Lecture 5: Fubini-Tonelli and Radon-Nikodym 1 of 13

Course: Theory of Probability ITerm: Fall 2013Instructor: Gordan Zitkovic

Lecture 5Theorems of Fubini-Tonelli and

Radon-Nikodym

Products of measure spaces

We have seen that it is possible to define products of arbitrary collec-tions of measurable spaces - one generates the -algebra on the prod-uct by all finite-dimensional cylinders. The purpose of the presentsection is to extend that construction to products of measure spaces,i.e., to define products of measures.

Let us first consider the case of two measure spaces (S,S , S) and(T, T , T). If the measures are stripped, the product S T is endowedwith the product -algebra S T = ({A B : A S , B T }).The family P = {A B : A S , B T } serves as a good startingpoint towards the creation of the product measure S T . Indeed, ifwe interpret of the elements in P as rectangles of sorts, it is natural todefine

(S T)(A B) = S(A)T(B).

The family P is a -system (why?), but not necessarily an algebra,so we cannot use Theorem 2.9 (Caratheodorys extension theorem) todefine an extension of S T to the whole S T . It it not hard,however, to enlarge P a little bit, so that the resulting set is an algebra,but that the measure S T can still be defined there in a naturalway. Indeed, consider the smallest algebra that contains P . It is easyto see that it must contain the family S defined by

A = {nk=1 Ak Bk : n N, Ak S , Bk T , k = 1, . . . , n}.

Problem 5.1. Show that A is, in fact, an algebra and that each elementC A can be written in the form

C = nk=1 Ak Bk,

for n N, Ak S , Bk T , k = 1, . . . , n, such that A1 B1, . . . , An Bnare pairwise disjoint.

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Lecture 5: Fubini-Tonelli and Radon-Nikodym 2 of 13

The problem above allows us to extend the definition of the setfunction S T to the entire A by

(S T)(C) =n

k=1

S(Ak)T(Bk),

where C = nk=1 Ak Bk for n N, Ak S , Bk T , k = 1, . . . , n is arepresentation of C with pairwise disjoint A1 B1, . . . , An Bn.

At this point, we could attempt to show that the so-defined setfunction is -additive on A and extend it using the Caratheodory ex-tension theorem. This is indeed possible - under the additional as-sumption of -finiteness - but we will establish the existence of prod-uct measures as a side-effect in the proof of Fubinis theorem below.

Lemma 5.1 (Sections of measurable sets are measurable). Let C be anS T -measurable subset of S T. For each x S the section Cx = {y T : (x, y) C} is measurable in T .

Proof. In the spirit of most of the measurability arguments seen so farin these notes, let C denote the family of all C S T such thatCx is T -measurable for each x S. Clearly, the rectangles A B,A S , B T are in A because their sections are either equal to orB, for each x S. Remember that the set of all rectangles generatesS T . The proof of the theorem will, therefore, be complete once itis established that C is a -algebra. This easy exercise is left to thereader.

Problem 5.2. Show that an analogous result holds for measurablefunctions, i.e., show that if f : S T R is a S T -measurable func-tion, then the function x 7 f (x, y0) is S-measurable for each y0 T,and the function y 7 f (x0, y) is T -measurable for each y0 T.

Proposition 5.2 (A simple Cavalieris principle). Let S and T be finitemeasures. For C S T , define the functions C : T [0, ) andC : S [0, ) by

C(y) = S(Cy), C(x) = T(Cx).

Then,

C L0+(T ), C L0+(S) and

C dT =

C dS. (5.1)

Proof. Note that, by Problem 5.2, the function x 7 1C(x, y) is S-measurable for each y T. Therefore,

1C(, y) dS = S(Cy) = C(y), (5.2)

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Lecture 5: Fubini-Tonelli and Radon-Nikodym 3 of 13

and the function C is well-defined.Let C denote the family of all sets in S T such that (5.1) holds.

First, observe that C contains all rectangles A B, A S , B T ,i.e., it contains a -system which generates S T . Therefore, by the- Theorem (Theorem 2.12), it will be enough to show that C is a-system. We leave the details to the reader1 1 Use representation (5.2) and the mono-

tone convergence theorem. Where is thefiniteness of the measures used?

Proposition 5.3 (Simple Cavalieri holds for -finite measures). The con-clusion of Proposition 5.2 continues to hold if we assume that S and T areonly -finite.

Proof (*). Thanks to -finiteness, there exists pairwise disjoint sequences{An}nN and {Bn}nN in S and T , respectively, such that n An = S,mBm = T and S(An) < and S(Bm) < , for all m, n N.

For m, n N, define the set-functions nS and mT on S and Trespectively by

nS(A) = S(An A), mT (B) = T(Bm B).

