# Infinite product measures

Jordan Bell
May 10, 2015

## 1 Introduction

The usual proof that the product of a collection of probability measures exists uses Fubini’s theorem. This is unsatisfying because one ought not need to use Fubini’s theorem to prove things having only to do with $\sigma$-algebras and measures. In this note I work through the proof given by Saeki of the existence of the product of a collection of probability measures.11 1 Sadahiro Saeki, A Proof of the Existence of Infinite Product Probability Measures, Amer. Math. Monthly 103 (1996), no. 8, 682–682. We speak only about the Lebesgue integral of characteristic functions.

## 2 Rings of sets and Hopf’s extension theorem

If $X$ is a set and $\mathscr{R}$ is a collection of subsets of $X$, we call $\mathscr{R}$ a ring of sets when (i) $\emptyset\in\mathscr{R}$ and (ii) if $A$ and $B$ belong to $\mathscr{R}$ then $A\cup B$ and $A\setminus B$ belong to $\mathscr{R}$. If $\mathscr{R}$ is a ring of sets and $A,B\in\mathscr{R}$, then $A\cap B=A\setminus(A\setminus B)\in\mathscr{R}$. Equivalently, one checks that a collection of subsets $\mathscr{R}$ of $X$ is a ring of sets if and only if (i) $\emptyset\in\mathscr{R}$ and (ii) if $A$ and $B$ belong to $\mathscr{R}$ then $A\triangle B$ and $A\cap B$ belong to $\mathscr{R}$, where $A\triangle B=(A\setminus B)\cup(B\setminus A)$ is the symmetric difference. One checks that indeed a ring of sets is a ring with addition $\triangle$ and multiplication $\cap$. If $\mathscr{S}$ is a nonempty collection of subsets of $X$, one proves that there is a unique ring of sets $\mathscr{R}(\mathscr{S})$ that (i) contains $\mathscr{S}$ and (ii) is contained in any ring of sets that contains $\mathscr{S}$. We call $\mathscr{R}(\mathscr{S})$ the ring of sets generated by $\mathscr{S}$.

If $\mathscr{A}$ is a ring of subsets of a set $X$, we call $\mathscr{A}$ an algebra of sets when $X\in\mathscr{A}$. Namely, an algebra of sets is a unital ring of sets. If $\mathscr{S}$ is a nonempty collection of subsets of $X$, one proves that there is a unique algebra of sets $\mathscr{A}(\mathscr{S})$ that (i) contains $\mathscr{S}$ and (ii) is contained in any algebra of sets that contains $\mathscr{S}$. We call $\mathscr{A}(\mathscr{S})$ the algebra of sets generated by $\mathscr{S}$.

For a nonempty collection $\mathscr{G}$ of subsets of a set $X$, we denote by $\sigma(\mathscr{G})$ the smallest $\sigma$-algebra of subsets of $X$ such that $\mathscr{G}\subset\sigma(\mathscr{G})$.

If $\mathscr{R}$ is a ring of subsets of a set $X$ and $\tau:\mathscr{R}\to[0,\infty]$ is a function such that (i) $\mu(\emptyset)=0$ and (ii) when $\{A_{n}\}$ is a countable subset of $\mathscr{R}$ whose members are pairwise disjoint and which satisfies $\bigcup_{n=1}^{\infty}A_{n}\in\mathscr{R}$, then

 $\tau\left(\bigcup_{n=1}^{\infty}A_{n}\right)=\sum_{n=1}^{\infty}\tau(A_{n}),$

we call $\tau$ a measure on $\mathscr{R}$. The following is Hopf’s extension theorem.22 2 Karl Stromberg, Probability for Analysts, p. 52, Theorem A3.6.

###### Theorem 1 (Hopf’s extension theorem).

Suppose that $X$ is a set, that $\mathscr{R}$ is a ring of subsets of $X$, and that $\tau$ is a measure on $\mathscr{R}$. If there is a countable subset $\{E_{n}\}$ of $\mathscr{R}$ with $\tau(E_{n})<\infty$ for each $n$ and such that $\bigcup_{n=1}^{\infty}E_{n}=X$, then there is a unique measure $\mu:\sigma(\mathscr{R})\to[0,\infty]$ whose restriction to $\mathscr{R}$ is equal to $\tau$.

