest that it may be possible to decompose signed measures in terms of absolutely continuous and singular measures. This indeed is the case; we begin by discussing a preliminary result in this direction. Theorem 2.3. Suppose A and JL are finite measures on (X,M). Then there exist finite measures Aa and As which satisfy the following properties:

(i) Aa <:: JL, As .L JL. (ii) A = Aa + As . Furthermore, the measures Aa, As are unique. Proof.

The set function

(A + JL)(E) = A(E) + JL(E) ,

E EM,

is a measure on (X,M) which satisfies

A(E) ::; (A + JL)(E) ,

all E in M .

2.

The Lebesgue and Radon-Nikodym Theorems

197

Whence, by (1.25) in Theorem 1.6, there exists a measurable function f: X ~ [0,1] such that

A(E)

= Lfd(A+J.t) = Lf dA + LfdJ.t,

EEM.

(2.2)

Next, let 9 = fXE' and observe that by (1.26), with I' = A + I' there, we have

fx fXE dA = fx fXEf d(A + 1'),

and consequently, (2.2) may be rewritten as

Now, this procedure may be iterated with 9 = f2XE' 9 = f3 XE , and so on, and (2.3) becomes

A(E) = L fn dA + L (f + ... + fn) dJ.t,

n = 1,2,...

(2.4)

Let B = {f = 1}, put

As(E)

= A(E n B),

E EM,

and observe that (2.4) becomes

A(E)

= As(E) + { fndA + {(f + ... + fn)dJ.t JEn(X\B) JE =I + J + K ,

(2.5)

say. Clearly, by LDCT, lim n .... oo J = 0. As for K, note that by MCT and (2.4) it follows that lim (f

n .... oo

+ ... + In) =

_f_ E L(J.t) 1- I

and consequently, J.t(B) = 0, and lim K n .... oo

=

(_f_ dJ.t .

JE 1- f

Now, by (2.6) we obtain that the measure Aa(A)

= 1_1_ dJ.t, JA 1-1

A EM,

(2.6)

XI.

198

Signed Measures

is absolutely continuous with respect to p,. Thus, returning to (2.5), we get A(E) = lim (I + J + K) = As(E) + Aa(E) , n-+oo

and (ii) holds. Next we show that As .L p,; since p,(B) = 0, this reduces to checking that As(X \ B) = dA = 0,

j

(X\B)nB

which is obviously true. Finally we show that the decomposition is unique; this is not hard. Suppose A = A~ + A~ is a decomposition of A that satisfies (ii) above, and let A~(X \ B') = p,(B') = 0, B' EM. We claim that

Aa(E)

= A~(E),

all E EM.

Indeed, since p,(B U B') = 0, by the absolute continuity of Aa and have

Aa(E n (B U B'))

= A~(E n (B U B')) = 0,

(2.7) A~

all E EM.

we

(2.8)

Moreover, since En (X \ (B U B')) is a subset of both X \ B and X \ B', we also have

= A~(E n (X \ (B U B'))) = 0, E EM. Whence, since Aa + As = A~ + A~, it readily follows that Aa(E n (X \ (B U B'))) = A~(E n (X \ (B U B'))) , E EM. (2.9) As(E n (X \ (B U B')))

Finally, combining (2.8) and (2.9), for each E E M we get

Aa(E) = Aa(E n (B U B')) + Aa(E n (X \ (B U B'))) = Aa(E n (X \ (B U B'))) = A~(E n (X \ (B U B'))) = A~(E n (B U B'))) + A~(E n (X \ (B U B'))) = A~(E), and (2.7) holds. Further, since Aa = A~ and since all the measures involved are finite, then As = A~ also holds. • We are now ready to prove a result in the spirit of Theorem 3.7 in Chapter X, it is appropriately known as the Lebesgue Decomposition Theorem.

2.

The Lebesgue and Radon-Nikodym Theorems

199

Theorem 2.4. Suppose J.L and A are O'-finite measures defined on (X,M). Then there exist O'-finite measures Aa,Aa defined on (X,M) such that:

(i) Aa <: J.L , Aa 1. J.L . (ii) A = Aa + As. Furthermore, the measures Aa and As are unique. Proof. The idea of the proof is to reduce the general hypothesis to the special case when the measures involved are finite, and then to invoke Theorem 2.3. First note that since J.L and A are O'-finite, we can write X as a pairwise disjoint union X =

UXk,

with

J.L(Xk) , A(Xk) <

00,

all k.

k

Next we localize the problem at hand by introducing the finite measures J.Lk and Ak defined on (X,M) by

For each k ~ 1, let Ak = Ak,a + Ak,s be the unique decomposition of Ak obtained in Theorem 2.3 with the property that

Moreover, since the Xk'S are pairwise disjoint it is also simple to verify that if we put

then A = Aa holds. •

+ As

is the unique decomposition of A for which (i) above

It is not hard to extend this result to signed measures. Indeed, we have

Theorem 2.5. Suppose J.L is a O'-finite measure and v is a signed measure defined on (X,M). If Ivi is O'-finite, then there exist signed measures Va, Vs defined on (X, M such that: (i) Va <: J.L , and Va 1. J.L.

(ii) V = Va

+ Va·

Furthermore, the signed measures Va and VB are unique.

XI.

200

Signed Measures

Proof. By the remarks following the Hahn decomposition theorem we can find O'-finite measures A, A' on (X, M) such that v = A - A'. Let now

A = Aa + As ,

A'

= A~ + A~

be the unique decomposition of A and A' obtained in Theorem 2.4 with the property that

It is not hard to check that if we put

then v =

Va

+ Vs is the (unique) decomposition of v that does the job.

•

These interesting results still do not address the question raised in considering (1.2) above. Now, in that example we had v <: Jl, and under this assumption we show that (a preliminary version of) the Radon-Nikodym Theorem holds. Theorem 2.6.

Suppose Jl and A are finite measures defined on

(X, M), and that A <: Jl. Then there exists a nonnegative integrable function h such that

A(E) =

~hdJl'

alIEEM.

(2.10)

h is called the Radon-Nikodym derivative of A with respect to Jl, and one writes h = dA/ dJl , or dA = h dJl . Furthermore, h is unique. Proof. The proof is identical to that of Theorem 2.3. First observe that since in the notation of Theorem 2.3 Jl(B) = 0, and since A <: Jl, we also have A(B) = O. But this readily implies that As(E) = 0 for all E EM, and consequently for any E in M we have

f

h= 1_

f

E L(Jl) . •

It is clear that we cannot expect (2.10) to hold in general. For instance, if A is the Lebesgue measure on R and Jl = 6 is the Dirac delta

2.

The Lebesgue and Radon-Nikodym Theorems

201

measure at 0, it is obvious that (2.10) cannot hold for any h E L(J.L); thus the assumption A < J.L is necessary. On the other hand, if J.L is the counting measure on [0,1], then A < J.L but still (2.10) is not true for any nonnegative measurable function h defined on [0,1]. The difficulty here is that the integral on the right-hand side of (2.10) is finite only when the set {h =I O} n E is at most countable, and [0,1] is an uncountable set of finite Lebesgue measure. A moment's thought will convince the reader that these are the only difficulties in extending Theorem 2.6 to a more general setting. Indeed, we have Theorem 2.7 (Radon-Nikodym Theorem). Let J.L be a O'-finite measure, and v a signed measure defined on (X,M). If Ivl is O'-finite and v < J.L, then there exists an extended real-valued measurable function h defined on X such that if E E M and Ivl(E) < 00,

v(E)

=

k

(2.11)

hdJ.L.

h is called the Radon-Nikodym derivative of v with respect to J.L, and writes h = dv/dJ.L, or dv = hdJ.L.

o~

Also h is unique, in the sense that if h' is another extended real-valued measurable function defined on X for which (2.11) holds, then h = h' J.L-a.e. Proof. The proof follows along the lines to that of Theorem 2.5; first we localize the problem at hand and reduce it to the particular case of finite measures. Write X as a pairwise disjoint union

X

= UXk ,

with

J.L(Xk),lvl(Xk) <

00 ,

all k.

k

Consider now the finite measures J.Lk and signed measures Vk defined on (X,M) by

= J.L(E n Xk), and vk(E) = v(E n Xk), k = 1,2, ... Further, write Vk = Ak - Ak , where the Ak'S and Ak's are finite measures J.Lk(E)

on (X, M) such that Vk = Ak - Ak for k = 1,2, ... It is also easy to check that under our assumptions we have

Ak, Ak

<

J.Lk,

and

Ak,B - Ak,B = 0,

all k.

XI.

202

Signed Measures

Now, by Theorem 2.6 there exist (unique) nonnegative functions hk' h~ in L(/1k) with the property that

A~(E) =

Ak(E) = k hk d/1k, and consequently for each k

k

h~ d/1k,

all E EM,

= 1,2, ... , we have

vk(E) = k(hk - hk)d/1k'

E EM.

It is now readily seen that the function h = L,k(h k - h~) has all the desired properties. •

The assumption that /1 is u-finite is essential for the validity of Theorem 2.7. For, suppose X is an uncountable set, and let M be the u-algebra of those subsets of X which are either at most countable or so that their complements are at most countable. For E E M put /1(E) = the number of elements of E if E is finite and /1(E) = 00 otherwise, and veE) = 0 or 1 according as to whether E is at most countable or not. Then v ~ /1, but no integral representation such as the one given in (2.11) is possible. Clearly /1 is not u-finite. The theorems of Lebesgue and Radon-Nikodym have many interesting applications, we discuss one next. If v is a signed Borel measure, we say that v is differentiable at x E R n , provided that Dv(x) = lim v(I(x,r)) exists. r-+O

II(x,r)1

Then the following is true. Proposition 2.8. Suppose v is a signed Borel measure so that Ivl is finite on bounded sets of Rn. Then v is differentiable a.e. More precisely, if v = Va + Vs is the Lebesgue decomposition of v with respect to the Lebesgue measure /1, Va ~ /1, Vs .1 /1, we have

Dvs = 0 a.e.

and

DVa = dva/ d/1 a.e.

Proof. Since Ivl is u-finite, Theorem 2.4 applies. Let h denote the Radon-Nikodym derivative of Va with respect to the Lebesgue measure and observe that

v(I(x,r)) _ vs(I(x,r)) II(x,r)1 - II(x,r)1 = A+B,

+

1

II(x,r))1

f iI(x,r)

hdy

2.

The Lebesgue and Radon-Nikodym Theorems

203

say. Since by the Lebesgue Differentiation Theorem we have limr-+o B = h a.e., we are reduced to showing that limr-+o A = 0 a.e.; this is not hard. First observe that since Va is the difference of two Borel measures whose variations are finite on bounded sets, also IVai enjoys this property. The proof now follows along the lines of Proposition 2.2. Since

Iva(I(x,r))1

~

Ival(I(x,r)),

it clearly suffices to show that (2.12) = 0 a.e. Let B be a Borel set such that Ival(B) = IRn\BI = 0, and for k = 1,2, ... ,

Dlval

let

Fk Since IFk that

n (Rn

={x E R

\ B)I

n' Ival(I(x,r)) 1} : li~-!~p II(x,r)1 > k

= 0, in order to prove

.

(2.12) it is enough to show (2.13)

Now, since IVai is a regular Borel measure, given e > 0, there exists an open set 0 ;2 B such that IVal(O) ~ e. Further, to each x E Fk n B associate an interval I(x,r) with the property that (2.14) Observe that if I( is an arbitrary compact subset of Fk n B, by a covering argument similar to that in the proof of the Hardy-Littlewood maximal theorem, specifically estimate (2.15) there, there exists a finite family I(xl,rI), ... ,I(xm,rm) of pairwise disjoint subintervals of 0 such that (2.15) But the intervals that appear on the right-hand side of (2.15) are special. In addition to being pairwise disjoint, they all satisfy (2.14). Therefore the sum on the right-hand side of (2.15) can be estimated by

Since e is arbitrary, this means that 11(1 = O. By the regularity of the Lebesgue measure it follows that (2.13) holds, and we have finished. •

XI.

