Infinite product measures

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

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

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

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

$$\tau \left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\tau (A_n),$$

we call $\tau$ a measure on $\mathrm{R}$. The following is Hopf’s extension theorem.

Theorem 1 (Hopf’s extension theorem).

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

3 Semirings of sets

If $X$ is a set and $𝒮$ is a collection of subsets of $X$, we call $𝒮$ a semiring of sets when (i) $\mathrm{\varnothing }\in 𝒮$, (ii) if $A$ and $B$ belong to $𝒮$ then $A\cap B\in 𝒮$, and (iii) if $A$ and $B$ belong to $𝒮$ then there are pairwise disjoint $C_1,\mathrm{\dots },C_n\in 𝒮$ such that

$$A\setminus B=\bigcup _{i=1}^n{C}_i.$$

If $𝒮$ is a semiring of subsets of a set $X$, we call $𝒮$ a semialgebra of sets when $X\in 𝒮$. One proves that if $𝒮$ is a semialgebra, then the collection $𝒜$ of all finite unions of elements of $𝒮$ is equal to the algebra generated by $𝒮$, and that each element of $𝒜$ is equal to a finite union of pairwise disjoint elements of $𝒮$.

4 Cylinder sets

Suppose that $\{({\mathrm{\Omega }}_i,{ℱ}_i,P_i):i\in I\}$ is a nonempty collection of probability spaces and let

$$\mathrm{\Omega }=\prod _{i\in I}{\mathrm{\Omega }}_i.$$

If $A_i\in {ℱ}_i$ for each $i\in I$ and $\{i\in I:A_i\ne {\mathrm{\Omega }}_i\}$ is finite, we call

$$A=\prod _{i\in I}{A}_i$$

a cylinder set. Let $𝒞$ be the collection of all cylinder sets. One checks that $𝒞$ is a semialgebra of sets.

Lemma 2.

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

$$\sum _{n=1}^{\mathrm{\infty }}P(A_n)=1$$

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

Proof.

Let $𝒜$ be the collection of all finite unions of cylinder sets. Because $𝒞$ is a semialgebra of sets, $𝒜$ is the algebra of sets generated by $𝒞$, and any element of $𝒜$ is equal to a finite union of pairwise disjoint elements of $𝒞$. Let $A\in 𝒜$. There are pairwise disjoint $B_1,\mathrm{\dots },B_j\in 𝒞$ whose union is equal to $A$. Suppose also that $\{C_i\}$ is a countable subset of $𝒞$ with pairwise disjoint members whose union is equal to $A$. Moreover, as $\mathrm{\Omega }\setminus A\in 𝒜$ there are pairwise disjoint $W_1,\mathrm{\dots },W_p\in 𝒞$ such that $\mathrm{\Omega }\setminus A={\bigcup }_{i=1}^p{W}_i$. On the one hand, $W_1,\mathrm{\dots },W_p,B_1,\mathrm{\dots },B_j$ are pairwise disjoint cylinder sets with union $\mathrm{\Omega }$, so

$$\sum _{i=1}^{j}P(B_i)+\sum _{i=1}^{p}P(W_i)=1.$$

On the other hand, $W_1,\mathrm{\dots },W_p,C_1,C_2,\mathrm{\dots }$ are pairwise disjoint cylinder sets with union $\mathrm{\Omega }$, so

$$\sum _{i=1}^{\mathrm{\infty }}P(C_i)+\sum _{i=1}^{p}P(W_i)=1.$$

Hence,

$$\sum _{i=1}^{j}P(B_i)=\sum _{i=1}^{\mathrm{\infty }}P(C_i);$$

this conclusion does not involve $W_1,\mathrm{\dots },W_p$. Thus it makes sense to define $\tau (A)$ to be this common value, and this defines a function $\tau :𝒜\to [0,1]$. For $C\in 𝒞$, $\tau (C)=P(C)$, i.e. the restriction of $\tau$ to $P$ is equal to $𝒞$.

If $\{A_n\}$ is a countable subset of $𝒜$ whose members are pairwise disjoint and $A={\bigcup }_{n=1}^{\mathrm{\infty }}{A}_n\in 𝒜$, for each $n$ let $C_{n,1},\mathrm{\dots },C_{n,j(n)}\in 𝒞$ be pairwise disjoint cylinder sets with union $A_n$. Then

$$\{C_{n,i}:n\ge 1,1\le i\le j(n)\}$$

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

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

But for each $n$,

$$\tau (A_n)=\sum _{i=1}^{j(n)}P(C_{n,i}),$$

so

$$\tau (A)=\sum _{n=1}^{\mathrm{\infty }}\tau (A_n).$$

This shows that $\tau :𝒜\to [0,1]$ is a measure. Then applying Hopf’s extension theorem, we get that there is a unique measure $\mu :\sigma (𝒜)\to [0,1]$ whose restriction to $𝒜$ 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 $\sigma (𝒜)=\sigma (𝒞)$. Because the restriction of $\tau$ to $𝒞$ is equal to $P$, the restriction of $\mu$ to $𝒞$ is equal to $P$. Now suppose that $\nu :\sigma (𝒜)\to [0,1]$ is a measure whose restriction to $𝒞$ is equal to $P$. For $A\in 𝒜$, there are disjoint $C_1,\mathrm{\dots },C_n\in 𝒞$ 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 $𝒜$ is equal to the restriction of $\mu$ to $𝒜$, from which it follows that $\nu =\mu$. ∎

5 Product measures

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

$$P(A)=\prod _{i\in I_A}{P}_i(A_i)=\prod _{i\in I}{P}_i(A_i),$$

for $A\in 𝒞$ and with $I_A=\{i\in I:A_i\ne {\mathrm{\Omega }}_i\}$, which is finite.