## 3 Semirings of sets

If $X$ is a set and $\mathscr{S}$ is a collection of subsets of $X$, we call $\mathscr{S}$ a semiring of sets when (i) $\emptyset\in\mathscr{S}$, (ii) if $A$ and $B$ belong to $\mathscr{S}$ then $A\cap B\in\mathscr{S}$, and (iii) if $A$ and $B$ belong to $\mathscr{S}$ then there are pairwise disjoint $C_{1},\ldots,C_{n}\in\mathscr{S}$ such that

 $A\setminus B=\bigcup_{i=1}^{n}C_{i}.$

If $\mathscr{S}$ is a semiring of subsets of a set $X$, we call $\mathscr{S}$ a semialgebra of sets when $X\in\mathscr{S}$. One proves that if $\mathscr{S}$ is a semialgebra, then the collection $\mathscr{A}$ of all finite unions of elements of $\mathscr{S}$ is equal to the algebra generated by $\mathscr{S}$, and that each element of $\mathscr{A}$ is equal to a finite union of pairwise disjoint elements of $\mathscr{S}$.33 3 V. I. Bogachev, Measure Theory, volume I, p. 8, Lemma 1.2.14.

## 4 Cylinder sets

Suppose that $\{(\Omega_{i},\mathscr{F}_{i},P_{i}):i\in I\}$ is a nonempty collection of probability spaces and let

 $\Omega=\prod_{i\in I}\Omega_{i}.$

If $A_{i}\in\mathscr{F}_{i}$ for each $i\in I$ and $\{i\in I:A_{i}\neq\Omega_{i}\}$ is finite, we call

 $A=\prod_{i\in I}A_{i}$

a cylinder set. Let $\mathscr{C}$ be the collection of all cylinder sets. One checks that $\mathscr{C}$ is a semialgebra of sets.44 4 S. J. Taylor, Introduction to Measure and Integration, p. 136, §6.1, Lemma.

###### Lemma 2.

Suppose that $P:\mathscr{C}\to[0,1]$ is a function such that

 $\sum_{n=1}^{\infty}P(A_{n})=1$

whenever $A_{n}$ are pairwise disjoint elements of $\mathscr{C}$ whose union is equal to $\Omega$. Then there is a unique probability measure on $\sigma(\mathscr{C})$ whose restriction to $\mathscr{C}$ is equal to $P$.

###### Proof.

Let $\mathscr{A}$ be the collection of all finite unions of cylinder sets. Because $\mathscr{C}$ is a semialgebra of sets, $\mathscr{A}$ is the algebra of sets generated by $\mathscr{C}$, and any element of $\mathscr{A}$ is equal to a finite union of pairwise disjoint elements of $\mathscr{C}$. Let $A\in\mathscr{A}$. There are pairwise disjoint $B_{1},\ldots,B_{j}\in\mathscr{C}$ whose union is equal to $A$. Suppose also that $\{C_{i}\}$ is a countable subset of $\mathscr{C}$ with pairwise disjoint members whose union is equal to $A$. Moreover, as $\Omega\setminus A\in\mathscr{A}$ there are pairwise disjoint $W_{1},\ldots,W_{p}\in\mathscr{C}$ such that $\Omega\setminus A=\bigcup_{i=1}^{p}W_{i}$. On the one hand, $W_{1},\ldots,W_{p},B_{1},\ldots,B_{j}$ are pairwise disjoint cylinder sets with union $\Omega$, so

 $\sum_{i=1}^{j}P(B_{i})+\sum_{i=1}^{p}P(W_{i})=1.$

On the other hand, $W_{1},\ldots,W_{p},C_{1},C_{2},\ldots$ are pairwise disjoint cylinder sets with union $\Omega$, so

 $\sum_{i=1}^{\infty}P(C_{i})+\sum_{i=1}^{p}P(W_{i})=1.$

Hence,

 $\sum_{i=1}^{j}P(B_{i})=\sum_{i=1}^{\infty}P(C_{i});$

this conclusion does not involve $W_{1},\ldots,W_{p}$. Thus it makes sense to define $\tau(A)$ to be this common value, and this defines a function $\tau:\mathscr{A}\to[0,1]$. For $C\in\mathscr{C}$, $\tau(C)=P(C)$, i.e. the restriction of $\tau$ to $P$ is equal to $\mathscr{C}$.