204

Signed Measures

3. PROBLEMS AND QUESTIONS In what follows, all measures are assumed to be defined on a O'-algebra M of subsets of a space X, even when this is not explicitly stated. 3.1 Referring to the notation of 4.8 in Chapter IV and 3.10 in Chapter IX, if v is a signed measure, describe v+ " v-, and v+ V v- . 3.2 Let p, be a measure and f an extended real-valued measurable function defined on X. If v is the signed measure given by

v(E)

= kfdP"

E EM,

describe v+, v- and Ivl in terms of f. What if p, is a signed measure and f has constant sign? What if both p, and f are allowed to have variable sign? 3.3 Let F denote the class of all measures on (X, M). Show that the relation ~ --< p, iff ~(E) ~

p,(E) ,

all E EM,

(3.1)

is a partial ordering on F. 3.4 Show that if (3.1) holds then Ll(p,) ~ Ll(~). Is the converse true? 3.5 A subset A E M is said to be positive with respect to a signed measure v if for each E ~ A, E E M, we have v(E) ~ O. Similarly, a measurable set C is said to be negative with respect to v if for each E ~ C, E E M, we have v(E) ~ O. Finally, a set N E M is said to be null with respect to v if it is simultaneously positive and negative with respect to v. Show that there may be sets of measure 0 which are not null, as there may be sets of positive measure which are not positive, and sets of negative measure which are not negative, all with respect to v. Also, investigate some of the properties of these classes of sets. For instance, show that every measurable subset of a positive set is positive, etc. 3.6 If v is a signed measure which does not assume the value 00, and if v(E) > -00, show that E contains a measurable subset A such that: (a) v(A) ~ v(E), and, (b) A is positive with respect to v. 3.7 If v is a signed measure, prove that E is null with respect to v iff Ivl(E) = O.

3.

Problems and Questions

205

3.8 If V is a signed measure and A, v are measures such that v = A-I', show that

A(E) ~ v+(E) ,

and

veE) ~ v-(E) ,

all E EM.

3.9 Suppose J.LF is the probability measure on R induced by the distribution function F, and that F(x) = J.LFCC-oo,X]). If A denotes the

Lebesgue measure on Rand dJ.LF/dA

F(x) =

= j, show that

f j(y)dI'F(Y)' J<-oo,x]

-00

< x < 00.

The nonnegative function j is known as the probability density of

F. Compute the probability density that corresponds to the distribution function 0 if x < a F( x) = { (x - a) / (b - a) if a < x < b 1 if x ~ b.

3.10 Suppose

0 if x < -1 F( x) = { 1 _ x 2 if x ;: -1,

and let I' be the signed measure on (R,8) that satisfies I'«x, y)) = F(y)-F(x). Find the Hahn decomposition of I' and for an arbitrary interval I = (x,y] of R find explicit formulas for I'+(I), I'-(I) and

II'I(I)· 3.11 Does there exist an increasing distribution function F on R such

that the induced Borel measure I'F is not absolutely continuous with respect to the Lebesgue measure on R? 3.12 If v is a finite measure and I' a measure, show that the following are equivalent: (a) v <: 1', and, (b) If the sequence {En} ~ M has the property that lim n -+ oo I'(En) = 0, then lim n -+ oo v(En) = O. 3.13 Show that v <: I' iff v+, v- <: I' iff

Ivl <: 1'.

3.14 If Vb v2 are signed measures, I' is a measure and vI, V2 <: 1', prove that VI + V2 <: I' and d(VI + v2)/dl' = dVI/dJ.L + dV2/dl'. 3.15 Given measures A,J.L, show that

dA/d(A + 1') + dJ.L/d(A + 1') = 1 (.\ + J.L)-a.e.

XI.

206

Signed Measures

3.16 If A,p, V are measures such that A ~ p and p ~ v, show that A ~ v and dAldv = (dAldp) . (dpldv) A-a.e. 3.17 Suppose p, v are O'-finite measures so that p that

~

v and v

~

p. Show

dv I dp = 1/( dpI dv) p-a.e. 3.18 Let p, v be O'-finite measures, and suppose p - v is a measure so that v ~ p - v. Show that p({dvldp = 1}) = o. 3.19 Let p, v be O'-finite measures, and suppose v

v({dvldp

~

= O}) = o.

p. Show that

3.20 Referring to 4.14 in Chapter VII, if p is a finite measure and A is a O'-finite measure on (Y,N), then show that there exists a function h E LI(A) such that Jx f 0 Tdp = Jy f dv = Jy fhdA for all fin

LI(v). 3.21 Let p, v be O'-finite measures and assume v f E LI(v), f(dvldp) E L(p) and f dv =

Jx

~

Jx

p. Show that for all f(dvldp)dp.

3.22 Consider the relation defined on classes of measures by A '" p iff A ~ p and p ~ A. Show that '" is an equivalence relation and describe the relation between LI(A) and LI(p). 3.23 Given a measure p, put

veE) =

{Ooo

if peE) = 0 if peE) > o.

Show that v is measure on (X, M), and that v

~

p. Also find

dv I dp. 3.24 Let (X, M,p) be a measure space, N eM a O'-algebra of subsets of X, and v = piN the restriction of p to N. Show that given f E L(p), there exists a uniquely determined, v-a.e. that is, It E L(v) such that fA f dp = fA It dv for every A EN. It is called the conditional expectation of f with respect to N. 3.25 Show that the Radon-Nikodym Theorem remains true if p is a O'-finite signed measure.

3.26 Let p, v be regular Borel measures. Show that if supp p 0, then p 1. v. 3.27 Show that if Vb v2 1. p, then

VI

+ V2 1. p.

n supp v

=

s.

Problems and Questions

207

3.28 Show that J.L 1. v iff 1J.L11. Ivl. 3.29 Let J.L, v be u-finite measures. Show that J.L 1. v iff there exists no nonzero measure A such that A ~ J.L and A ~ v. 3.30 Suppose A and J.L are finite measures. Prove there is a (measurable) partition {AI,A2,A3} of X such that:

(i) A(Al) = 0, (ii) J.L(A 2) = 0, and (iii) On A 3, A ~ J.L and J.L ~ A. Also, show there exists a finite positive measurable function h on A3 such that for every nonnegative measurable function f on A 3,

f fdA = f fhdJ.L lA3 l~

and

f

l~

f dJ.L =

f (J / h ) dA . lA3

3.31 Suppose v is a signed Borel measure with the property that Ivl is finite on bounded sets on Rn. Referring to 3.25 in Chapter VIII, show that if R = {R} is a regular family, then

v({x + R}) diam(R}-+O IRI lim

exists a.e.

3.32 Suppose v is as in 3.31 and that A is a Borel subset of R n so that Dv(x) ;::: A for all x E A. Prove that v(A) ;::: AlAI. 3.33 Assume J.L, J.Lm are finite Borel measures, m = 1,2, ... , such that lim m -+ oo J.Lm(A) = J.L(A) for all A E Bn , monotonically (either nondecreasing or nonincreasing). Prove that

lim DJ.Lm = DJ.L

m-+oo

a.e.

3.34 A complex-valued set function v of the form

veE) = vI(E) + iV2(E) ,

E EM,

where VI and V2 are signed measures is called a complex measure. VI and V2 are called the real and imaginary parts of v, respectively. This is an open ended question: Discuss the properties of complex measures. For instance, show that if Ivl is given by 4.12 in Chapter IV, then the set function Ivl is a measure called the (total) variation of v. Also prove that if 7] = sup{lv(A)I: A E M, ACE}, then 7]

5I v l(E) 5 47],

all E EM.

XI.

208

Signed Measures

3.35 Let Jt and v be complex measures. We say that v is absolutely continuous with respect to Jt, and we write v <:: Jt, if IJtI(E) = 0 implies veE) 0, for E E M.

=

Discuss properties of absolutely continuous measures. For instance, prove that if v = VI + iV2, then the following are equivalent: (i) v <:: Jt, (ii) vt, VI' vi, v; <:: Jt, and, (iii)lvl <:: Jt. 3.36 Let Jt and v be complex measures. We say that Jt arid v are singular, and we write Jt 1. v, if there exists a set A E M such that IJtI(A) = Ivl(X \ A) = O. Discuss properties of singular measures. 3.37 Can you think of a Lebesgue decomposition theorem in case

(X,M,Jt) is a O'-finite measure space and v is a complex measure on (X,M)?

CHAPTER

XII

1Y Spaces

In this chapter we introduce the Lebesgue spaces of p-integrable functions and study their basic properties. In considering the various results discussed here, the reader should keep in mind the three basic examples: The LP spaces of Lebesgue measurable functions on the line, or Rn, the LP spaces of Lebesgue measurable functions on a bounded interval, and the sequence i P spaces. 1.

THE LEBESGUE LP SPACES

Let (X,M,JL) be a measure space and I an extended real-valued measurable function defined on X. Then, for 0 < p < 00, I/IP is also measurable and the expression

II/lIp =

(Ix I/IP dJL) IIp,

0

< p < 00,

(1.1)

whether finite or not, is well-defined and is called the "p norm" of I. The case p = 1 has been studied in Chapter VIII, and the natural question to consider is to what extent those results can be extended to values of p other than 1. First a definition. The Lebesgue class LP(X,JL), or plainly LP(X) or LP(JL), is defined as

LP(X,JL) = {I measurable: II/lIp < oo},

0


00.

(1.2)

Our immediate goals are to show that LP(JL) is a linear class and to introduce a metric that will turn LP(JL) into a complete metric space.

209

XII.

210

LP Spaces

It is not hard to verify that LP(p) is a linear class. Indeed, given E LP(p) and a real scalar ,x, first note that I + ,xg is measurable. Furthermore, since for nonnegative real numbers a, b we have

I, 9

(1.3) it follows that I/(x) + ,xg(x)IP ~ 2P(I/(x)iP consequently, we also have

+ 1,xIPlg(x)IP),x

E X, and

As for the metric, inspired by (1.1) in Chapter VIII, a natural choice is

dp(J,g) =

III - gllp,

0

< p < 00.

Since dp(J, g) = 0 implies I = 9 p-a.e, strictly speaking, the elements of LP(p) are equivalence classes of functions defined on X, where we agree that I = 9 means that I = 9 p-a.e. Now, since the cases 0 < p < 1 and 1 < p < 00 are essentially different, we treat them separately; we do the former case first. So, fix 0 < p < 1, and note that (1.3) can be improved to (1.4) with equality occurring in (1.4) iff ab = O. To see this observe that if ab = 0 there is nothing to prove. On the other hand, if ab =f. 0, and since (1.4) is to hold for any a, b > 0, we may replace a by ab there and consider the equivalent inequality

¢( t)

= (1 + t)P -

1 - tP ~ 0 ,

all t

~

0,

(1.5)

with ¢(t) = 0 ifft = O. But this is easily checked: Indeed, since 0 < p < 1, we have ¢'(t) = p(l + t)p-l - pt p- l < 0 for t > 0, and consequently, ¢(t) < ¢(O) = 0, which gives (1.5), including the remark concerning equality. With this observation out of the way, let A, B be disjoint measurable subsets of X neither of which is null, and let I = XA' 9 = XB. Then, putting a = J.L(A)I/p and b = J.L(B)I/p in (1.4) we obtain

dp(J,g) =

(L II - glP dP)

= (J.L(A)I/P)P

l/p

= (J.L(A)

+ J.L(B»I/P

+ (J.L(B)I/P)P//P ~ J.L(A)I/P + J.L(B)I/P,

1.

The Lebesgue LP Spaces

211

or, in other words,

Thus, the inequality opposite to the triangle inequality holds, and dp cannot be a distance function on LP(J.L). This is not, however, a serious difficulty. Indeed, by (1.4), the expression

satisfies the requirements of a metric, and essentially all the results discussed in Chapter VIII are true: Endowed with this metric LP(J.L) is a complete metric space and Co(Rn) is dense in LP(Rn). It is interesting to point out that LP integrable functions are not necessarily integrable, or locally integrable for that matter. For instance, when 0 < p < 1, by 4.46 in Chapter VII, the function Ixl-n E LP(I(O,l)), but it is not locally integrable in any neighbourhood of the origin. In order to deal with the case 1 < p < 00 we need a preliminary result which is essential in what follows. We have already encountered a particular instance of this result, the Cauchy-Schwarz inequality, in 3.1 in Chapter VIII. We say that 1 < p, q < 00 are conjugate indices provided that l/p+ l/q = 1.

Conjugate indices are also related by the expressions p + q = pq and q = p/(p-1). An important property that the conjugate indices p = 1/17, q = 1/(1 - 17), 0 < 17 < 1, satisfy, is this: For any a, b ~ 0 we have

aTJb1 - TJ ~ 17a + (1 - 17)b.

(1.6)

Furthermore, equality holds in (1.6) iff a = b. To see this note that if ab = 0 there is nothing to prove. Otherwise, if ab :j:. 0, and since (1.6) is to hold for any a, b > 0, we may replace a by ab there and consider the equivalent assertion

¢( t) = t'1 - "It - (1 - 17) with ¢(t)

~

0,

all t

~

0,

= 0 iff t = 1. Note that since 0 < 17 < 1, we have ¢'(t)

= 17t '1- 1 -

17

> 0 if 0 < t < 1 if t = 1 < 0 if t > 1.

