Arbitrary Products of Measure Spaces

Identification Lemma: Let $\left\{ D_{p}\right\}_{p \in P}$ denote a disjoint family of nonempty subsets of $I$ with $\bigcup_{p \in P} D_{p}=I$. Define $\Phi : \Omega \rightarrow \underset {p \in P}{\times } \Omega_{D_{p}}$ for every $\omega =(\omega_{i})_{i \in I} \in \Omega$ by the rule $\Phi (\omega )=\left((\omega_{i})_{i \in D_{p}}\right)_{p \in P}$. Then: (a) $\Phi$ is one-to-one and onto $\underset {p \in P}{\times } \Omega_{D_{p}}$. Let $\mathcal{F}^{*}$ denote the $\sigma$-field of subsets of $\underset {p \in P}{\times } \Omega_{D_{p}}$ generated by sets of the form $\underset {p \in P}{\times } A_{D_{p}}$, where $A_{D_{p}} \in \mathcal{F}_{D_{p}}$ for each $p \in P$ and $A_{D_{p}} \neq \Omega_{D_{p}}$ for at most finitely many $p \in P$. (b) For every set in $\mathcal{F}$ there is a unique corresponding set in $\mathcal{F}^{*}$, and vice versa; i.e., $\Phi$ establishes a one-to-one correspondence between $\mathcal{F}$ and $\mathcal{F}^{*}$.

Identification Lemma: Let $\left\{ D_{p}\right\}_{p \in P}$ denote a disjoint family of nonempty subsets of $I$ with $\bigcup_{p \in P} D_{p}=I$. Define $\Phi : \Omega \rightarrow \underset {p \in P}{\times } \Omega_{D_{p}}$ for every $\omega =(\omega_{i})_{i \in I} \in \Omega$ by the rule $\Phi (\omega )=\left((\omega_{i})_{i \in D_{p}}\right)_{p \in P}$. Then: (a) $\Phi$ is one-to-one and onto $\underset {p \in P}{\times } \Omega_{D_{p}}$. Let $\mathcal{F}^{*}$ denote the $\sigma$-field of subsets of $\underset {p \in P}{\times } \Omega_{D_{p}}$ generated by sets of the form $\underset {p \in P}{\times } A_{D_{p}}$, where $A_{D_{p}} \in \mathcal{F}_{D_{p}}$ for each $p \in P$ and $A_{D_{p}} \neq \Omega_{D_{p}}$ for at most finitely many $p \in P$. (b) For every set in $\mathcal{F}$ there is a unique corresponding set in $\mathcal{F}^{*}$, and vice versa; i.e., $\Phi$ establishes a one-to-one correspondence between $\mathcal{F}$ and $\mathcal{F}^{*}$.

Claim 4: Define the collection $\mathscr {H}=\left\{ B \subseteq \Omega : B=\Phi_{F}^{-1}\left(A_{F} \times \Omega_{F^{c}}\right) \text{ for some finite } F \subseteq I \text{ and } A_{F} \in \mathcal{F}_{F}\right\}$. Then: (a) $\mathscr {H}$ is a field on $\Omega$ with $\mathscr {G} \subseteq \mathscr {H} \subseteq \mathcal{F}$, where $\mathscr {G}$ is the field on $\Omega$ generated by the sets $\underset {i \in I}{\times } A_{i}$ with $A_{i} \in \mathcal{F}_{i}$ for each $i \in I$ and $A_{i} \neq \Omega_{i}$ for at most finitely many $i \in I$; (b) $\sigma (\mathscr {H})=\mathcal{F}$; (c) there exists a unique finitely additive measure $\mu$ on the field $\mathscr {H}$ such that $\mu \left(\Phi_{F}^{-1}\left(A_{F} \times \Omega_{F^{c}}\right)\right)=\mu_{F}(A_{F})$ for every finite $F \subseteq I$ and $A_{F} \in \mathcal{F}_{F}$. Subclaim 1 (used in proving (a)): Let $A_{F_{1}} \in \mathcal{F}_{F_{1}}$. There exists $C_{F_{1} \cup F_{2}} \in \mathcal{F}_{F_{1} \cup F_{2}}$ with the property that $\Phi_{F_{1}}^{-1}\left(A_{F_{1}} \times \Omega_{F_{1}^{c}}\right)=\Phi_{F_{1} \cup F_{2}}^{-1}\left(C_{F_{1} \cup F_{2}} \times \Omega_{\left(F_{1} \cup F_{2}\right)^{c}}\right)$. Subclaim 2 (used in showing that $\mu$ is well defined, given finite $F_{1}, F_{2} \subseteq I$ and $F_{3}=F_{1} \cup F_{2}$): Let $A_{F_{1}} \in \mathcal{F}_{F_{1}}$. Then $\Phi_{F_{1}}^{-1}\left(A_{F_{1}} \times \Omega_{F_{1}^{c}}\right)=\Phi_{F_{3}}^{-1}\left[\Phi_{F_{1}, F_{3}}^{-1}\left(A_{F_{1}} \times \Omega_{F_{3} \backslash F_{1}}\right) \times \Omega_{F_{3}^{c}}\right]$.

