8.8: Product Measures. Iterated Integrals
- Page ID
- 22219
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.
(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.
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.
(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
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 \(s : \mathcal{C} \rightarrow E^{1}.\)
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 \(\mathcal{C} \subseteq \mathcal{P} \subseteq \mathcal{P}^{*}\)).
\(\mathcal{P}\) 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.
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.
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.
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.\(\quad \square\)
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.\(\quad \square\)
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 \(X \times Y\).