= { =0

XII.

210

LP Spaces

It is not hard to verify that LP(J-L) is a linear class. Indeed, given f,g E LP(J-L) and a real scalar .,x, first note that f + .,Xg is measurable. Furthermore, since for nonnegative real numbers a, b we have

( a + b)P ::; 2P (a P + bP) ,

0
it follows that If(x) + .,Xg(x)IP ::; 2P(lf(x)iP consequently, we also have

(1.3)

00 ,

+ 1.,XIPlg(x)IP),x

E X, and

As for the metric, inspired by (1.1) in Chapter VIII, a natural choice is

dp(f,g) = IIf - gllp,

0
00.

Since dp(f, g) = 0 implies f = 9 J-L-a.e, strictly speaking, the elements of LP(J-L) are equivalence classes of functions defined on X, where we agree that f = 9 means that f = 9 J-L-a.e. Now, since the cases 0 < p < 1 and 1 < p < 00 are essentially different, we treat them separately; we do the former case first. So, fix 0 < p < 1, and note that (1.3) can be improved to

(a

+ b)P ::; aP + bP,

a, b ~ 0 ,

(1.4)

with equality occurring in (1.4) iff ab = o. To see this observe that if ab = 0 there is nothing to prove. On the other hand, if ab ::f:. 0, and since (1.4) is to hold for any a, b > 0, we may replace a by ab there and consider the equivalent inequality


+ t)P -

1 - t P ::; 0 ,

all t

~

0,

(1.5)

with 0, and consequently,
dp(f,g)

=

(Ix If - glP dJ-L)

= ((J-L(A)l/P)P

l/p

= (J-L(A) + J-L(B))l/ P

+ (p.(B)l/P)p

riP ~

p.(A)l/P + p.(B)l/P ,

1.

The Lebesgue LP Spaces

211

or, in other words,

Thus, the inequality opposite to the triangle inequality holds, and dp cannot be a distance function on LP(p.). This is not, however, a serious difficulty. Indeed, by (1.4), the expression

d~(J,g) =

LIf -

glP dp.

satisfies the requirements of a metric, and essentially all the results discussed in Chapter VIII are true: Endowed with this metric LP(p.) is a complete metric space and Co(Rn) is dense in LP(Rn). It is interesting to point out that LP integrable functions are not necessarily integrable, or locally integrable for that matter. For instance, when 0 < p < 1, by 4.46 in Chapter VII, the function Ixl- n E LP(I(O,l)), but it is not locally integrable in any neighbourhood of the origin. In order to deal with the case 1 < p < 00 we need a preliminary result which is essential in what follows. We have already encountered a particular instance of this result, the Cauchy-Schwarz inequality, in 3.1 in Chapter VIII. We say that 1 < p, q < 00 are conjugate indices provided that l/p+ l/q = 1.

Conjugate indices are also related by the expressions p + q = pq and q = p/(p - 1). An important property that the conjugate indices p = 1/1], q = 1/(1 - 1]), 0 < 1] < 1, satisfy, is this: For any a, b ~ 0 we have

aTlb 1 -

TI

~ 1]a

+ (1 -

1])b.

(1.6)

Furthermore, equality holds in (1.6) iff a = b. To see this note that if ab = 0 there is nothing to prove. Otherwise, if ab ::j:. 0, and since (1.6) is to hold for any a, b > 0, we may replace a by ab there and consider the equivalent assertion

¢( t) = tTl - 1]t - (1 - 1])

~

with ¢(t) = 0 iff t = 1. Note that since 0

¢'(t)

= 1]tTl - 1 -

all t

~

0,

< 1] < 1, we have

>O

1]

0,

= { =0

ifO
if t = 1 < 0 if t > 1.

XII.

212

LP Spaces

Thus 'I/J(t) decreases to 0 as t increases to 1, and then increases to 00 as t ~ 00, and 'I/J(t) = 0 iff t = 1. This givs (1.6), including the remark concerning equality. A useful reformulation of (1.6), known as Young's inequality, is this: If a, b > 0 and p, q are conjugate indices, then aP bq ab < - + (1.7) -

P

q

and equality holds in (1.7) iff a P = bq • We are now ready to prove Holder's inequality. Theorem 1.1. Let (X,M,p) be a measure space, p,q conjugate indices, and suppose 1 E LP(p), 9 E Lq(p). Then Ig is integrable, and

Ix 11g1

dp

~ IIll1plIgllq·

(1.8)

Moreover, equality holds in (1.8) iff there exist nonnegative constants A, B, AB i 0, such that (1.9) Proof. If the right-hand side of (1.8) is 0, then either 1 = 0 p-a.e. or else 9 = 0 p-a.e., and we have equality in (1.8). If, on the other hand, IIll1plIgllq i 0, by replacing 1 and 9 by I/11111p and g/llgllq if necessary, we may assume that IIllip = IIgllq = 1. Now, since 1 and 9 are finite p-a.e., from (1.7) it follows that

1111g1 ~ 111 P + Iglq, p

q

p-a.e.,

(1.10)

and consequently, integrating over X we obtain

f 11g1 dp ~ ! f 111 P dp +! f Iglq dp = 1, lx p lx q lx which is precisely (1.8). A moment's thought will convince the reader that equality can occur in (1.8) iff it occurs in (1.10) p.-a.e. But, by (1.7), this is true iff 111 P = Iglq p-a.e., and this gives (1.9) when IIllip = IIgllq = 1. As for arbitrary functions 1 and g, we normalize them as above and note that equality holds iff 111P /lIll1~ = Iglq /lIgll~ p.-a.e. Whence a possible choice for the (nonunique) constants is A = IIgll~ and B = IIll1~. • We are now in a position to prove one of the essential results in the theory of LP spaces, which is due to Minkowski (1864-1909), namely, Minkowski's inequality.

1.

The Lebesgue IJ' Spaces

213

Theorem 1.2. Suppose (X,M,JL) is a measure space and I,g E LP(JL), 1 ~ p < 00. Then

(1.11) As for equality in (1.11) there are two separate cases, depending on whether p = 1 or not. If 1 < p < 00 the condition is: There exist constants A,B, AB =I- 0, such that AI = Bg JL-a.e. On the other hand, if p = 1 the condition is: There exists a nonnegative measurable function h such that Ih = 9 JL-a.e. on the set {Ig =I- o}. Proof. (1.11), in case p = 1, was already established in Chapter VIII. As for p > 1, since I,g are finite JL-a.e., so is 1+ g, and we have that II + glP is less than or equal to

II + gill + glP-l

~

1I11I + glP-l + Igill + glP-l

JL-a.e.

(1.12)

Whence integrating (1.12) over X we get

III + gll~ ~

L1I11I +glP-l + LIgill +glP-l dJL

dJL

(1.13)

=1+J,

say. To estimate 1 and J we apply Holder's inequality with indices p and its conjugate q = p/(p - 1), and note that 1

~ IIllIp

-L )(P-l)/P f ( Jx II + gl(P-l)p_l dJL = 1IIIIpili + gll~-l,

and similarly,

J ~

IIgllplll + gll~-l .

(1.14)

(1.15)

Whence substituting (1.14) and (1.15) into (1.13) it follows that

which is equivalent to (1.11). Finally, when 1 < p < 00, it is clear that we only have equality in (1.11) provided we have equality in (1.12) and in the estimates of 1 and J in (1.14) and (1.15), respectively. Now, equality holds in (1.12) if I and 9 are of the same sign JL-a.e. Also, by (1.9), equality holds in the estimates of 1 and J if for some constants A,B we have AIIIP = II + glq = BlglP JL-a.e. Whence combining these remarks it follows that equality holds if, as asserted, AI = Bg JL-a.e.

XII.

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LP Spaces

When p = 1 we must have equality in II + gl ~ III + Igl J.L-a.e., and this occurs when I and 9 are of the same sign J.L-a.e. Then the function h = g/ I will do the job. • An interesting consequence of Minkowski's inequality is that dp is a metric on LP(J.L), 1 < p < 00. The only property that offers any difficulty is the triangle inequality, and it is obtained as follows: If I,g,h E LP(J.L), we have

dp(J,g) = III - gllp = II(J - h) + (h - g)lIp ~ III - hll p + IIh - gllp = dp(J, h) + dp(h,g). Furthermore,

Theorem 1.3 (F.Riesz-Fischer). Let (X,M,J.L) be a measure space. Then, the distance function dp turns LP(J.L) into a complete metric space, 1 < p < 00. Proof. We fix 1 < p < 00, and assume that {In} is a Cauchy sequence of functions in LP(J.L). We must show that there is a function I E LP(J.L) so that limn-+oo dp(Jn,J) = O. First observe that since {In} is Cauchy we can find an increasing sequence nk+l > nk such that

dp(Jn,Jnk) ~ Put now

Ina

all n ~ nk,

1/2k,

k = 1,2, ...

= 0, and let m

gm =

L lInk - Ink_II,

m = 1,2, ...

k:::::l

Clearly the sequence {gm} is nondecreasing, let 9 denote its limit. Furthermore, since by (a simple extension of) Minkowski's inequality it follows that IIgmllp ~ 1 for all m, by MCT we get that gP dJ.L ~ l.Thus, in particular, 9 is finite J.L-a.e., and the series with terms Ink - Ink_I' k = 1,2, ... , converges absolutely to a finite sum J.L-a.e. Let then

Ix

m

I = m-+oo lim "(Jnk - Ink-I) ; L...J k:::::l

I is measurable and finite J.L-a.e. Moreover, since the sum on the righthand side above telescopes to Inm' it readily follows that lim Ink = I

k-+oo

J.L-a.e.

1.

The Lebesgue LP Spaces

215

We want to show that the convergence is also in the metric of LP(Jl). First observe that since

III ~ 9,

and

I/n,,1

~ 9n" ~ 9,

all k,

and since 9 E LP(Jl), by LDCT we get lim {lIn" - liP dJl

k-+oo

JX

= k-+oo lim d~(Jn", J) = o.

To complete the proof we invoke the well-known fact that if a Cauchy sequence in a metric space has a convergent subsequence, then the sequence itself converges to the same limit. • As in the case of integrable functions, the metric structure of LP( X, Jl) permits us to establish the following properties: (i) Simple functions are dense in LP(X,Jl) , 1 < p < 00. (ii) Co(R n ), cgo(Rn) are dense in LP(Rn ), 1 < p < 00. (iii) cg(I) = {I E cg(Rn): I vanishes off I} is dense in LP(I), 1 < p < 00, 0 ~ k ~ 00. (iv) The class of sequences {en} which eventually vanish is dense in lP, 0< P < 00. (v) The translates of LP(Rn ) functions are continuous in the norm, i.e., if IE LP(Rn ), then limlhl-+o 11/(· + h) - 1(·)lIp = 0, 1 < p < 00. Since the proof of these statements follows along the lines of that of the corresponding results for p = 1, we leave their verification to the reader. Finally, in the scale of Lebesgue spaces there are two limiting cases left to consider: The case p = 0 and the case p = 00. The idea here is to let p --+ 0 and p --+ 00 in the expressions corresponding to II/II~ and II/lIp respectively, and study what happens. In the limiting case p = 0, it is clear that the limit exists and it equals 11/110 = i{Jt=o} dJl. The class LO(X, Jl), consisting of those extended realvalued finite Jl-a.e. measurable functions I defined on X so that Jl( {I '# OJ) < 00, enjoys many ofthe properties ofthe LP(X,Jl) spaces and is very useful in applications; we discuss it no further here. As for the case p = 00, to fix ideas consider I = [0,1] and a measurable function I defined on 1. If 1 < p < q < 00 are given, observe that r = q/p > 1 and r' = r/(r - 1) = q/(q - p) are conjugate indices, and consequently, by HOlder's inequality, and assuming that the quantities involved are finite, we have

i I/IP ~ (i Iflpr dx

dx ) l/r

(i

dX) l/r' =

(i Ifl

q

dX) q/p ,

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216

LP Spaces

or equivalently,

II/l1p:5 II/l1q,

1< p

< q < 00.

Whence, the p norms of I are nondecreasing and lim q-+ oo II/l1q exists. Suppose this limit is finite, this is the case of interest to us, and call it L. By Chebychev's inequality we have

~I{I/I > ~}Il/q :5

II/l1q :5

all ~ >

L,

o.

(1.16)

Moreover, since lim I{I/I > ~}Il/q = 1 whenever q-+oo

I{I/I > ~}I =I 0,

by (1.16) it readily follows that

I{I/I > ~}I

= 0,

all ~ > L.