Claim 7: Let $\left\{ (\Omega_{i}, \mathcal{F}_{i}, \mu_{i})\right\}_{i=1}^{\infty }$ denote a sequence of measure spaces with $\mu_{i}(\Omega_{i})=1$ for each $i \in \mathbb {N}$. There exists a unique measure $\mu$ on $\mathcal{F}$ such that $\mu \left(\Phi_{F}^{-1}\left(A_{F} \times \Omega_{F^{c}}\right)\right)=\mu_{F}(A_{F})$ whenever $F \subseteq \mathbb {N}$ is finite and $A_{F} \in \mathcal{F}_{F}$. In the course of showing that the finitely additive measure $\mu$ from Claim 4 is in fact countably additive on $\mathscr {H}$, one fixes a nonincreasing sequence $\left\{ E_{m}\right\}$ of $\mathscr {H}$-sets with $\bigcap_{m} E_{m}=\varnothing$, writes $E_{m}=\Phi_{\mathbb {N}_{m}}^{-1}\left(A_{\mathbb {N}_{m}} \times \Omega_{\mathbb {N}_{m}^{c}}\right)$, and, assuming toward a contradiction that $\mu (E_{m})$ does not tend to $0$, constructs a point $\xi =(\xi_{1}, \xi_{2}, \ldots ) \in \Omega$. Subclaim 3: It cannot be the case that for some $n \in \mathbb {N}$ and some large $m>n$ we have $\; \mathbb {1}_{A_{\mathbb {N}_{m}}}\left(\xi_{1}, \ldots , \xi_{n}, \omega_{n+1}, \ldots , \omega_{m}\right)=0$ for each $\omega_{n+1} \in \Omega_{n+1}, \ldots , \omega_{m} \in \Omega_{m}$.

Note that Claim 4 of this section does not require the individual component spaces to have measure one, and hence we may construct a measure $\mu_{F}$ as given in the definition following Claim 4 for this slightly more general situation. Claim 4: Define the collection $\mathscr {H}=\left\{ B \subseteq \Omega : B=\Phi_{F}^{-1}\left(A_{F} \times \Omega_{F^{c}}\right) \text{ for some finite } F \subseteq I \text{ and } A_{F} \in \mathcal{F}_{F}\right\}$. Then: (a) $\mathscr {H}$ is a field on $\Omega$ with $\mathscr {G} \subseteq \mathscr {H} \subseteq \mathcal{F}$, where $\mathscr {G}$ is the field on $\Omega$ generated by the sets $\underset {i \in I}{\times } A_{i}$ with $A_{i} \in \mathcal{F}_{i}$ for each $i \in I$ and $A_{i} \neq \Omega_{i}$ for at most finitely many $i \in I$; (b) $\sigma (\mathscr {H})=\mathcal{F}$; (c) there exists a unique finitely additive measure $\mu$ on the field $\mathscr {H}$ such that $\mu \left(\Phi_{F}^{-1}\left(A_{F} \times \Omega_{F^{c}}\right)\right)=\mu_{F}(A_{F})$ for every finite $F \subseteq I$ and $A_{F} \in \mathcal{F}_{F}$.

