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 \(X \times Y\).