(1.17)

Functions that satisfy (1.17) are called "essentially bounded". In contrast to bounded functions, essentially bounded functions may assume infinite values, but only on null sets. These observations motivate our next definition. Let (X,M,JL) be a measure space and I an extended real-valued measurable function defined on X. The expression

11/1100 =

inf {A>

0 :JL({I/I > ~}) = O} ,

(1.18)

whether finite or not, is well-defined and is called the JL-essential sup,of III, or the "00 norm" of I. The Lebesgue class Loo(X,JL) consists of those measurable functions I with 11/1100 < 00, and it is also denoted by Loo(JL) or Loo(X). Note that since

JL({I/I>

11/1100 + lin})

we also have

JL( {III>

= 0,

n

1l/1I00}) = 0,

= 1,2, ... , (1.19)

and the infimum in (1.18) is attained. The next step in the study of the Loo spaces is to establish which properties of the LP spaces remain valid in this setting, and which do not. For instance, since convergence in the LOO norm corresponds to uniform convergence, if 1 = [0,1], only continuous functions can be approximated in Loo(l) by continuous functions, and consequently, Go(l) is not dense in Loo(l). On the other hand, on a positive note we have

1.

The Lebesgue If' Spaces

Theorem 1.4.

217

Endowed with the metric

doo(J, g)

= 111 - gil 00 ,

(1.20)

LOO(X,J.l) becomes a complete metric space. Proof. To show that doo is a metric the only property that offers any difficulty is the triangle inequality. The proof of this depends on the following variant of Minkowski's inequality:

111 + glloo ::; 1111100 + IIglioo .

(1.21)

To see this observe that

and since by (1.19) the sets on the right-hand side above are null, so is the set on the left-hand side, and (1.21) follows. The completeness of LOO(J.l) with the metric doo follows along the lines of Theorem 1.3 and is therefore left for the reader to verify. • Another important property of Loo(J.l) is that 1 and 00 are conjugate indices, in the sense that 1/1+1/00 = 1. A justification for this convention is given by the following extension of Holder's inequality.

Proposition 1.5.

Let (X,M,J.l) be a measure space, and suppose E L(J.l), and we have

1 E L(J.l), 9 E LOO(J.l). Then Ig

Ix 11g1 Equality holds in (1.22) iff

dJ.l ::;

111111 IIg II 00 .

(1.22)

Igl = IIglioo J.l-a.e.

Proof. By (1.19) we may assume the integral in (1.22) to be extended over the set { Igl ::; IIglloo}' and in this case (1.22) holds trivially. As for the case of equality in (1.22), observe that since

Igl/lIglioo ::; 1 the relation

J.l-a.e.,

Ix 111(lgl/lIgIl00) Ix 111 dJ.l =

can only be true if, as asserted,

Igl/lIglioo =

dJ.l,

1 J.l-a:e.

•

XII.

218

LP Spaces

2. FUNCTIONALS ON LP The next topic we consider is that of mappings defined on LP(X, ft), and the simplest case is that of scalar-valued mappings. For instance, point-evaluation is a well-defined mapping on Co(R n ), and since this class is dense in LP(Rn ), < p < 00, it is natural to consider whether pointevaluation may be extended, in some natural way, to LP(Rn ). Since functions in these classes need only be defined ft-a.e. it is not intuitively clear how to construct such an extension, or whether, in fact, one such extension exists. The answer to these questions is postponed to Chapter XIV, where the Hahn-Banach Theorem is discussed. In this chapter we take a different approach, one suggested by Holder's inequality. For example, referring to (1.22), given 9 E LOO(ft), let Lg be the mapping on L(ft) given by

°

(2.1) Clearly Lg is well-defined, and it satisfies the following properties: (a) For I,h E L(ft) and a scalar A we have Lg(f + Ah) = Lgi + ALgh, (b) ILgl1 ~ IIgliooll/lll for all I E L(ft), and, (c) If limn-+ oo d1 (fn,!) = 0, then limn-+ oo Lgln = Lgi. In fact, Lg is a prototype of those mappings L on the Lebesgue LP spaces which satisfy the following properties: (i) L is a well-defined scalar-valued mapping. (ii) (Linearity) For each scalar A and I,g E LP(ft),

L(f + Ag)

= LI + ALg.

(iii) (Continuity) If limn-+ oo dp(fn,j) = 0, then limn-+oo Lin = LI. (iv) (Boundedness) There is a constant k such that

ILII ~

kll/ll p ,

all

IE LP(ft)·

More precisely, a mapping L which satisfies (i) above is called a functional. L is said to be a linear, continuous or bounded functional if (ii), (iii) or (iv), respectively, hold. Although not apparent at a first glance, the concepts of continuity and boundedness are equivalent. Proposition 2.1. Suppose L is a linear functional on LP(ft). Then L is continuous iff L is bounded.

2.

Functionals on LP

219

Proof. To prove the necessity, suppose that L is continuous but unbounded. Then, for each n, there is In E LP(J-L) such that

Now, the sequence 9n = (l/nll/nllp)/n' n = 1,2, ... , satisfies lim dp(9n' 0) = 0,

n-+oo

and

IL9nl

~ 1,

all n,

thus contradicting the continuity of L. As for the sufficiency, observe that if I, In E LP(J-L), n = 1,2, ... , by the linearity and boundedness of L we have

ILln - LII = IL(Jn -

1)1 ~ k II/n - Ilip .

Whence, if dp(Jn, I) ~ 0, the right-hand side in the above inequality tends to 0 with n, as does the left-hand side there, and the desired conclusion follows. •

on

It is rather straightforward to construct bounded linear functionals 0 < p ~ 1. Indeed, if (m n) E too, the mapping

[P,

L(cn ) =

L mncn ,

(cn) E t P ,

n

is readily seen to be such a functional. In fact, by (1.4) it is clear that

and (iv) above holds with k = lI(m n)lIoo there. The following result, then, is a bit surprising.

Theorem 2.2 (M.M. Day). Let I = [0,1], and suppose L is a continuous linear functional on LP(I), 0 < p < 1. Then L is the zero functional, i.e., for every IE LP(I) we have LI = O. Proof. Suppose, to the contrary, that L is not the zero functional. Then, by rescaling if necessary, we may assume that there is a function I E LP(J-L) such that (2.2) LI = 1, II/lIp # O. Observe that as a function of a; in I,

Ix[o.,x] is continuous in the metric of

LP(J-L), and consequently, since L is continous we have that
= L (IX[o.,xD,

a;

E I,

XII.

220

LP Spaces

is a continuous real-valued function defined on I which satisfies <1>(0) = 0, and <1>(1) = LI = 1. Since is continuous there is x E (0,1) such that (x) = 1/2; further, since I = Ix[o,x] + IX[x,I] (in LP(IL)), by the linearity of L it readily follows that

(2.3) Moreover, since

(2.4) one of the summands on the left-hand side of (2.4) does not exceed 1I/1I~/2. Call 91 one of the functions in (2.4), either Ix[o,x] or IX[x,I]' such that 119III~ ~ 1I/1I~/2, and rewrite this estimate

(2.5) Put now

It = 291 and observe that, by combining (2.3) and (2.5), we get

Lit

= 1,

and

IIltll~ ~ 2(P-I)II/II~.

Repeating the above argument with function h, say, such that

Lh

= 1,

and

It in place of I above we obtain a

IIhll~ ~ 2(P-I)lIltll~ ~ 22(P-I)II/II~.

It is now apparent that iterating this inequality we get a sequence {In} LP(I) which satisfies

LIn = 1, Since 0 < p

and

II/nll~ ~ 2n(p-I)II/II~·

~

(2.6)

< 1, (2.6) implies that the In's satisfy LIn = 1,

lim ~(Jn,O) = 0,

n-+oo

which is impossible if L is continuous. This contradiction was derived from (2.2), and hence L is the zero functional. • The situation is quite different when p ~ 1: Not only are there plenty of functionals on LP(IL), but it is also possible to characterize them. We begin with a definition. The norm IILII of a bounded linear functional L on LP(IL), 1 ~ p ~ 00, is defined by the quantity

IILII = sup

IIJllp~O

ILII -11/11p •

(2.7)

2.

Functionals on LP

221

For instance, as pointed out in (ii) above, the functional Lg on L(p) given by (2.1) satisfies IILgll ~ IIglioo. In fact, if pis O'-finite, or more generally semifinite, we have equality here. To see this, given c > 0, let E E M be a set of positive measure so that Igl

> IIgll - c , p-a.e. on

E.

Moreover, since p is O'-finite, or semi finite , we may also assume that 00 and consequently, the function I = (sgng)XE E L(p). Finally, since

p(E) <

LgI =

l

(sgng)g dp = llgl dp

~ (lIglloo - c)

p(E) = (lIglioo - c) 11/111,

it follows that which gives the desired conclusion since c is arbitrary. It is natural to pose the analogous question for Lg, 0 ::J 9 E Lq(p), considered as a bounded linear functional on LP(p), l/p + l/q = 1. First observe that by Holder's inequality we have

(2.8) and indeed Lg is a well-defined continuous linear functional on LP(p) with IILgll ~ IIgliq. To see that we actually have equality in the norms, if 1 < p < 00, put I = (sgng)lglq/p, and note that II/II~ = IIgll~ < 00 and Ig = Iglq/P(sgng) = Iglq. Thus, for this particular LP function I we have LgI = Iglq dp, and consequently,

Ix

IILgll ~ ILgII/ll/llp

= IIgll;/lIgll;/p = IIgllq,

as we wanted to show. The case p = 00 is even simpler, as the limiting process gives the right answer: If 0 ::J 9 E L(p) now, put 1= sgng, and observe that 11/1100 = 1 and

LgI

=

L

(sgng)g dp

= IIglh .

Whence, as asserted, IILgll ~ LgI = IIglll. A natural question to consider is whether every bounded linear functional L on LP(p) is of the form L = Lq, for some 9 E Lq(p), 1/p+1/q = 1. In Chapter XIV we will see that this is not the case for p = 00, but how about for finite p's? In order to address this question we consider a converse to HOlder's inequality.

XII.

222

LP Spaces

Proposition 2.3. Let (X,M,p) be a measure space, 1 ~ p ~ l/p + l/q = 1, and suppose I E LP(p). Then if 1 ~ p < 00, we have

00,

(2.9) If p is CT-finite it is also true that

11/1100 =

sup ILg/I· IIglh :5 1

(2.10)

Proof. We may assume that I :j:. 0 on a set of p positive measure, for otherwise there is nothing to prove. Now, if 1 ~ p < 00, by Holder's inequality it follows that the sup on the right-hand side of (2.9) is less than or equal to II/lIp. Furthermore, putting 9 = (1/1I/11~-1)(sgn J)l/lp-1, we see at once that IIgllq = 1, and that

Lgi =

1I/1~~-1

LI/I

P-

1(sgn J)I dp = II/lIp,

and (2.9) holds. As for (2.10), again by Holder's inequality it suffices to show that the sup on the right-hand side there is at least 11/1100. Now, since p is CT-finite, given c > 0, we can find a set E E M such that

III > 11/1100 Then the function 9 satisfies

c,

p-a.e. on E,

0 < peE) <

00 .

= (1/ p(E))(sgn J)XE is integrable, has norm 1, and

Lgi =

p(~)

k

I(sgn J) dp 2:

Since c is arbitrary, (2.10) holds.

11/1100 -

c.

•

An interesting and useful variant of this result is Theorem 2.4. Let (X,M,p) be a measure space, 1 ~ p ~ 00, l/p + l/q = 1, and let I be an extended real-valued measurable function defined on X with the property that for every simple function 4> defined on X we have (2.11) Then, if p is CT-finite, it follows that the constant in (2.11).

I

E LP(p) and II/lIp ~ k, where k is

2.

Functionals on LP

223

Proof. Write X = Un Xn as an increasing union of sets of finite measure, and begin by constructing a sequence {tPn} of simple functions such that

ItPn I ::; III ,

and

lim

n-+oo

tPn = I

everywhere.

(2.12)

Further, let
Jxfill dJL =

lim n-+oo

Jxf I In dJL ::; k,

as we wished to show. Next, if 1 < p < 00, put In = l
(i) IIn
Ix l
P

dJL =

=

Ix I/n
IIx I In

dJL ::;

dJLI ::;

Ix I/n/l

dJL

kll/nll q ::; kll
It is clear that unless I = 0 JL-a.e., and in this case there is nothing to prove, we may also assume that 0 < lI
With this computation out of the way, observe that by Fatou's Lemma we have

XII.

