# 8.8: Product Measures. Iterated Integrals

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Let $$(X, \mathcal{M}, m)$$ and $$(Y, \mathcal{N}, n)$$ be measure spaces, with $$X \in \mathcal{M}$$ and $$Y \in \mathcal{N}.$$ Let $$\mathcal{C}$$ be the family of all "rectangles," i.e., sets

$A \times B,$

with $$A \in \mathcal{M}, B \in \mathcal{N}, m A<\infty,$$ and $$n B<\infty$$.

Define a premeasure $$s : \mathcal{C} \rightarrow E^{1}$$ by

$s(A \times B)=m A \cdot n B, \quad A \times B \in \mathcal{C}.$

Let $$p^{*}$$ be the $$s$$-induced outer measure in $$X \times Y$$ and

$p : \mathcal{P}^{*} \rightarrow E^{*}$

the $$p^{*}$$-induced measure ("product measure," $$p=m \times n$$) on the $$\sigma$$-field $$\mathcal{P}^{*}$$ of all $$p^{*}$$-measurable sets in $$X \times Y$$ (Chapter 7, §§5-6).

We consider functions $$f : X \times Y \rightarrow E^{*}$$ (extended-real).

I. We begin with some definitions.

## Definitions

(1) Given a function $$f : X \rightarrow Y \rightarrow E^{*}$$ (of two variables $$x, y$$), let $$f_{x}$$ or $$f(x, \cdot)$$ denote the function on $$Y$$ given by

$f_{x}(y)=f(x, y);$

it arises from $$f$$ by fixing $$x$$.

Similarly, $$f^{y}$$ or $$f(\cdot, y)$$ is given by $$f^{y}(x)=f(x, y)$$.

(2) Define $$g : X \rightarrow E^{*}$$ by

$g(x)=\int_{Y} f(x, \cdot) dn,$

and set

$\int_{X} \int_{Y} f dn dm=\int_{X} g dm,$

also written

$\int_{X} dm(x) \int_{Y} f(x, y) dn(y).$

This is called the iterated integral of $$f$$ on $$Y$$ and $$X,$$ in this order.

Similarly,

$h(y)=\int_{X} f^{y} dm$

and

$\int_{Y} \int_{X} f dm dn=\int_{Y} h dn.$

Note that by the rules of §5, these integrals are always defined.

(3) With $$f, g, h$$ as above, we say that $$f$$ is a Fubini map or has the Fubini properties (after the mathematician Fubini) iff

(a) $$g$$ is $$m$$-measurable on $$X$$ and $$h$$ is $$n$$-measurable on $$Y$$;

(b) $$f_{x}$$ is $$n$$-measurable on $$Y$$ for almost all $$x$$ (i.e., for $$x \in X-Q$$, $$m Q=0); f^{y}$$ is $$m$$-measurable on $$X$$ for $$y \in Y-Q^{\prime}, n Q^{\prime}=0;$$ and

(c) the iterated integrals above satisfy

$\int_{X} \int_{Y} f dn dm=\int_{Y} \int_{X} f dm dn=\int_{X \times Y} f dp$

(the main point).

For monotone sequences

$f_{k} : X \times Y \rightarrow E^{*} \quad(k=1,2, \ldots),$

we now obtain the following lemma.

## Lemma $$\PageIndex{1}$$

If $$0 \leq f_{k} \nearrow f$$ (pointwise) on $$X \times Y$$ and if each $$f_{k}$$ has Fubini property (a), (b), or (c), then $$f$$ has the same property.

Proof

For $$k=1,2, \ldots,$$ set

$g_{k}(x)=\int_{Y} f_{k}(x, \cdot) dn$

and

$h_{k}(y)=\int_{X} f_{k}(\cdot, y) dm.$

By assumpsion,

$0 \leq f_{k}(x, \cdot) \nearrow f(x, \cdot)$

pointwise on $$Y.$$ Thus by Theorem 4 in §6,

$\int_{Y} f_{k}(x, \cdot) \nearrow \int_{Y} f(x, \cdot) dn,$

i.e., $$g_{k} \nearrow g$$ (pointwise) on $$X,$$ with $$g$$ as in Definition 2.

Again, by Theorem 4 of §6,

$\int_{X} g_{k} dm \nearrow \int_{X} g dm;$

or by Definition 2,

$\int_{X} \int_{Y} f dn dm=\lim _{k \rightarrow \infty} \int_{X} \int_{Y} f_{k} dn dm.$

Similarly for

$\int_{Y} \int_{X} f dm dn$

and

$\int_{X \times Y} f dp.$

Hence $$f$$ satisfies (c) if all $$f_{k}$$ do.

