8.8: Product Measures. Iterated Integrals

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 $a, b \geq 0$.

(ii) If, in addition,

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

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 $X \times Y$.