224

V Spaces

and (2.11) holds. As for the case p = 00, let c > 0 be given, and put E = {If I > k +c}. If p,(E) > 0, since p, is u-finite, E contains a subset B, say, of positive, finite measure. Setting 1



= p,(B) (sgn f)XB ,

11<1>111

= 1,

by (2.11) we have

k?

Ix

fdp, >

p,(~)

k+ (k

which is impossible. Thus, we have p,(E)

c) dp, = k + c,

= 0, and IIflloo ::; k.

•

We are now in a position to describe the bounded linear functionals on LP(p,). Theorem 2.5 (F. Riesz Representation Theorem). Let (X,M,p,) be a measure space, 1 ::; p < 00, and lip + 1/q = 1. Then, if p, is u-finite, to each continuous linear functional L on LP(p,) there corresponds a unique function 9 E Lq(p,) such that IILII = IIgllq, and

Lf

= Lgf =

Ix

fgdp"

all f E LP(p,).

Proof. Suppose first that p, is a finite measure and introduce the set function (2.13) v(E) = LXE' E E M. Since L is linear it readily follows that v is an additive set function on M. We claim that actually v is a signed measure and that v <: It. Clearly v(0) = O. Moreover, since L is bounded, v only assumes finite values and for each E E M we have

Iv(E)1 = ILXEI ::;

IILII p,(E)l/P < 00.

(2.14)

Thus, to verify that v is a signed measure, it only remains to check that v is u-additive. Let, then, {En} be a sequence of pairwise disjoint measurable subsets of X, and put E = Un En. By (2.13) and (2.14), and since E \ U:'=l En = U~=m+I En, we get

Iv(E) - L:'=l V (En)I = Iv(E \ U:'=lEn)I = ILXUoo n=m+1 E n I ::; ilL II J.t(U~=m+IEn)l/P .

(2.15)

2.

Functionals on LP

225

Now, since the En's are pairwise disjoint and JI.(E)

JI.(U~=m+IEn) =

,",00

L....n =m+l

JI.(En ) _ 0,

< 00, we have as m _

00,

and consequently, the right-hand side of (2.15) tends to 0 as m Whence, it readily follows that

00.

and v is a signed measure on (X,M). Moreover, if JI.(E) = 0, from (2.14) we get that Iv(E)1 ~ IILIIJI.(E)l/P = 0, and consequently, v <: JI.. We are now in a position to invoke the Radon-Nikodym Theorem. Let 9 = dv/dJl. be the Radon-Nikodym derivative of v with respect to Jl.j 9 is uniquely determined and locally integrable, we want to show that also 9 E Lq(JI.) and that IIgllq = IILII. First recall that, in particular, we have

veE)

= 19dJl.,

(2.16)

E EM.

Now, if I = L: CnXEn is a simple function defined on X, by (2.13) and (2.16) we get

LI

=L =

CnLXEn

=L

cnv(En)

=L

Cn In 9 dJl.

Ix (LCnXEn) gdJl. Ix IgdJl.. =

(2.17)

Moreover, since L is bounded, by (2.17) we obtain

Ilxl9dJl.I

~ IILllll/llp,

and by Theorem 2.4 it follows that 9 E Lq(JI.) and IIgllq ~ IILII. In order to show that L = L9 , we must still prove that (2.17) is true for arbitrary f's in LP(JI.). Given I E LP(JI.), let {In} be a sequence of simple functions such that In - I and I/nl ~ III. By (2.17) we have

LIn =

L

IngdJl.,

n = 1,2, ...

(2.18)

and consequently, the limit of the right-hand side of (2.18) and that of the left-hand side there, if they exist, must be equal.

LP Spaces

XII.

226

As for the left-hand side, note that since L is bounded, we have

ILfn - Lfl ~ IILllllfn - flip

-+

as n

0,

-+ 00 ,

and consequently, lim Lfn

n-+oo

= Lf.

(2.19)

On the other hand, by the linearity of the integral and HOlder's inequality, we also have

Ii

fngdJL -

i

fgdJLI

~

i

~

IIfn - fllplIgllq

fllgl dJL

Ifn -

0,

-+

as n

-+ 00 ,

and consequently, lim { fngdJL n-+oo

ix

= (

ix fgdJL.

(2.20)

Whence combining (2.18), (2.19) and (2.20), we get

Lf =

i

fgdJL,

(2.21)

all f E LP(JL).

To complete the proof in the general case write X = Un Xn as an increasing union of sets of finite measure, and consider the restriction Ln of the functional L to LP(Xn , JL). Each Ln is a linear functional, and since

the Ln's are also bounded. Thus, by the first part of the proof, we can find functions gn E Lq(Xn,JL), n = 1,2, ... , such that

(2.22) and

(2.23) Now, since Xn ~ Xm for all n ~ m, we also have LP(Xn,JL) ~ LP(Xm,JL). Furthermore, since for each f E LP(Xn,JL) we have Lnf = Lmf for all m ~ by (2.23) it follows that for such functions we have Ign dJL = fgm dJL, or

Ix

n,

Ix

Ix

I(gn - gm) dJL = 0,

all m

~ n.

(2.24)

2.

Functionals on LP

227

In particular, since J.L(Xn ) < 00, the functions sgn (9n - 9m) E LP(Xn,J.L) for all m ~ n, and by (2.24) we get that 19n - 9ml = J.L-a.e. on X n • In other words, 9n = 9m Jlra.e. on Xn for all m ~ n, and consequently, the function 9 on X given by

°

is well-defined and it satisfies 191 = limn->oo 19n1 J.L-a.e. Thus, by Fatou's Lemma and (2.22) it follows that (2.25) and 9 is an Lq(J.L) function with norm less than or equal to IILII. Next note that since for each I E LP(J.L) we have

In = IXxn

---+

I J.L-a.e.,

and

Ilnl ~ III J.L-a.e.,

by LDCT it follows that limn->oo IIln - Ilip = 0, and consequently, by the continuity of L we obtain lim Lin = LI. n->oo Moreover, since by (2.23) Lin = IXn In9n dJ.L = Ix In9dJ.L, by Holder's inequality we also have limn->oo Ix In9 dJ.L = Ix 19 dJ.L, and consequently,

as we wanted to show. It thus only remains to verify that IILII = 11911q, and by (2.25) we only need to check that IILII ~ 11911q. But since L = L g , this is an easy consequence of Holder's inequality. • Two natural questions arise from this result: How can we go about representing the bounded linear functionals on LOO (J.L) , and, is the assumption concerning the u-finiteness of J.L necessary? The former question will be addressed in Chapter XIV, and the latter question has two answers, to wit: If 1 < p < 00, it is not necessary that J.L be u-finite, and if p = 1, it is. We do the case p = 1 first. Let X = (0,1), let M be the u-algebra of those subsets E of X which are either countable or such that X \ E is countable, and assume J.L is the

XII.

228

LP Spaces

counting measure on (X,M); p. is. not O'-finite. If v denotes the counting measure on (X, P(X)), put

LI

=

Ix

IX(o,l/2) dv,

IE Ll(p.).

Since for I E Ll(p.) the set {I "=I O} is p. and v O'-finite, it is clear that ILII ~ 111111, and L is a bounded linear functional on Ll(p.) of norm less than or equal to 1. It is intuitively clear that if LI = LgI, then 9 must be the function X(O,l/2)' which is measurable with respect to the 0'algebra P(X), but not measurable with respect to the O'-algebra M; the verification of this observation is left to the reader. On the other hand, the situation is quite different if 1 < p < 00, for then X(O,l/2) r;. Lq(v) for any q < 00. Finally, to see that in the case 1 < p < 00 the O'-finiteness of p. is not needed in the Riesz representation theorem, note that, in the notation of that theorem, given a O'-finite subset E of X, there is a unique function 9 = 9E vanishing off E so that 9E E Lq(E,p.) and

LI =

k

19 dp.,

all I E LP(E,p.).

Furthermore, if LIE denotes the restriction of L to LP(E,p.), we also have

119Ellq ~ IILIEII ~ liLli,

all O'-finite E.

Also (a simple variant of) the argument in (2.24) above gives that if El ;2 E are O'-finite subsets of X, then we have 9EI = 9E p.-a.e. on E, and 119Ellq ~ 119El liq. Let now TJ be the finite quantity TJ = sup {1I9Ellq : E is a O'-finite subset of X} ,

and let {En} be a sequence of O'-finite subsets of X with the property that lim n ..... oo 119E"lIq = TJ· Observe that if E = Un En, then E is also a O'-finite subset of X, and since

119Ellq ~ 119EJq,

all n,

it readily follows that 119Ellq = TJ. Now, life outside E is uneventful. Indeed, let A be a O'-finite subset of X, and put Al = (A \ E) U E. Then Al is also a O'-finite subset of X, and since q < 00 and

{ 19A 1 1qdp. = { 19A1q dp. + { 19E1q dp. JAI JA\E JE = { IgAlq dp. + TJq ~ TJq , JA\E

3.

Weak Convergence

229

it readily follows that gA = 0 J1,-a.e. on A \ E. This is all we need to know: If I is an arbitrary function in LP(J1,) , then the set A = {I ¥= O} is a-finite, cf. 4.11 in Chapter VII, and

LI

L

=

L

=

f Ig dJ1, + f IgE dJ1, = f IgE dJ1" JA\E A JAnE Jx

I 9 A dJ1,

=

I 9 A dJ1,

which is the desired representation of L.

3. WEAK CONVERGENCE Assume (X,M,J1,) is a measure space, and let 1,ln E LP(J1,), n = 1,2, ... , 1 ~ p < 00. We say that the sequence {In} converges weakly to I in LP(J1,), if, with lip + 1/q = 1, we have

lim

n_oo

f IngdJ1,= Jx f IgdJ1" Jx

allgELQ(J1,).

(3.1)

We now give a few examples to show that there is no connection between weak convergence and any of the other forms of convergence, unless further assumptions are made on either the sequence itself or the measure space involved. For instance, in £P, consider the sequence {en} consisting of those sequences en = (0, ... ,1,0, ... ) with 1 in the nth place and zeroes elsewhere. If 1 < p < 00, and x = (XI, ... ,X n , ... ) E £q, then the functional Lx has the property that

(3.2) and since by the Riesz representation theorem these are all the functionals on £P, the sequence {en} converges weakly to O. Nevertheless, since lien - em lip = 21 / p for all n ¥= m, neither the sequence itself nor any of its subsequences converges to O. Neither does {en} converge to 0 in measure, nor uniformly, nor even in the pointwise sense. Note however that {en} does not converge weakly to 0 in £1; this is clear since the sequence x = (1,1, ... ) is bounded and Lxen = 1 for all n. Now, in the case of £1 we have the following interesting result. Proposition 3.1 (Schur). x in £1, then

If the sequence {xn} converges weakly to

lim IIx n

n .....oo

-

xIII =

o.

(3.3)

XII.

230

LP Spaces

Proof. By considering, if necessary, the sequence {xn - x} we may assume that {xn} converges weakly to O. Suppose that limn-+ oo Ilxnlll i- 0; passing to a subsequence, if needed, we may assume that IIxnlll ~ TJ > 0 for a.ll n. In this case also (1/llxnlldxn converges weakly to 0, and so we may as well assume that IIxnlll = 1 for a.ll n. In addition, if Xn = (Xn,b ... ,xn,m, ... ), n = 1,2, ... , by the weak convergence it follows that lim Xn m

n-+oo

'

= 0,

(3.4)

a.ll m.

Observe that since IIXl111 = 1, we can find ml so that L::!:llxl,ml > 3/4. Further, by (3.4) it readily follows that there exists an index n2 > 1 such that L::!:1 IX n2 ,m I < 1/4, and consequently, since IIx~ lit = 1, we can find an index m2 > ml so that L::;m1 +1 Ix~,m I > 3/4. The pattern is now clear: Having chosen sequences mo = 0 < ml < m2 < ... < mk and nl = 1 < n2 < ... < nk, choose first nk+1 with the property that

mk

L IXnk+ ,ml < 1/4, 1

m=1 and then

mk+1

so that

mk+1

L

Ixnk+l1m l > 3/4.

m=mk+1 Consider now the sequence Y E [00 with terms

Since IYm I : : ; 1 for all m, the functional Lyon

[I

satisfies

00

ILy(xnk)1 =

L xnk,mYm

m=1

~ m=mk+1 1: IXnk,ml1

(I: + m=1

f )

IXnk,mYml

m=mk+1 +1

mk+1 = 2

L

IXnk,ml-lIxnklh> 2·3/4 -1 =1/2,

k = 1,2, ...

m=mk+1 Thus, lim sup ILy(xn)1 ~ 1/2, which contradicts the weak convergence of {xn} to o. •

3.