Next, let $$f_{k}$$ have property (b); so $$(\forall k) f_{k}(x, \cdot)$$ is $$n$$-measurable on $$Y$$ if $$x \in X-Q_{k}$$ ($$m Q_{k}=0$$). Let

$Q=\bigcup_{k=1}^{\infty} Q_{k};$

so $$m Q=0,$$ and all $$f_{k}(x, \cdot)$$ are $$n$$-measurable on $$Y,$$ for $$x \in X-Q.$$ Hence so is

$f(x, \cdot)=\lim _{k \rightarrow \infty} f_{k}(x, \cdot).$

Similarly for $$f(\cdot, y).$$ Thus $$f$$ satisfies (b).

Property (a) follows from $$g_{k} \rightarrow g$$ and $$h_{k} \rightarrow h. \quad \square$$

Using Problems 9 and 10 from §6, the reader will also easily verify the following lemma.

## Lemma $$\PageIndex{2}$$

(i) If $$f_{1}$$ and $$f_{2}$$ are nonnegative, $$p$$-measurable Fubini maps, so is $$af_{1}+b f_{2}$$ for $$a, b \geq 0$$.

$\int_{X \times Y} f_{1} d p<\infty \text { or } \int_{X \times Y} f_{2} d p<\infty,$

then $$f_{1}-f_{2}$$ is a Fubini map, too

## Lemma $$\PageIndex{3}$$

Let $$f=\sum_{i=1}^{\infty} f_{i}$$ (pointwise), with $$f_{i} \geq 0$$ on $$X \times Y$$.

(i) If all $$f_{i}$$ are $$p$$-measurable Fubini maps, so is $$f$$.

(ii) If the $$f_{i}$$ have Fubini properties (a) and (b), then

$\int_{X} \int_{Y} f dn dm=\sum_{i=1}^{\infty} \int_{X} \int_{Y} f_{i} dn dm$

and

$\int_{Y} \int_{X} f dm dn=\sum_{i=1}^{\infty} \int_{Y} \int_{X} f_{i} dm dn.$

II. By Theorem 4 of Chapter 7, §3, the family $$\mathcal{C}$$ (see above) is a semiring, being the product of two rings,

$\{A \in \mathcal{M} | mA<\infty\} \text { and }\{B \in \mathcal{N} | nB<\infty\}.$

(Verify!) Thus using Theorem 2 in Chapter 7, §6, we now show that $$p$$ is an extension of $$s : \mathcal{C} \rightarrow E^{1}.$$

## Theorem $$\PageIndex{1}$$

The product premeasure s is $$\sigma$$-additive on the semiring $$\mathcal{C}.$$ Hence

(i) $$\mathcal{C} \subseteq \mathcal{P}^{*}$$ and $$p=s<\infty$$ on $$\mathcal{C},$$ and

(ii) the characteristic function $$C_{D}$$ of any set $$D \in \mathcal{C}$$ is a Fubini map.

Proof

Let $$D=A \times B \in \mathcal{C};$$ so

$C_{D}(x, y)=C_{A}(x) \cdot C_{B}(y).$

(Why?) Thus for a fixed $$x, C_{D}(x, \cdot)$$ is just a multiple of the $$\mathcal{N}$$-simple map $$C_{B},$$ hence $$n$$-measurable on $$Y.$$ Also,

$g(x)=\int_{Y} C_{D}(x, \cdot) dn=C_{A}(x) \cdot \int_{Y} C_{B} dn=C_{A}(x) \cdot nB;$

so $$g=C_{A} \cdot n B$$ is $$\mathcal{M}$$-simple on $$X,$$ with

$\int_{X} \int_{Y} C_{D} dn dm=\int_{X} g dm=nB \int_{X} C_{A} dm=nB \cdot m A=sD.$

Similarly for $$C_{D}(\cdot, y),$$ and

$h(y)=\int_{X} C_{D}(\cdot, y) dm.$

Thus $$C_{D}$$ has Fubini properties (a) and (b), and for every $$D \in \mathcal{C}$$

$\int_{X} \int_{Y} C_{D} dn dm=\int_{Y} \int_{X} C_{D} dm dn=sD.$

To prove $$\sigma$$-additivity, let

$D=\bigcup_{i=1}^{\infty} D_{i} \text { (disjoint), } D_{i} \in \mathcal{C};$

so

$C_{D}=\sum_{i=1}^{\infty} C_{D_{i}}.$

(Why?) As shown above, each $$C_{D_{i}}$$ has Fubini properties (a) and (b); so by (1) and Lemma 3,

$sD=\int_{X} \int_{Y} C_{D} dn dm=\sum_{i=1}^{\infty} \int_{X} \int_{Y} C_{D_{i}} dn dm=\sum_{i=1}^{\infty} sD_{i},$

as required.