Write $E$ as the limit of the sequence $\left\{ A_{1} \times \cdots \times A_{n} \times \Omega_{n+1} \times \Omega_{n+2} \times \cdots \right\}_{n=1}^{\infty }$. This sequence is in $\mathcal{F}$ and is decreasing to $\underset {i=1}{\overset {\infty }{\times }} A_{i}$. Use the fact that measures are continuous from above and invoke the product property of $\mu$.

Claim 7: Let $\left\{ (\Omega_{i}, \mathcal{F}_{i}, \mu_{i})\right\}_{i=1}^{\infty }$ denote a sequence of measure spaces with $\mu_{i}(\Omega_{i})=1$ for each $i \in \mathbb {N}$. There exists a unique measure $\mu$ on $\mathcal{F}$ such that $\mu \left(\Phi_{F}^{-1}\left(A_{F} \times \Omega_{F^{c}}\right)\right)=\mu_{F}(A_{F})$ whenever $F \subseteq \mathbb {N}$ is finite and $A_{F} \in \mathcal{F}_{F}$.

Framework: Let $(\Omega_{i}, \mathcal{F}_{i}, \mu_{i})$ denote a measure space with $\mu_{i}(\Omega_{i})=1$ for each $i$ in an index set $I$, taken to be $I=\mathbb {N}$ in this section. Let $\mathbb {N}_{m}=\left\{ 1, \cdots , m\right\}$ for each $m \in \mathbb {N}$. Let $f \in L^{1}(\Omega , \mathcal{F}, \mu )$, where $\mathcal{F}$ and $\mu$ are the product $\sigma$-field and measure discussed and constructed in the previous two sections of this chapter. For each $\omega \in \Omega$ and $n \in \mathbb {N}$, write

[Note for each $\omega \in \Omega$ that $f_{n}(\omega )$ does not depend on $\omega_{n+1}, \omega_{n+2}, \cdots$.] For each $\omega \in \Omega$ and $n \in \mathbb {N}$, we write

[Note for each $\omega \in \Omega$ that $f_{n}^{\prime }(\omega )$ does not depend on $\omega_{1}, \cdots , \omega_{n}$.]

Claim 1: $\lim_{n} f_{n}(\omega )=f(\omega )$ for $\mu$-a.e. $\omega \in \Omega$, where $f_{n}$ is as defined in the Framework of this section (see Exercise 11.3.1).

Kolmogorov's Zero-One Law: Let $\omega$ and $\xi$ denote elements of $\Omega$. We say that $\omega$ and $\xi$ are ultimately equal, written $\omega =_{u} \xi$, if and only if there is $N$ such that $\omega_{n}=\xi_{n}$ for all $n \geq N$. Now let $U \in \mathcal{F}$ have the following property: for every $\omega \in \Omega$, $\omega \in U$ if and only if every $\xi \in \Omega$ with $\xi =_{u} \omega$ is in $U$. Then $\mu (U) \in \left\{ 0,1\right\}$.

Claim 4: $\lim_{n} f_{n}^{\prime }(\omega )$ exists and equals the number $\int_{\Omega } f \, d\mu$ for $\mu$-a.e. $\omega \in \Omega$, where $f_{n}^{\prime }$ is as defined in the Framework of this section (see Exercise 11.3.1).