If $\{A_{n}\}$ is a countable subset of $\mathscr{A}$ whose members are pairwise disjoint and $A=\bigcup_{n=1}^{\infty}A_{n}\in\mathscr{A}$, for each $n$ let $C_{n,1},\ldots,C_{n,j(n)}\in\mathscr{C}$ be pairwise disjoint cylinder sets with union $A_{n}$. Then

 $\{C_{n,i}:n\geq 1,1\leq i\leq j(n)\}$

is a countable subset of $\mathscr{C}$ whose elements are pairwise disjoint and with union $A$, so

 $\tau(A)=\sum_{n=1}^{\infty}\sum_{i=1}^{j(n)}P(C_{n,i}).$

But for each $n$,

 $\tau(A_{n})=\sum_{i=1}^{j(n)}P(C_{n,i}),$

so

 $\tau(A)=\sum_{n=1}^{\infty}\tau(A_{n}).$

This shows that $\tau:\mathscr{A}\to[0,1]$ is a measure. Then applying Hopf’s extension theorem, we get that there is a unique measure $\mu:\sigma(\mathscr{A})\to[0,1]$ whose restriction to $\mathscr{A}$ is equal to $\tau$. It is apparent that the $\sigma$-algebra generated by a semialgebra is equal to the $\sigma$-algebra generated by the algebra generated by the semialgebra, so $\sigma(\mathscr{A})=\sigma(\mathscr{C})$. Because the restriction of $\tau$ to $\mathscr{C}$ is equal to $P$, the restriction of $\mu$ to $\mathscr{C}$ is equal to $P$. Now suppose that $\nu:\sigma(\mathscr{A})\to[0,1]$ is a measure whose restriction to $\mathscr{C}$ is equal to $P$. For $A\in\mathscr{A}$, there are disjoint $C_{1},\ldots,C_{n}\in\mathscr{C}$ with $A=\bigcup_{i=1}^{n}C_{i}$. Then

 $\nu(A)=\sum_{i=1}^{n}\nu(C_{i})=\sum_{i=1}^{n}P(C_{i})=\sum_{i=1}^{n}\mu(C_{i}% )=\mu(A),$

showing that the restriction of $\nu$ to $\mathscr{A}$ is equal to the restriction of $\mu$ to $\mathscr{A}$, from which it follows that $\nu=\mu$. ∎

## 5 Product measures

Suppose that $\{(\Omega_{i},\mathscr{F}_{i},P_{i}):i\in I\}$ is a nonempty collection of probability spaces. The product $\sigma$-algebra is $\sigma(\mathscr{C})$, the $\sigma$-algebra generated by the cylinder sets. We define $P:\mathscr{C}\to[0,1]$ by

 $P(A)=\prod_{i\in I_{A}}P_{i}(A_{i})=\prod_{i\in I}P_{i}(A_{i}),$

for $A\in\mathscr{C}$ and with $I_{A}=\{i\in I:A_{i}\neq\Omega_{i}\}$, which is finite.

###### Lemma 3.

Suppose that $I$ is the set of positive integers. If $\{A_{n}\}$ is a countable subset of $\mathscr{C}$ with pairwise disjoint elements whose union is equal to $\Omega$, then

 $\sum_{n=1}^{\infty}P(A_{n})=1.$
###### Proof.