It is easy to check that all nS and mT , m, n N are finite measures

on S and T , respectively. Moreover, S(A) = n=1 nS(A), T(B) =m=1

mT (B). In particular, if we set

nC(y) =

nS(Cy) and

mC (x) =

mT (Cx), for all x S and y S, we have

C(y) = S(Cy) =

n=1

nS(Cy) =

n=1

nC(y), and

C(x) = T(Cx) =

m=1

mT (Cx) =

m=1

mC (x),

for all x S, y T.We can apply the conclusion of Proposition 5.2 to all pairs (S,S , nS)

and (T, T , mT ), m, n N, of finite measure spaces to conclude that allelements of the sums above are measurable functions and that so areC and C.

Similarly, the sequences of non-negative functions ni=1 nC(y) and

mi=1 iC(x) are non-decreasing and converge to C and C. Therefore,

by the monotone convergence theorem,

C dT = limn

n

i=1

iC dT , and

C dS = limn

n

i=1

iC dS.

On the other hand, we have

nCdmT =

mC d

nS, by Proposition 5.2.

Summing over all n N we haveC dmT =

nN

mC d

nS =

mC dS,

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Lecture 5: Fubini-Tonelli and Radon-Nikodym 4 of 13

where the last equality follows from the fact (see Problem 5.3 below)that

nN

f dnS =

f dS,

for all f L0+. Another summation - this time over m N - completesthe proof 2. 2 The argument of the proof above un-

covers the fact that integration is a bilin-ear operation, i.e., that the mapping

( f , )

f d,

is linear in both arguments.

Problem 5.3. Let {An}nN be a measurable partition of S, and let themeasure n be defined by n(A) = (A An) for all A S . Showthat for f L0+, we have

f d = nN

f dn.

Proposition 5.4 (Finite products of measure spaces). Let (Si,Si, i), i =1, . . . , n be -finite measure spaces. There exists a unique measure3 - denoted 3 The measure 1 n is called

the product measure, and the mea-sure space (S1 Sn,S1 Sn, 1 n) the product (measurespace) of measure spaces (S1,S1, 1),. . . , (Sn,Sn, n).

by 1 n - on the product space (S1 Sn,S1 Sn) withthe property that

(1 n)(A1 An) = 1(A1) . . . n(An),

for all Ai Si, i = 1, . . . , n. Such a measure is necessarily -finite.

Proof. To simplify the notation, we assume that n = 2 - the generalcase is very similar. For C S1 S2, we define

(1 2)(C) =

S2C d2,

where C(y) = 1(Cy) and Cy = {x S1 : (x, y) C}. It follows fromProposition 5.3 that 1 2 is well-defined as a map from S1 S2 to[0, ]. Also, it is clear that (1 2)(A B) = 1(A)2(B), for allA S1, B S2. It remains to show that 1 2 is a measure. We startwith a pairwise disjoint sequence {Cn}nN in S1 S2. For y S2, thesequence {(Cn)y}nN is also pairwise disjoint, and so, with C = nCn,we have

C(y) = 1(Cy) = nN

2

((Cn)y

)=

nNCn(y), y S2.

Therefore, by the monotone convergence theorem (see Problem 3.13for details) we have

(1 2)(C) =

S2C d2 =

nN

S2

Cn d = nN

(1 2)(Cn).

Finally, let {An}nN, {Bn}nN be sequences in S1 and S2 (respec-tively) such that 1(An) < and 2(Bn) < for all n N andn An = S1, nBn = S2. Define {Cn}nN as an enumeration of thecountable family {Ai Bj : i, j N} in S1S2. Then (12)(Cn) < and all n N and nCn = S1 S2. Therefore, 12 is -finite.

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Lecture 5: Fubini-Tonelli and Radon-Nikodym 5 of 13

Now that we know that product measures exist, we can state andprove the important theorem which, when applied to integrable func-tions bears the name of Fubini, and when applied to non-negativefunctions, of Tonelli. We state it for both cases simultaneously (i.e., onL01) in the case of a product of two measure spaces. An analogoustheorem for finite products can be readily derived from it. When thevariable or the underlying measure space of integration needs to bespecified, we write

S f (x) (dx) for the Lebesgue integral

f d.

Theorem 5.5 (Fubini, Tonelli). Let (S,S , S) and (T, T , T) be two -finite measure spaces. For f L01(S T) we have

f d(S T) =

S

(T

f (x, y) T(dy))

S(dx)

=

T

(S

f (x, y) S(dx))

T(dy)(5.3)

Proof. All the hard work has already been done. We simply need tocrank the Standard Machine. Let H denote the family of all functionsin L0+(ST) with the property that (5.3) holds. Proposition 5.3 impliesthat H contains the indicators of all elements of S T . Linearity ofall components of (5.3) implies that H contains all simple functions inL0+, and the approximation theorem 3.10 implies that the whole L0+ isin H. Finally, the extension to L01 follows by additivity.

Since f is always in L01, we have the following corollary

Corollary 5.6. For f L0(S T), we have

f L01(S T) if and only if

S

(T

f(x, y) T(dy))

S(dx) < .

Example 5.7.