Weak Convergence

231

In a different direction, an interesting example is the sequence {In} ~ LP([O,1j), 1 ~ p, given by In = nX[O,lln)' which converges to 0 in measure and a.e., and does not converge to 0 weakly in LP([O,1j) for any p 2: 1. (Just consider the functional induced by X[O,I)') Finally, the sequence {In} ~ LP(R), 1 ~ p, given by In = (1/n)X[1,en), converges uniformly to 0, yet it does not converge weakly to 0 in LP( R) for any p 2: 1. (The functional induced by (1/X)X[1,00)(X) will do.) There are additional assumptions that we may impose on weakly convergent sequences to ensure they they also are convergent. Again we discuss the t P case; the result is also true for general LP(J-t) spaces, but the proof is more complicated. Proposition 3.2. Suppose the sequence {xn} converges weakly to x in t P , 1 ~ P < 00, and that, in addition,

(3.5) Then we also have

(3.6) Proof.

As before it follows that if

x = (xt, ... ,x m, ... ),

Xn = (xn,t, ... ,xn,m,"')'

n = 1,2, ...

then lim Xn m = Xm ,

n~oo

'

(3.7)

all m .

Also, by (3.5) and (3.7), for each fixed M we have

00 )I/P (00 ) IIp ( P P n~~ m~ IXn,mI = m~ Ixnl

(3.8)

Whence, for each fixed M we get, M-I

)llP

IIXn - xl!p ~ ( ~ IXn,m - xml P

(00

+ m~ IXn,m -

) IIp

xml P

=A+B, say. It is not hard to estimate B. By Minkowski's inequality and (3.8) it follows that for all sufficiently large n, B$

Ctlz•.mIPr + CtlzmlP)"P $3Ctl•.IP)"P.

XII.

232

V Spaces

which, since x E i P , can be made arbitrarily small provided M is sufficiently large. Once M is fixed, it is clear that, on account of (3.7), A also can be made arbitrarily small provided n is large enough. Thus (3.6) holds, and we have finished. • As noted above, LP spaces do not have the Bolzano-Weierstrass property: There are bounded sequences for which no convergent subsequence may be found. The concept of weak convergence is also relevant in this context. Theorem 3.3. Let {/k} be a bounded sequence in LP(Rn), 1 < P < 00, with bound M. Then there exist a subsequence k m -+ 00 and a function / E LP(Rn) with II/lip ~ M, such that {/k m} converges weakly to / in LP(Rn). Proof. We divide the proof into a number of steps, and begin by showing that there is a subsequence k m -+ 00 with the property that limm -+ oo /kmg dx exists, provided 9 is any function in a fixed countable /kgh family {gh} ~ Lq(Rn), 1/p+l/q = 1. This is not hard: Let Ck,h = for k, h ~ 1, and note that by HOlder's inequality we have

Inn

Inn

(3.9) Fix h = 1 now. By (3.9), (Ck,l) is a bounded sequence and consequently, there is a subsequence kl -+ 00 such that limkl-+OO Ckl,1 exists. Repeating this argument with (CkI.l) in place of (Ck,t) above we obtain a new subsequence k2 -+ 00, say, such that limk:l-+oo Ck2,h exists for h = 1,2. These are the first steps of the by now familiar Cantor diagonal process which ensures the existence of a subsequence k m -+ 00 so that limkm-+oo Ckm,h exists for each h. We choose now for the 9h's a dense family in Lq(Rn ), which, since Lq(Rn ) is separable, is clearly possible, and define the functional L on the gh's by means of the expression Lg = limm -+oo /km 9 dx. Now, L is decidedly linear over these gh's, i.e., L(9hl + ).g~) = Lghl + )'Lg~ for all scalars )., and by HOlder's inequality it also satisfies ILgl ~ MlIgliq. We claim that L can be extended linearly and continuously to all of Lq(Rn). Indeed, to each 9 E Lq(Rn ) there corresponds a sequence of gh's such that limh-+oo IIg - ghllq = 0 and lim IIghllq = IIgliq. Now, for these gh's -the sequence of scalars (Lgh) is Cauchy, and consequently, convergent. Putting Lg = limh-+oo Lgh, L turns out to be a well-defined linear functional on Lq(Rn), and since

Inn

ILgl ~ lim sup ILg - Lghl + lim sup ILghl ~ MlIgllq ,

4.

Problems and Questions

233

L is also bounded and has norm IILII ~ M. By Theorem 2.5 there exists a function IE LP(Rn ) with II/lIp ~ M such that Lg LJg; the function I satisfies all the required conditions. •

=

The proof of this interesting result relies on the fact that the functional

L, originally defined on a subset of Lq(Rn), can be extended to all of Lq(Rn) without an increase of its "norm." A more general setting where this is also true is described in Chapter XIV.

4. PROBLEMS AND QUESTIONS In what follows (X,M,JL) denotes a measure space, and we don't find it necessary to stress this point at each instance. 4.1 Suppose 0 < p < q ~ 00. Give examples of functions I defined on R such that I E Lr(R) iff (a) p < r < q, (b) p ~ r ~ q, and, (c) r =p. 4.2 Let I be a bounded interval of R. By means of an example show that, in general, no
4.3 The following inequality is often referred to as Holder's inequality, prove it and identify the cases of equality: If I E LP(JL) and 9 E Lq(JL), 1 < p, q < 00, 1/p+1/q = 1, then IIx Ig dJLI ~ II/lIplIgllq· 4.4 The following is an extension of Holder's inequality to more than two indices: Suppose I E LP(JL), 9 E Lq(JL), h E Lr(JL), 1 < p,q,r < 00, l/p + l/q + l/r = 1. Prove that Igh E L(JL) and that

Can you think of a further extension to more than three functions? 4.5 Let 1= [0,11"]. Show that

II x- I / 4 sin x dx ~ 11"3/4.

4.6 Show that if for some 0 < p < 00, I E LP(JL) n LOO(JL), then for all p < q < 00, IE Lq(JL) and II/l1q ~ 1I/11~/qIl/Il~p/q. 4.7 If for some 0 < p,q < 00, IE LP(JL) n Lq(JL), then I E Lr(JL) for all p < r < q, and II/l1r ~ 1I/11~-'lIl/Il~, where 0 < TJ < 1 is given by l/r = (1 - TJ)/p + TJ/q. 4.8 Let I = [0,11"] and I E L2(I). Is it possible to have simultaneously Il(/(x) - sinx)2dx ~ 4/9 and !1(/(x) - cosx)2dx ~ 1/9?

XII.

234

LP Spaces

4.9 Suppose an extended real-valued function I defined on Rn satisfies the following two properties: (a) There is a p, 1 ~ p < 00, such that I E LP( I), for every bounded interval I in Rn, and, (b) I II liP ~ clIlp-l II I/IP, for every bounded interval I and a constant 0 < c < 1 independent of I. Show that I = 0 a.e. 4.10 Let 0 < p < q ~ 00. Prove that LP(J.t) is not contained in Lq(J.t) iff X contains sets of arbitrarily small, positive, I' measure. 4.11 Let 0 < p < q < 00. Show that Lq(J.t) is not contained in LP(J.t) iff X contains sets of arbitrarily large, finite, I' measure. What can you say about the case q = oo? 4.12 Prove that if limn_oo II/nllp = 0, 1 ~ P ~ 00, then there exist a subsequence {Ink} and a nonnegative function h E LP(J.t) such that link I ~ h J.t-a.e., and limk_oo Ink = 0 J.t-a.e. 4.13 Prove that if II/n - Ilip --+ 0 and IIgn - gllq lip + 1/q = 1, then II/ngn - Igll1 --+ o.

--+

0, 1 ~ p, q ~

00,

4.14 The LP(X,J.t) spaces are not, in general, separable when 0 < p ~ 00. Give an example of a measure space (X,M,J.t) so that LP(J.t) is not separable, and prove that the Lebesgue spaces LP( Rn) are separable for 0 < p < 00. On the other hand, Loo(Rn) is not separable. When is Loo(J.t) separable? 4.15 Suppose

I E LP(Rn ), 0 < p < 00, and compute lim f I/(y + h) Ihl-oo Jll!'

+ l(y)IPdy.

4.16 Suppose I, In E LP(J.t), n = 1,2, ... , satisfy limn _ oo In = I J.t-a.e., and limn_oo II In lip = II/lIp, 0 < p < 00. Prove that lim II/n n-oo

Ilip = o.

4.17 Is the conclusion of 4.16 true if we replace J.t-a.e. convergence by convergence in measure? 4.18 Let (X,M,J.t) be a finite measure space, 0 < r < p, and {In} a sequence of V(J.t) functions such that II/nllp ~ k for all n and limn _ oo In = I J.t-a.e. Prove that limn _ oo II/n - Ilir = 0, and that the conclusion may fail if either J.t(X) = 00 or r = p. 4.19 (Vitali's Convergence Theorem) Let {In} be a sequence of LP(J.t) functions, 1 ~ P < 00, and assume limn _ oo In = I J.t-a.e. Show that

4.

Problems and Questions

235

I E LP(p) and limn..... oo II/n - Ilip = 0 iff: (a) For each c > 0, there exists a set Ae such that p(Ae) < 00 and IX\A. I/nl Pdp < c for all n, and, (b) liml'(E) .....O IE I/nlPdp = 0, uniformly in n. 4.20 Verify that for every measurable function

I

f I/IP dp = f ptp-1p( {III> t} )dt, Jx J[O,oo)

0
< 00.

4.21 Suppose the nonnegative functions I,g E LP(p), 1 < p < 00, satisfy the relation p( {g > ,x}) ~ I{g>>.} I dp for all ,x > O. Show that IIglip ~ p'lI/l1p, where lip + lip' = 1.

f

4.22 Suppose 0 < r < p < q ~ 00, and let I E LP(p). Show that I can be written I = 9 + h, where 9 E F(p) and h E Lq(p). Further, given t > 0, we can choose 9 and h so that IIgll~ ~ tr-PII/II: and IIhll~ ~ tq-PII/II:·

4.23 Suppose I E wk-L(Rn) is such that I{I f:. O}I < 00. Prove that IE LP(Rn) for each 0 < p < 1. Also, if IE wk-L(Rn)nLOO(Rn), show that I E LP(Rn) for 1 < p < 00.

4.24 The Hardy-Littlewood maximal operator takes LP(Rn ) functions into LP(Rn) functions, 1 < p < 00. More precisely, show there is a constant C = cn,p such that

4.25 Show that if I E LP(Rn ), 1 ~ p ~ 00, then the integral of I differentiates to I( x) for almost every x in Rn. 4.26 Given an interval I = [a,b] in the line, show that a necessary and sufficient condition for a function F to be the integral of IE LP(I), 1 < p < 00, is that the sums

formed for every partition {a ~ Xo < ... < The sup of these sums is then IIlf(x)IPdx.

Xn

~

b} be bounded.

4.27 Suppose p is a finite measure and v <:: p. Prove that the RadonNikodym derivative h = dvJdp E LP(p) iff there is a constant c such that for all measurable at most countable partitions {En} of X, we

XII.

236

LP Spaces

have L:~=l v(En)P / J.L(En)p-l ~ c. What is the relation between c and IIhllp? 4.28 If J.L is a finite measure, and the sequence {In} of nonnegative funcIn dJ.L = ." and liminf I~ dJ.L < 00 for tions satisfies lim sup some 1 < p < 00, show that J.L( {lim sup In ~ .,,}) > O.

Ix

Ix

4.29 Show that each function I E LP(J.L), 0 < p < ing property: lim,X-+oo .VJ.L( {III> A}) = O.

00,

satisfies the follow-

4.30 Suppose I is an extended real-valued function defined on a probability measure space (X,M,J.L). Show that ess sup I

= inf{A: J.L( {x EX: I( x) ~ A}) = I} .

4.31 If I is an extended real-valued function defined on X, define its essential infimum by ess inf 1= sup{A:J.L({x E X :/(x)

< A}) = O}.

Explore the properties of this quantity. In particular show that if I ~ 0 J.L-a.e., then ess inf 1= 1/111/11100. 4.32 Suppose that limn-+ oo II/n - Ilip = 0, 0 < p < 00, and that 9,9n are uniformly bounded measurable functions, n = 1,2, ... , such that limn-+oo 9n = 9 J.L-a.e. Prove that limn-+oo II/n9n - 1911p = O. 4.33 Suppose I, In E LP(J.L) , n = 1,2, ... ,1 < p < 00. Show that In converges weakly to I in LP(J.L) iff (a) sup II/nllp < 00, and, (b) limn -+oo In dJ.L = I dJ.L for all E E M.