For each $k\geq 1$, there is some $i_{k}$ and $A_{k,1}\in\mathscr{F}_{1},\ldots,A_{k,i_{k}}\in\mathscr{F}_{i_{k}}$ such that

 $A_{k}=\prod_{i=1}^{\infty}A_{k,i},$

with $A_{k,i}=\Omega_{i}$ for $i>i_{k}$. Let $m\geq 1$, let $x=(x_{i})\in A_{m}$, and let $n\geq 1$. If $n=m$,

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i>i_{m}}P_{i% }(A_{n,i})\right)=1=\delta_{m,n}.$

If $m\neq n$ and $y_{i}\in\Omega_{i}$ for each $i>i_{m}$ and we set $y_{i}=x_{i}$ for $1\leq i\leq i_{m}$, then because $A_{m}$ and $A_{n}$ are disjoint and $y\in A_{m}$, we have $y\not\in A_{n}$ and therefore there is some $i$, $1\leq i\leq i_{n}$, such that $y_{i}\not\in A_{n,i}$. Thus

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i>i_{m}}\chi% _{A_{n,i}}(y_{i})\right)=\prod_{i=1}^{\infty}\chi_{A_{n,i}}(y_{i})=0.$ (1)

Either $i_{n}\leq i_{m}$ or $i_{n}>i_{m}$. In the case $i_{n}\leq i_{m}$ we have $A_{n,i}=\Omega_{i}$ for $i>i_{m}$ and thus

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i>i_{m}}\chi% _{A_{n,i}}(y_{i})\right)=\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i}),$

hence by (1),

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i>i_{m}}P_{i% }(A_{n,i})\right)=\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})=0=\delta_{m,n}.$

In the case $i_{n}>i_{m}$, we have $A_{n,i}=\Omega_{i}$ for $i>i_{n}$ and thus

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i>i_{m}}\chi% _{A_{n,i}}(y_{i})\right)=\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)% \left(\prod_{i=i_{m}+1}^{i_{n}}\chi_{A_{n,i}}(y_{i})\right),$

hence by (1) we have that for $y_{i}\in\Omega_{i}$, $i>i_{m}$,

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i=i_{m}+1}^{% i_{n}}\chi_{A_{n,i}}(y_{i})\right)=0.$

Therefore, integrating over $\Omega_{i}$ for $i=i_{m}+1,\ldots,i_{n}$,

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i=i_{m}+1}^{% i_{n}}P_{i}(A_{n,i})\right)=0,$

so

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i>i_{m}}P_{i% }(A_{n,i})\right)=0=\delta_{m,n}.$

We have thus established that for any $m\geq 1$, $x\in A_{m}$, and $n\geq 1$,

 $\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(\prod_{i>i_{m}}P_{i% }(A_{n,i})\right)=\delta_{m,n}.$ (2)

 $\sum_{n=1}^{\infty}P(A_{n})<1,$

i.e.

 $\sum_{n=1}^{\infty}\prod_{i=1}^{\infty}P_{i}(A_{n,i})<1.$ (3)

If

 $\sum_{n=1}^{\infty}\chi_{A_{n,1}}(x_{1})\prod_{i=2}^{\infty}P_{i}(A_{n,i})=1$

for all $x_{1}\in\Omega_{1}$, then integrating over $\Omega_{1}$ we contradict (3). Hence there is some $x_{1}\in\Omega_{1}$ such that

 $\sum_{n=1}^{\infty}\chi_{A_{n,1}}(x_{1})\prod_{i=2}^{\infty}P_{i}(A_{n,i})<1.$

Suppose by induction that for some $j\geq 1$, $x_{1}\in\Omega_{1},\ldots,x_{j}\in\Omega_{j}$ and

 $\sum_{n=1}^{\infty}\left(\prod_{i=1}^{j}\chi_{A_{n,i}}(x_{i})\right)\left(% \prod_{i=j+1}^{\infty}P_{i}(A_{n,i})\right)<1.$

If

 $\sum_{n=1}^{\infty}\left(\prod_{i=1}^{j+1}\chi_{A_{n,i}}(x_{i})\right)\left(% \prod_{i=j+2}^{\infty}P_{i}(A_{n,i})\right)=1$

for all $x_{j+1}\in\Omega_{j+1}$, then integrating over $\Omega_{j+1}$ we contradict (3). Hence there is some $x_{j+1}\in\Omega_{j+1}$ such that

 $\sum_{n=1}^{\infty}\left(\prod_{i=1}^{j+1}\chi_{A_{n,i}}(x_{i})\right)\left(% \prod_{i=j+2}^{\infty}P_{i}(A_{n,i})\right)<1.$

Therefore, by induction we obtain that for any $j$, there are $x_{1}\in\Omega_{1},\ldots,x_{j}\in\Omega_{j}$ such that