Assertion (i) now follows by Theorem 2 in Chapter 7, §6. Hence

$sD=pD=\int_{X \times Y} C_{D} dp;$

so by formula (1), $$C_{D}$$ also has Fubini property (c), and all is proved.$$\quad \square$$

Next, let $$\mathcal{P}$$ be the $$\sigma$$-ring generated by the semiring $$\mathcal{C}$$ (so $$\mathcal{C} \subseteq \mathcal{P} \subseteq \mathcal{P}^{*}$$).

## Lemma $$\PageIndex{4}$$

$$\mathcal{P}$$ is the least set family $$\mathcal{R}$$ such that

(i) $$\mathcal{R} \supseteq \mathcal{C}$$;

(ii) $$\mathcal{R}$$ is closed under countable disjoint unions; and

(iii) $$H-D \in \mathcal{R}$$ if $$D \in \mathcal{R}$$ and $$D \subseteq H, H \in \mathcal{C}$$.

This is simply Theorem 3 in Chapter 7, §3, with changed notation.

## Lemma $$\PageIndex{5}$$

If $$D \in \mathcal{P}$$ ($$\sigma$$-generated by $$\mathcal{C}),$$ then $$C_{D}$$ is a Fubini map.

Proof

Let $$\mathcal{R}$$ be the family of all $$D \in \mathcal{P}$$ such that $$C_{D}$$ is a Fubini map. We shall show that $$\mathcal{R}$$ satisfies (i)-(iii) of Lemma 4, and so $$\mathcal{P} \subseteq \mathcal{R}.$$

(ii) Let

$D=\bigcup_{i=1}^{\infty} D_{i} \text { (disjoint),} \quad D_{i} \in \mathcal{R}.$

Then

$C_{D}=\sum_{i=1}^{\infty} C_{D_{i}},$

and each $$C_{D_{i}}$$ is a Fubini map. Hence so is $$C_{D}$$ by Lemma 3. Thus $$D \in \mathcal{R}$$, proving (ii).

(iii) We must show that $$C_{H-D}$$ is a Fubini map if $$C_{D}$$ is and if $$D \subseteq H, H \in \mathcal{C}.$$ Now, $$D \subseteq H$$ implies

$C_{H-D}=C_{H}-C_{D}.$

(Why?) Also, by Theorem 1, $$H \in \mathcal{C}$$ implies

$\int_{X \times Y} C_{H} d p=p H=s H<\infty,$

and $$C_{H}$$ is a Fubini map. So is $$C_{D}$$ by assumption. So also is

$C_{H-D}=C_{H}-C_{D}$

by Lemma 2(ii). Thus $$H-D \in \mathcal{R},$$ proving (iii).

By Lemma 4, then, $$\mathcal{P} \subseteq \mathcal{R}.$$ Hence $$(\forall D \in \mathcal{P}) C_{D}$$ is a Fubini map.$$\quad \square$$

We can now establish one of the main theorems, due to Fubini.

## Theorem $$\PageIndex{2}$$ (Fubini I)

Suppose $$f : X \times Y \rightarrow E^{*}$$ is $$\mathcal{P}$$-measurable on $$X \times Y$$ ($$\mathcal{P}$$ as above) rom. Then $$f$$ is a Fubini map if either

(i) $$f \geq 0$$ on $$X \times Y,$$ or

(ii) one of

$\int_{X \times Y}|f| dp, \int_{X} \int_{Y}|f| dn dm, o r \int_{Y} \int_{X}|f| dm dn$

is finite.

In both cases,

$\int_{X} \int_{Y} f dn dm=\int_{Y} \int_{X} f dm dn=\int_{X \times Y} f dp.$

Proof

First, let

$f=\sum_{i=1}^{\infty} a_{i} C_{D_{i}} \quad\left(a_{i} \geq 0, D_{i} \in \mathcal{P}\right),$

i.e., $$f$$ is $$\mathcal{P}$$-elementary, hence certainly $$p$$-measurable. (Why?) By Lemmas 5 and 2, each $$a_{i} C_{D_{i}}$$ is a Fubini map. Hence so is $$f$$ (Lemma 3). Formula (2) is simply Fubini property (c).