IE

IE

4.34 Suppose {In} is a bounded set in LP(J.L), 1 < p < 00. Show that if limn -+ oo In = I J.L-a.e., then also In converges weakly to I in LP(J.L). 4.35 Show that the conclusion of 4.34 holds with the assumption of the pointwise J.L-a.e.convergence replaced by convergence in measure. 4.36 Let (X,M,J.L) be a measure space and {In} a sequence of LP(J.L) functions so that liminf II In lip < 00 for some 0 < p < 00. Suppose further that limn -+ oo In = I J.L-a.e. What can then be said about II/lIp? Well, prove that ;!.~ (II In - III~

-

II/nll~

-

II/II~) = O.

CHAPTER

XIII

Fubini '8 Theorem

In this chapter we deal with the questions of "exchanging the order of integration" in a double integral, and of evaluating an integral as an iterated one.

1. ITERATED INTEGRALS The Lebesgue theory of integration developed in Chapter VII is independent of the dimension of the ambient space, and this is an attractive feature. Nevertheless, the computation of an integral in Rn is usually carried out as a succession of n one-dimensional integrals. The following is a familiar situation: Suppose f is a continuous function on a rectangle I = [a, b] X [e, d] in R2. Then it is true that

Lf(x,Y)dA

=

lb Ld

f(x,y)dydx

=

Ld lb

f(x,y)dxdy,

where all integrals are taken in the sense of Riemann. Simple examples indicate that, in general, this is not always the case. Consider, for instance, 1= (0,1) x (0,1) and

x2 _ y2

f(x,y)

= (x 2 + y2)2'

(x,y) E I.

J;

Since as a simple computation shows f(x,y) dy = 1/(1 + x 2), we have 1 1 Jo Jo f(x,y)dydx = 7r/4. Similarly, J; J; f(x,y) dx dy = -7r/4. Now, a couple ofthings go wrong with f. First notice that for 0 < x < 1, we have f(x,y) dy = 1/2x, and consequently, If(x,y)ldy ~ 1/2x.

J;

J;

XIII.

238

f; f;

It is then apparent that II( x ,y) Idy dx = {(x,y) E I: Ixl > V3lyl } we have

I/(x,y)1 ~

1

00.

Fubini's Theorem

Also, since on the region

1

2(x2 + y2) ,

by x.xx in Cahpter VII, it is clear that I ~ L(I). Motivated by this example we introduce the following definition: Let It ~ R n and 12 ~ R m be intervals in R n and R m respectively, and consider the "rectangle" I = II X 12 ~ R n X R m = Rn+m. An integrable function I E L(I) is said to satisfy, or have, "property F", provided the following three properties hold: (i) For almost every x E II, in the sense of the n-dimensional Lebesgue measure, I(x,y) is a measurable and integrable function of y on 12 , with respect to the m-dimensional Lebesgue measure. (ii) As a function of x E II, fh I( x ,y) dy is measurable and Lebesgue integrable. (iii)

f 1= f f

11

111 112

I(x,y) dy dx .

What we hope to prove is that every I E L(I) has property F. This we achieve in a series of lemmas, each involving a basic step required in approximating arbitrary integrable functions. First an observation: By setting I = 0 off I if necessary we may assume that II = Rn, 12 = Rm and I = Rn+m.

Lemma 1.1. Suppose {Ik} is a non decreasing sequence of functions each of which has property F, and suppose that limk-+oo!k = IE L(Rn+m). Then I also has property F. Proof. Let Nk be a null set (in Rn) such that !k(x,y) is measurable and integrable as a function of y E Rm for x ¢ Nk, and let N = Uk Nk; clearly INI = O. Now, for x ¢ N, Ik(X,y) increases (in the wide sense) to I(x,y), and consequently, for those x's, I(x,.) is measurable. Thus by MeT it follows that lim k-+oo

JfRm Ik(X,y) dy = JfRm I(x,y) dy.

(1.1)

Moreover, since each Ik has property F, each integral on the left-hand side of (1.1) is a measurable, integrable function of x and consequently,

1.

Iterated Integrals

239

fRm f(x,y) dy is also a measurable function of x. Again by MCT, (1.1)

gives lim {

(

k-+oo iRn i~

fk(x,y) dy dx = {

(

f(x,y) dy dx.

iRn iRm

(1.2)

On the other hand, also by MCT, lim {

k-+oo iRn+m

fk

= iRn+m ( f·

(1.3)

Now, since the fk'S have property F, for each k the integral on the lefthand side of (1.2) is equal to that on the left-hand side of (1.3) and consequently, we get

Since

f

E L(Rn+m),

f

has property F.

•

Corollary 1.2. Suppose {fk} is a nonincreasing sequence of functions each of which has property F, and suppose that limk-+oo /k = f E L(Rn+m). Then f also has property F. Lemma 1.3. Suppose E = nr=10k, Ok open, is a Gs subset of Rn+m so that lOki < 00 for some k. Then XE satisfies property F. Proof. We proceed in steps. Note first that if II and 12 are open, bounded intervals in Rn and Rm respectively, and if I = II X 12 , then XI(x,y) = XII (x)Xh(y), and XI has property F. Also note that if A is any subset of the boundary of I, we then have

I{y E R m : (x ,y) E A} I = 0,

a.e. x in Rn.

Roughly speaking, this relation is true except, possibly, along the "endpoints" of II. It then follows that

which together with JRn+rn XA =

IAI

= 0, implies that XA has property

F. II

Consider next a subset j of Rn+m consisting of a rectangle I = 12 , It, 12 open in RR and R m respectively, plus some portion A of

X

XIII.

240

Fubini's Theorem

the boundary of I. Since Xi = XI + XA' and since as is readily verified any finite linear combination offunctions that satisfy property F also has property F, then by the above results it follows that Xi has property F. Now suppose 0 is an open set in R n +m with finite measure; we can write 0 = Uk i k as the pairwise disjoint union of subsets i k of Rn+m consisting of open rectangle Ik and a subset of their boundary. Thus, Xo = limn -+oo Lk=l Xi,.' and since each sum on the right-hand side above has property F, by Lemma 1.1 Xo also satisfies property F. We are finally ready to handle XE' Under our conditions we may assume that

n 00

E =

lOll < 00.

Ok,

k=l

Then XE = limN-+oo Xnf=l Ok' and each of the sets nf=IOk is open and has finite measure. Therefore, each of the functions on the right-hand side above has property F, and, by Corollary 1.2, XE has property F. • Lemma 1.4.

Let N ~ Rn+m,

INI = O. Then XN

has property F.

Proof. By Corollary 1.5 in Chapter V there is a Gs set E ~ N such that lEI = INI = O. Now, XE has property F, and, in particular,

I JRm I XE(x,y) dydx = lEI = O. Jnn Thus, by Corollary 2.3 in Chapter VII, we get that

fnm XE(x,y)dy = Ib E R m :(x,y) E E}I = 0,

x a.e. in R n ,

which in turn implies that

Inn Inm XN(x,y)dydx = Since also JRn+m XN

F.

•

lib

E R m : (x,y) E

= INI = 0, it readily follows

Lemma 1.5. Suppose E Then XE has property F.

~

N}I dx = O.

that XN has property

Rn+m is Lebesgue measurable,

lEI < 00.

1.

Iterated Integrals

Proof.

241

By 1.5 in Chapter V we have E = H \ N,

HaGs set, and N null.

where H = n~l Ok, and some Ok has finite measure. Now, since XE = XH - XN' and since by Lemmas 1.2 and 1.4 the functions XH and XN have property F, it follows that XE also has property F. • The stage is now set for a preliminary version of Fubini's theorem. Theorem 1.6. Proof.

Suppose f E L(Rn+m). Then f has property F.

Since f = f+ - f- , f+, f- integrable, we may assume that

f is nonnegative. The conclusion follows now by a familiar limiting argument. Indeed, let {
o ~
and

lim
k-+oo

f

a.e.

Now, since each
= {y E R n : (x,y) E E},

all x E R n ,

and

EY = {x E R n : (x ,y) E E} ,

all y E R m

•

Although not intuitively apparent, the sections of a measurable subset of Rn+m are a.e. measurable in the respective ambient spaces. This is a particular instance of our next result, which extends the following interesting consequence of the preliminary version of Fubini's theorem: If f(x,y) E L(Rn+m), then f(x,y) is a Lebesgue measurable function of y E Rm, for almost every x E Rn. We now show that the same is true if f is merely measurable; the result about the sections follows by considering f = XE' which is the first step of the proof. Proposition 1.7. Suppose f is a measurable function defined on Rn+m. Then, for almost every x E Rn, f(x,y) is a measurable function of y E Rm. Symmetrically, for almost every y E R m , f(x,y) is a measurable function of x E RR.

XIII.

242

Fubini's Theorem

=

Proof. Assume first that J XE' where E a measurable subset of Rn+m. Then write E = HUN, where H is of type Fq in Rn+m and INI o. Then Ex Hx U N x , where Hx is an Fq subset of Rm, and by Lemma 1.4, INxl 0 for almost every x ERn. Therefore, Ex is measurable for almost every x E Rn. Similarly for EY. If J is any measurable function on Rn+m now, given A real, consider the set E(A) ((x,y) E Rn+m :J(x,y) > A}. Since E(A) is measurable in Rn+m, the section E(A)x is measurable in Rm for almost every x in Rnj the exceptional null subset of Rn depends, of course, on A. The union N of these exceptional sets for all rational A is also a null set, and the set {y E R m : J( x ,y) > A} is measurable, provided that A is rational and x ft N. By 4.1 in Chapter VI the same is true for all A, and consequently J( x ,y) is a measurable function of y E R m , for almost every x E Rn. The other statement is proved in an analogous fashion. •

=

=

=

=

We now state Fubini's theorem in its general form. Theorem 1.8. Let J(x,y) be a measurable function defined on a measurable subset E of Rn+m. Then, (i) For almost every x E Rn, J(x,y) is a measurable function of yon Ex. (ii) For almost every y E Rm, J(x,y) is a measurable function of x on EY. (iii) If J E L(E), then for almost every x in Rn, J(X,.) E L(Ex). Moreover, fEz J(x,y) dy is an integrable function of x and

f J= f f

JE

JRn JEz

J(x,y)dydx.

(iv) If J E L(E), then for almost every y in R m , J(., y) E L(EY). Moreover, fEll J(x,y) dy is an integrable function of y and

f

JE

J=

f

f

JRm JEll

J(x,y)dxdy.

Proof. Let i be the function equal to J on E and equal to 0 elsewhere in Rn+m. Since J is measurable on E and E is measurable, i is measurable on Rn+m. Therefore, by Proposition 1.7, /( x ,y) is a measurable function of y for almost every x in Rn. Moreover, since Ex is measurable for almost every x in Rn, it follows that J(x,y) is measurable as a function of yon Ex, for almost every x. This proves (i)j the proof of (ii) is analogous.

1.

Iterated Integrals

243

Now, if f is integrable on E, then we get

f f=f

iE

iRn+m

j

j E L(Rn+m) and by Theorem 1.6

=f f

iRn iRm

j( x ,y) dy dx .

(1.4)

On the other hand, Ex is measurable for almost every x in Rn, and for those x's we get

f

iRm

j(x,y) dy

=f

iE:z:

f(x,y) dy.

(1.5)

Whence integrating (1.5) over R n , and combining the resulting expression with (1.4), it readily follows that (iii) holds. The proof of (iv) is analogous. • Next we consider whether a converse to Fubini's theorem is true. For instance, suppose f is measurable on R n +m , and the iterated integrals of f exist and are equal. Does it, then, follow that f is integrable? The answer to this question is no, and the counterexample is constructed along the lines of Figure 5 below.

the = 1

Figure 5

tPli: =

-1

XIII.

244

Fubini's Theorem

Indeed, let J = [0,1] x [0,1], divide J into four equal squares, separate the bottom left square, call it J}, divide the right upper square into four equal squares, call the bottom left square J 2 , and so on. On each square Jk, k = 1,2, ... , define a function tPk as follows: Divide each Jk into four equal squares, and let tPk equal -Ion the interior of left bottom and right upper squares, equal 1 on interior ofthe right bottom and left upper squares, and otherwise. Put now 00 1 I(x,y) = -IJIXk(X,y).

°

I: k=l

k

There is no problem with convergence here since at most one of the terms of the above series is nonzero. Also I, being the sum of measurable functions, is itself measurable. Now, since for all x E [0,1] we have itO,l] I(x,y)dy = 0, it is clear that

f

f

I( x ,y) dy dx

J[O,l] J[O,l]

=

°.