 $\sum_{n=1}^{\infty}\left(\prod_{i=1}^{j}\chi_{A_{n,i}}(x_{i})\right)\left(% \prod_{i=j+1}^{\infty}P_{i}(A_{n,i})\right)<1.$ (4)

Write $x=(x_{1},x_{2},\ldots)\in\Omega$. Because $\Omega=\bigcup_{m=1}^{\infty}A_{m}$, there is some $m$ for which $x\in A_{m}$. For $j=i_{m}$, (4) states

 $\sum_{n=1}^{\infty}\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(% \prod_{i>i_{m}}P_{i}(A_{n,i})\right)<1.$

But (2) tells us

 $\sum_{n=1}^{\infty}\left(\prod_{i=1}^{i_{m}}\chi_{A_{n,i}}(x_{i})\right)\left(% \prod_{i>i_{m}}P_{i}(A_{n,i})\right)=\sum_{n=1}^{\infty}\delta_{m,n}=1,$

 $\sum_{n=1}^{\infty}P(A_{n})=1,$

proving the claim. ∎

###### Lemma 4.

Suppose that $I$ is an uncountable set. If $\{A_{n}\}$ is a countable subset of $\mathscr{C}$ with pairwise disjoint elements whose union is equal to $\Omega$, then

 $\sum_{n=1}^{\infty}P(A_{n})=1.$
###### Proof.

For each $n$, there are $A_{n,i}\in\mathscr{F}_{i}$ with $A_{n,i}=\Omega_{i}$, and $I_{n}=\{i\in I:A_{i}\neq\Omega_{i}\}$ is finite. Then $J=\bigcup_{n=1}^{\infty}I_{n}$ is countable. Let $\Omega_{J}=\prod_{i\in J}\Omega_{i}$, let $\mathscr{C}_{J}$ be the collection of cylinder sets corresponding to the probability spaces $\{(\Omega_{i},\mathscr{F}_{i},P_{i}):i\in J\}$, and define $P_{J}:\mathscr{C}_{J}\to[0,1]$ by

 $P_{J}(B)=\prod_{i\in J_{B}}P_{i}(B_{i})=\prod_{i\in J}P_{i}(B_{i}),$

for $B\in\mathscr{C}_{J}$ and with $J_{B}=\{i\in J:B_{i}\neq\Omega_{i}\}$, which is finite. $P_{J}$ satisfies

 $P_{J}(B)=P\left(B\times\prod_{i\in I\setminus J}\Omega_{i}\right),\qquad B\in% \mathscr{C}_{J}.$

Let $B_{n}=\prod_{i\in J}A_{n,i}$, i.e. $A_{n}=B_{n}\times\prod_{i\in I\setminus J}A_{n,i}$. Then $\{B_{n}\}$ is a countable subset of $\mathscr{C}_{J}$ with pairwise disjoint elements whose union is equal to $\Omega_{J}$, and applying Lemma 3 we get that

 $\sum_{n=1}^{\infty}P_{J}(B_{n})=1,$

and therefore

 $\sum_{n=1}^{\infty}P(A_{n})=1.$

Now by Lemma 2 and the above lemma, there is a unique probability measure $\mu$ on $\sigma(\mathscr{C})$ whose restriction to $\mathscr{C}$ is equal to $P$. That is, when $\{(\Omega_{i},\mathscr{F}_{i},P_{i}):i\in I\}$ are probability spaces and $\mathscr{C}$ is the collection of cylinder sets corresponding to these probability spaces, with $\Omega=\prod_{i\in I}\Omega_{i}$ and $P:\mathscr{C}\to[0,1]$ defined by

 $P(A)=\prod_{i\in I}P(A_{i})$

for $A=\prod_{i\in I}A_{i}\in\mathscr{C}$, then there is a unique probability measure $\mu$ on the the product $\sigma$-algebra such that $\mu(A)=P(A)$ for each cylinder set $A$. We call $\mu$ the product measure, and write

 $\bigotimes_{i\in I}\mathscr{F}_{i}=\sigma(\mathscr{C})$

and

 $\prod_{i\in I}P_{i}=\mu.$