Now take any $$\mathcal{P}$$-measurable $$f \geq 0.$$ By Lemma 2 in §2,

$f=\lim _{k \rightarrow \infty} f_{k} \text { on } X \times Y$

for some sequence $$\left\{f_{k}\right\} \uparrow$$ of $$\mathcal{P}$$-elementary maps, $$f_{k} \geq 0.$$ As shown above, each $$f_{k}$$ is a Fubini map. Hence so is $$f$$ by Lemma 1. This settles case (i).

Next, assume (ii). As $$f$$ is $$\mathcal{P}$$-measurable, so are $$f^{+}, f_{-},$$ and $$|f|$$ (Theorem 2 in §2). As they are nonnegative, they are Fubini maps by case (i).

So is $$f=f^{+}-f^{-}$$ by Lemma 2(ii), since $$f^{+} \leq|f|$$ implies

$\int_{X \times Y} f^{+} d p<\infty$

by our assumption (ii). (The three integrals are equal, as $$|f|$$ is a Fubini map.)

Thus all is proved.$$\quad \square$$

III. We now want to replace $$\mathcal{P}$$ by $$\mathcal{P}^{*}$$ in Lemma 5 and Theorem 2. This works only under certain $$\sigma$$-finiteness conditions, as shown below.

## Lemma $$\PageIndex{6}$$

Let $$D \in \mathcal{P}^{*}$$ be $$\sigma$$-finite, i.e.,

$D=\bigcup_{i=1}^{\infty} D_{i} \text { (disjoint)}$

for some $$D_{i} \in \mathcal{P}^{*},$$ with $$pD_{i}<\infty$$ $$(i=1,2, \ldots).$$

Then there is $$a K \in \mathcal{P}$$ such that $$p(K-D)=0$$ and $$D \subseteq K$$.

Proof

As $$\mathcal{P}$$ is a $$\sigma$$-ring containing $$\mathcal{C},$$ it also contains $$\mathcal{C}_{\sigma}.$$ Thus by Theorem 3 of Chapter 7, §5, $$p^{*}$$ is $$\mathcal{P}$$-regular.

For the rest, proceed as in Theorems 1 and 2 in Chapter 7, §7.$$\quad \square$$

## Lemma $$\PageIndex{7}$$

If $$D \in \mathcal{P}^{*}$$ is $$\sigma$$-finite (Lemma 6), then $$C_{D}$$ is a Fubini map.

Proof

By Lemma 6,

$(\exists K \in \mathcal{P}) \quad p(K-D)=0, D \subseteq K.$

Let $$Q=K-D,$$ so $$p Q=0,$$ and $$C_{Q}=C_{K}-C_{D};$$ that is, $$C_{D}=C_{K}-C_{Q}$$ and

$\int_{X \times Y} C_{Q} d p=p Q=0.$

As $$K \in \mathcal{P}, C_{K}$$ is a Fubini map. Thus by Lemma 2(ii), all reduces to proving the same for $$C_{Q}.$$

Now, as $$p Q=0, Q$$ is certainly $$\sigma$$-finite; so by Lemma 6,

$(\exists Z \in \mathcal{P}) \quad Q \subseteq Z, p Z=p Q=0.$

Again $$C_{Z}$$ is a Fubini map; so

$\int_{X} \int_{Y} C_{Z} d n d m=\int_{X \times Y} C_{Z} d p=p Z=0.$

As $$Q \subseteq Z,$$ we have $$C_{Q} \leq C_{Z},$$ and so

\begin{aligned} \int_{X} \int_{Y} C_{Q} dn dm &=\int_{X}\left[\int_{Y} C_{Q}(x, \cdot) dn\right] dm \\ & \leq \int_{X}\left[\int_{Y} C_{Z}(x, \cdot) dn\right] dm=\int_{X \times Y} C_{Z} dp=0. \end{aligned}

Similarly,

$\int_{Y} \int_{X} C_{Q} dm dn=\int_{Y}\left[\int_{X} C_{Q}(\cdot, y) dm\right] dn=0.$

Thus setting

$g(x)=\int_{Y} C_{Q}(x, \cdot) dn \text { and } h(y)=\int_{X} C_{Q}(\cdot, y) dm,$

we have

$\int_{X} g dm=0=\int_{Y} h dn.$

Hence by Theorem 1(h) in §5, $$g=0$$ a.e. on $$X,$$ and $$h=0$$ a.e. on $$Y.$$ So $$g$$ and $$h$$ are "almost" measurable (Definition 2 of §3); i.e., $$C_{Q}$$ has Fubini property (a).