For part (b): $\mathcal{F}^{(n)}$ denotes the $\sigma$-field $\left\{ A_{\mathbb {N}_{n}} \times \Omega_{\mathbb {N}_{n}^{c}}: A_{\mathbb {N}_{n}} \in \mathcal{F}_{\mathbb {N}_{n}}\right\}$ constructed in the proof of Claim 1 of this section, and $\mathcal{F}_{0}=\bigcap_{n=1}^{\infty } \mathcal{F}^{(n)}$ denotes the tail $\sigma$-field of $\Omega$ -- the $\sigma$-field on which Kolmogorov's Zero-One Law is applied.

Use either the Zero-One Law or Claim 4. Claim 4: $\lim_{n} f_{n}^{\prime }(\omega )$ exists and equals the number $\int_{\Omega } f \, d\mu$ for $\mu$-a.e. $\omega \in \Omega$, where $f_{n}^{\prime }$ is as defined in the Framework of this section (see Exercise 11.3.1).

Since $\mu (\Omega )=1$, we have $L^{p}(\Omega , \mathcal{F}, \mu ) \subseteq L^{1}(\Omega , \mathcal{F}, \mu )$ by Claim 1 of Section 8.5, so $f \in L^{1}(\Omega , \mathcal{F}, \mu )$. Using the Fubini Theorem, the mapping

is in $L^{1}(\Omega_{1}, \mathcal{F}_{1}, \mu_{1})$. This function is $\mathcal{F}_{1}/\mathcal{B}^{*}$-measurable, and since the mapping

does not depend on $\omega_{2}$, it is $\mathcal{F}/\mathcal{B}^{*}$-measurable. Next, verify that

(The second inequality follows by Claim 1 of Section 8.5.) This shows that $Sf$ is in $L^{p}(\Omega , \mathcal{F}, \mu )$ and $\left\| Sf\right\|_{p} \leq \left\| f\right\|_{p}$.

Framework: Let $(\Omega_{i}, \mathcal{F}_{i}, \mu_{i})$ denote an infinite collection of measure spaces with $\mu_{i}(\Omega_{i})=1$ for each $i \in I$. For every finite subset $F \subseteq I$, we use the identification conventions of Claim 9 of Section 11.2 to identify the infinite product space $(\Omega , \mathcal{F}, \mu )$ with $(\Omega_{F} \times \Omega_{F^{c}}, \mathcal{F}_{F} \otimes \mathcal{F}_{F^{c}}, \mu_{F} \otimes \mu_{F^{c}})$. We take $1 \leq p<+\infty$ and $f \in L^{p}(\Omega , \mathcal{F}, \mu )$. We define $f_{F^{c}}$ on $\Omega$ by writing

Therefore, $f_{F^{c}}$ is a function of the form $Sf$ as given in Exercise 11.3.8 (with $F$ playing the role of the index set there identified with $\Omega_{1}$, and $F^{c}$ the role identified with $\Omega_{2}$). We next define $f_{F}$ on $\Omega$ by writing

Claim 3: Let $F \subseteq \mathbb {N}$ be nonempty and finite, and for each $i \in F$ let $f_{i}$ denote a function in $L^{1}(\Omega_{i}, \mathcal{F}_{i}, \mu_{i})$. Let $g$ denote the mapping on $\Omega$ given for every $\omega \in \Omega$ by the rule $\omega \mapsto \prod_{i \in F} f_{i}(\omega_{i})$. Then

$L^{2}$ Strong Law of Large Numbers: For each $k \in \mathbb {N}$, let $g_{k}$ denote a function in $L^{2}(\Omega_{k}, \mathcal{F}_{k}, \mu_{k})$ such that $\int_{\Omega_{k}} g_{k} \, d\mu_{k}=0$. Let $f_{k}$ denote the function on $\Omega$ that assigns each $\omega \in \Omega$ to $g_{k}(\omega_{k})$. Assume that $\sum_{k=1}^{\infty } \frac{\left\| g_{k}\right\|_{2}^{2}}{k^{2}}$ converges. Then $\lim_{n} \frac{1}{n} \sum_{k=1}^{n} f_{k}(\omega )=0$ for $\mu$-a.e. $\omega \in \Omega$.