A similar argument gives that the other iterated integral is 0, and so both iterated integrals exist and are equal. Is I integrable, and, if so, is its integral equal to o? Well, since ItPk(x,y)1 = 1 for (x,y) E Jk, it follows that 1 f ItPkl = 00, fill = J[O,l] x [0,1] k IJkl

I:- JJk

and the integral of I is not defined on [0,1] X [0,1]. In deciding what goes wrong with I, we note that it changes signs. In fact, this is the only difficulty. Theorem 1.9 (Tonelli). Suppose function defined on Rn+m. Then,

f

JRn+m

1= f

f

Jnn JR!"

I

is a nonnegative measurable

I(x,y) dy dx = f

f I(x,y) dx dy.

JRm JRn

(1.6)

The identity in (1.6) is understood as follows: If anyone of the three expressions there is infinite, so are the other two, and if anyone is finite, the other two are also finite and all three are then equal. Proof. Let {/k} be a nondecreasing sequence of integrable functions that converges to I a.e. For instance, if Xk(X,y) denotes the characteristic function of Il(O,k) x 12 (0,k), then the sequence

h(x,y)

= min{/(x,y), k}Xk(X,y) ,

k

= 1,2, ...

1.

Iterated Integrals

245

will do. :Now",,,, apply Fubini's theorem to each !k and note that outside of a null subset Nk of R n, JRm !k(X,y) dy is well-defined and

r

JRn+m

fk =

rr

JRn JRm

fk(x,y)dydx

k = 1,2,...

Let N = Uk Nki N is a null subset of R n and for x E R n it follows that lim k-+oo

\

(1.7)

N, by MeT,

Jr Rm !k(X,y) dy = Jr Rm !(x,y) dy.

Thus, again by MeT, lim k-+oo

r JRm r !k(X,y) dydx = JRnr JRm r !(x,y) dydx. JRn

(1.8)

On the other hand, also by MeT, we get (1.9) Whence, combining (1.7), (1.8), and (1.9) we see that

r

JRn+m

!=

r r !(x,y)dydx,

JRn JRm

which is one of the identities in (1.6). The other is obtained in a similar fashion. • An interesting application of Tonelli's theorem is the following remark concerning the evaluation of the measure of a set E in R n +m in terms of its sections. Since for each fixed x E R n , XE(x,y) i- 0 iff XEJy) i- 0, we have

A similar argument also gives that lEI = Note that in particular we have that in Rn iff IEYI = 0 for y a.e. in Rm.

JRm IEYI dy .

E is

null iff

IExl = 0 for

x

a.e.

XIII.

246

Fubini's Theorem

2. CONVOLUTIONS AND APPROXIMATE IDENTITIES Given integrable functions I, 9 defined on Rn, we have already noted that the pointwise product I 9 is not necessarily integrable. In other words, the expression I(y)g(y) dy is not necessarily finite. It may come as a surprise that a closely related expression, namely the convolution 1* g( x) of I and 9 at x, is nevertheless finite a.e. on Rn. The precise definition is as follows: Suppose I,g E L(Rn), and consider the integral

IRn

I*g(x)= [

iRn I(x-y)g(y)dy,

xERn.

(2.1)

Suppose for the moment that I,g ~ 0, and note that for any fixed x E Rn, the integrand in (2.1) is a nonnegative measurable function, so that 1* g( x) is a well-defined real number or 00. But, is there any x for which 1* g(x) < oo? A way to go about answering this question is to show that actually the function I * 9 is integrable, for then it will be finite a.e. As for the integrability, observe that

III * gill

= [

[ I(x iNn iRn

y)g(y) dydx,

and this expression corresponds to one of the iterated integrals of the function I(x - y)g(y). But it is the other iterated integral, namely,

iNn iNn I(x -

y)g(y)dxdy = =

iNn iNn I(x -

y) dxg(y) dy

1I/11111g111 < 00 ,

which is easy to handle. To pass from one to the other we wish to invoke Tonelli's theorem, and to do this we need to prove first that the function h( x ,y) defined by

h(x,y) = I(x - y)g(y) ,

X,y E R n ,

is actually measurable in Rn x Rn = R2n. Now, since 9 is measurable, given a scalar ,x, the set

((x,y) E R 2n :g(y) > ,x} = R n x {y E R n : g(y) > ,x} is a measurable subset of R2n, and 9 is also measurable when considered as a function defined on R2n.

2.

Convolutions and Approximate Identities

247

So, we are reduced to showing that if I is a measurable function defined on R n, then h(x,y) = I(x - y) is a measurable function on R n x R n = R2n. To clarify the ideas behind this result we do the case n = 1, where the main difficulties are already apparent; we leave it to the reader to think of the general proof. We proceed step by step and assume first that I = XI' where 1= (a,b) is an open interval. In this case the level sets

((x,y) E R2: I(x - y) > A},

A real,

are empty if A ~ 1, and the open strip {(x,y) : a < x - y < b} otherwise. In either case, the level sets are measurable. Next, if 1= Xo, 0 open, then write 0 = Uk h as the pairwise disjoint union of the open intervals h, and note that the levels sets are now {/(x-y) > A} = Uk{XI" (x-y) > A}, and hence they are also measurable. If I = Xa, GaGs subset in R, then we have G = k Ok, Ok open, and consequently I(x - y) is the limit I(x - y) = limk-+oo Xo,,(x - y) of measurable functions, and hence measurable. The same is true if I = XN' N a null subset of Rn. Indeed, let G be a Gs subset in R such that N ~ G, IGI = 0, and let 9 = {(x,y) E R2: x - y E G}. As pointed out above, we have 191 = 0 iff 19x1 = 0 for all x E R. Now,

n

9x = {y

E R: x - y E G} = {y E R: y = x - g,g E G} = x - G,

and by the translation invariance ofthe Lebesgue measure we have Ix - GI = IGI = 0 for all x E R. Thus 191 = O. Consider now

N = {(x,y)

E R2:x - YEN} ~

19x1 =

9.

By the completeness of the Lebesgue measure it follows that also INI = 0, and so XN( x - y) is measurable. Now, if 1= XE' where E is a measurable subset of R, write E = G\N, with GaGs subset of Rand N a null set, and note that in this case I(x - y) = Xa(x - y) - XN(x - y), is also measurable. Whence simple functions 4> also enjoy the property that 4>( x - y) is measurable, and the same is true of the limits of simple functions, to wit, arbitrary measurable functions. This is precisely what we set to prove. Returning to the properties of convolutions, we have Theorem 2.1.

Suppose I,g E L(Rn). Then

I I/(x JRB

y)llg(y)1 dy <

00,

x a.e. in Rn.

(2.2)

XIII.

FUbini's Theorem

y)g(y) dy.

(2.3)

248

For those x's that satisfy (2.2) we define

1* g(x)

= iR!' I(x -

Then 1* 9 E L(Rn), and (2.4)

Proof. First observe that the function h on R 2n defined by h(x,y) = I/(x - y)llg(y)1 is measurable and nonnegative. Thus by Tonelli's theorem we have the equality of the iterated integrals of h, and consequently,

r iRnr I/(x iRn

y)llg(y)1 dy dx =

r iRnr I/(x iRn

y)llg(y)1 dx dy

=

iR!' iR!' I/(x -

y)1 dx Ig(y)1 dy.

By the translation invariance of the Lebesgue measure it readily follows that the inner-most integral above is equal to IIIIII, and the right-hand side is IIgllll1/111. Now, since II * g(x)1 ~ I/(x - y)llg(y)1 dy, (2.4) holds. •

IRn

What more can we say about convolutions? For instance, a natural question to consider is whether the operation of convolution has a unit, i.e., whether there exists an integrable function I such that 1* 9 = 9 for any 9 E L(Rn). Suppose such a function I exists, and given 0 f:. x E R n , let 0 < 1] < Ixl. Put now g(y) = (1/21])n X1 (0,1/)(Y) and note that

O=g(x)=1*9(X)=II(01 )1 ,1]

r

il(O,1/)

I(x-y)dy.

(2.5)

Now, by the Lebesgue Differentiation Theorem, the limit as 1] ~ 0 of the above expression is I(x) a.e., and consequently, we get that I = 0 a.e. Thus, since 0 * 9 = 0, such a function I cannot exist. Nevertheless, if we put x = 0 in (2.5), since g(O) = 1/11(0,1])1, again by the Lebesgue Differentiation Theorem it follows that 00

= lim 11(01 1/-+0

,1]

Still one more property: Since integrable, we get

)1 i r1(0,1/) I( -y) dy =

I

~

1(0).

0, by taking 9 to be nonnegative and

2.

Convolutions and Approximate Identities

249

IRn

and so we have f = 1. Combining these remarks we note that the integrable function f, which does not exist, satisfies the following properties: It is 0 a.e. away from the origin, where it assumes the value 00, and it has integral 1. In other words, f corresponds to the Dirac measure 00. The convolution is also well-defined in the more general context of the Lebesgue LP-spaces, 1 < p ~ 00. More precisely, we have Theorem 2.2. Suppose 9 E L(Rn), f E LP(Rn), 1 < p ~ 00. Then the convolution f * g(x) is a well-defined LP(Rn) function that satisfies (2.6) Proof. Since If *g( x)1 ~ If I*Igl( x) we may assume that f and 9 are nonnegative. Since the case p = 00 follows at once from the definition of convolution, we assume that 1 < p < 00. Let 1 < q < 00 be the conjugate index to p, i.e., l/p + l/q = 1, and note that by Holder's inequality, Tonelli's theorem, and the translation invariance of the integral we have

Ln (Ln ~ Ln (Inn

IIf * gll~ =

f(x - y)g(y)l/Pg(y)l/qdY)P dx f( x - y)P g(y )P·l/P dY)

(Inn g(y )q.l/q y/q dx

= IIglhllfll~lIglli/q = IIglli/q+lllfll~ = IIgllillfll~, and, as asserted, (2.6) holds.

•

Thus the convolution inherited the integrability properties of fj next we show that it also inherits the "smoothness" properties of g. Smooth integrable functions aboundj e- 1x12 is one such example. In applications it is important to have at hand a smooth integrable function 4> that vanishes off Ixl ~ 1. To construct such a function, let "p be defined on the line by

= {e 1 / t

t<0 t ~ O. Since all the derivatives of"p exist when t =J 0 and converge to 0 when "p(t)

o

-+ 0, and since by a simple application of l'Hopital's rule it follows that all derivatives of"p vanish when t = 0, "p E COO(R). The function 4> is now defined by letting 4>(x) = ¢(lxI 2 - 1). Before we continue we need a simple observation concerning convolutions: By the translation invariance of the Lebesgue measure it readily

t

XIII.

250

follows that at those x's where

Fubini's Theorem

I * g( x) is defined, we also have

I*g(x)= iR"I(Y)9(X-Y)dY . We are now ready to prove Theorem 2.3. Suppose I E LP(Rn), 1 :::; P :::; m ~ 1. Then 1* l/J E LP(Rn ) n cm(Rn).

00,

and

l/J E C[f(Rn),

Proof. Since, by Theorem 2.2, 1* l/J E LP(Rn), only the smoothness of the convolution needs to be proved. We first show that 1* l/J is continuous. Indeed, given x,h ERn, note that

II * l/J(x + h) - 1* l/J(x)1 =

lin

I(y)(l/J(X + h - y) -l/J(X - Y))dyl

JRn II(y)IIl/J(x + h -

:::; f

y) -l/J(X - y)1 dy

:::; IIllIplll/J(X + h - .) -l/J(X - ')lIq, (2.7) conjugate indices. Now, if q < 00, since l/J E Lq(Rn) and

where p,q are since translations are continuous in Lq, the right-hand side of (2.7) goes to oas Ihl -+ 0, and so does the left-hand side; thus I*l/J(x) is continuous. On the other hand, if q = 00, by the first inequality above it readily follows that II * l/J(x + h) - 1* l/J(x)1 :::; 111111 sup Il/J(' + h) -l/J(-)I·

Now, by the uniform continuity of l/J we get that the right-hand side of the above inequality goes to 0 with h, and so does the left-hand side there. As for the smoothness, let h = (h, 0, ... ,0) denote the vector with scalar h i- 0 in the first position and zeros elsewhere, and put

"'( h)_l/J(x+h-y)-l/J(x- y ) _ (al/J) ( _ ) .,.., x,y, h aX! x y. By the conditions of the theorem it is clear that for each fixed x E R n , ~(x,y,h) -+ 0, uniformly and boundedly in y as h -+ O. Whence

lim h-+O

f I(y)~(x,y,h) dy = 0, JJln

Moreover, since the integral in (2.8) equals

x E Rn.

(2.8)

2.

Convolutions and Approximate Identities

251

it readily follows that

au *
VXl

o