Similarly, one establishes (b), and (3) yields Fubini property (c), since

$\int_{X} \int_{Y} C_{Q} dn dm=\int_{Y} \int_{X} C_{Q} dm dn=\int_{X \times Y} C_{Q} dp=0,$

as required.$$\quad \square$$

## Theorem $$\PageIndex{3}$$ (Fubini II)

Suppose $$f : X \times Y \rightarrow E^{*}$$ is $$\mathcal{P}^{*}$$-measurable on $$X \times Y$$ and satisfies condition (i) or (ii) of Theorem 2.

Then $$f$$ is a Fubini map, provided $$f$$ has $$\sigma$$-finite support, i.e., $$f$$ vanishes outside some $$\sigma$$-finite set $$H \subseteq X \times Y$$.

Proof

First, let

$f=\sum_{i=1}^{\infty} a_{i} C_{D_{i}} \quad\left(a_{i}>0, D_{i} \in \mathcal{P}^{*}\right),$

with $$f=0$$ on $$-H$$ (as above).

As $$f=a_{i} \neq 0$$ on $$A_{i},$$ we must have $$D_{i} \subseteq H;$$ so all $$D_{i}$$ are $$\sigma$$-finite. (Why?) Thus by Lemma 7, each $$C_{D_{i}}$$ is a Fubini map, and so is $$f.$$ (Why?)

If $$f$$ is $$\mathcal{P}^{*}$$-measurable and nonnegative, and $$f=0$$ on $$-H,$$ we can proceed as in Theorem 2, making all $$f_{k}$$ vanish on $$-H.$$ Then the $$f_{k}$$ and $$f$$ are Fubini maps by what was shown above.

Finally, in case (ii), $$f=0$$ on $$-H$$ implies

$f^{+}=f^{-}=|f|=0 \text { on }-H.$

Thus $$f^{+}, f^{-},$$ and $$f$$ are Fubini maps by part (i) and the argument of Theorem 2.$$\quad \square$$

Note 1. The $$\sigma$$-finite support is automatic if $$f$$ is $$p$$-integrable (Corollary 1 in §5), or if $$p$$ or both $$m$$ and $$n$$ are $$\sigma$$-finite (see Problem 3). The condition is also redundant if $$f$$ is $$\mathcal{P}$$-measurable (Theorem 2; see also Problem 4).

Note 2. By induction, our definitions and Theorems 2 and 3 extend to any finite number $$q$$ of measure spaces

$\left(X_{i}, \mathcal{M}_{i}, m_{i}\right), \quad i=1, \ldots, q.$

One writes

$p=m_{1} \times m_{2}$

if $$q=2$$ and sets

$m_{1} \times m_{2} \times \cdots \times m_{q+1}=\left(m_{1} \times \cdots \times m_{q}\right) \times m_{q+1}.$

Theorems 2 and 3 with similar assumptions then state that the order of integrations is immaterial.

Note 3. Lebesgue measure in $$E^{q}$$ can be treated as the product of $$q$$ one- dimensional measures. Similarly for $$L S$$ product measures (but this method is less general than that described in Problems 9 and 10 of Chapter 7, §9).

IV. Theorems 2(ii) and 3(ii) hold also for functions

$f : X \times Y \rightarrow E^{n}\left(C^{n}\right)$

if Definitions 2 and 3 are modified as follows (so that they make sense for such maps): In Definition 2, set

$g(x)=\int_{Y} f_{x} dn$

if $$f_{x}$$ is $$n$$-integrable on $$Y,$$ and $$g(x)=0$$ otherwise. Similarly for $$h(y).$$ In Definition 3, replace "measurable" by "integrable."

For the proof of the theorems, apply Theorems 2(i) and 3(i) to $$|f|.$$ This yields

$\int_{Y} \int_{X}|f| dm dn=\int_{X} \int_{Y}|f| dn dm=\int_{X \times Y}|f| dp.$

Hence if one of these integrals is finite, $$f$$ is $$p$$-integrable on $$X \times Y,$$ and so are its $$q$$ components. The result then follows on noting that $$f$$ is a Fubini map (in the modified sense) iff its components are. (Verify!) See also Problem 12 below.

V. In conclusion, note that integrals of the form

$\int_{D} f dp \quad\left(D \in \mathcal{P}^{*}\right)$

reduce to

$\int_{X \times Y} f \cdot C_{D} dp.$

Thus it suffices to consider integrals over $$X \times Y$$.

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