Lemma 3.

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

$$\sum _{n=1}^{\mathrm{\infty }}P(A_n)=1.$$

Proof.

For each $k\ge 1$, there is some $i_k$ and $A_{k,1}\in {ℱ}_1,\mathrm{\dots },A_{k,i_k}\in {ℱ}_{i_k}$ such that

$$A_k=\prod _{i=1}^{\mathrm{\infty }}{A}_{k,i},$$

with $A_{k,i}={\mathrm{\Omega }}_i$ for $i>i_k$. Let $m\ge 1$, let $x=(x_i)\in A_m$, and let $n\ge 1$. 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 $m\ne n$ and $y_i\in {\mathrm{\Omega }}_i$ for each $i>i_m$ and we set $y_i=x_i$ for $1\le i\le i_m$, then because $A_m$ and $A_n$ are disjoint and $y\in A_m$, we have $y\notin A_n$ and therefore there is some $i$, $1\le i\le i_n$, such that $y_i\notin A_{n,i}$. Thus

$$\left(\prod _{i=1}^{i_m}{\chi }_{A_{n,i}}(x_i)\right)\left(\prod _{i>i_m}{\chi }_{A_{n,i}}(y_i)\right)=\prod _{i=1}^{\mathrm{\infty }}{\chi }_{A_{n,i}}(y_i)=0.$$ (1)

Either $i_n\le i_m$ or $i_n>i_m$. In the case $i_n\le i_m$ we have $A_{n,i}={\mathrm{\Omega }}_i$ 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 $A_{n,i}={\mathrm{\Omega }}_i$ 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 $y_i\in {\mathrm{\Omega }}_i$, $i>i_m$,

$$\left(\prod _{i=1}^{i_m}{\chi }_{A_{n,i}}(x_i)\right)\left(\prod _{i=i_m+1}^{i_n}{\chi }_{A_{n,i}}(y_i)\right)=0.$$

Therefore, integrating over ${\mathrm{\Omega }}_i$ for $i=i_m+1,\mathrm{\dots },i_n$,

$$\left(\prod _{i=1}^{i_m}{\chi }_{A_{n,i}}(x_i)\right)\left(\prod _{i=i_m+1}^{i_n}{P}_i(A_{n,i})\right)=0,$$

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 $m\ge 1$, $x\in A_m$, and $n\ge 1$,

$$\left(\prod _{i=1}^{i_m}{\chi }_{A_{n,i}}(x_i)\right)\left(\prod _{i>i_m}{P}_i(A_{n,i})\right)={\delta }_{m,n}.$$ (2)

Suppose by contradiction that

i.e.

(3)

If

$$\sum _{n=1}^{\mathrm{\infty }}{\chi }_{A_{n,1}}(x_1)\prod _{i=2}^{\mathrm{\infty }}{P}_i(A_{n,i})=1$$

for all $x_1\in {\mathrm{\Omega }}_1$, then integrating over ${\mathrm{\Omega }}_1$ we contradict (3). Hence there is some $x_1\in {\mathrm{\Omega }}_1$ such that

Suppose by induction that for some $j\ge 1$, $x_1\in {\mathrm{\Omega }}_1,\mathrm{\dots },x_j\in {\mathrm{\Omega }}_j$ and

If

$$\sum _{n=1}^{\mathrm{\infty }}\left(\prod _{i=1}^{j+1}{\chi }_{A_{n,i}}(x_i)\right)\left(\prod _{i=j+2}^{\mathrm{\infty }}{P}_i(A_{n,i})\right)=1$$

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

Therefore, by induction we obtain that for any $j$, there are $x_1\in {\mathrm{\Omega }}_1,\mathrm{\dots },x_j\in {\mathrm{\Omega }}_j$ such that

(4)

Write $x=(x_1,x_2,\mathrm{\dots })\in \mathrm{\Omega }$. Because $\mathrm{\Omega }={\bigcup }_{m=1}^{\mathrm{\infty }}{A}_m$, there is some $m$ for which $x\in A_m$. For $j=i_m$, (4) states

But (2) tells us

$$\sum _{n=1}^{\mathrm{\infty }}\left(\prod _{i=1}^{i_m}{\chi }_{A_{n,i}}(x_i)\right)\left(\prod _{i>i_m}{P}_i(A_{n,i})\right)=\sum _{n=1}^{\mathrm{\infty }}{\delta }_{m,n}=1,$$

a contradiction. Therefore,

$$\sum _{n=1}^{\mathrm{\infty }}P(A_n)=1,$$

proving the claim. ∎

Lemma 4.

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

$$\sum _{n=1}^{\mathrm{\infty }}P(A_n)=1.$$

Proof.

For each $n$, there are $A_{n,i}\in {ℱ}_i$ with $A_{n,i}={\mathrm{\Omega }}_i$, and $I_n=\{i\in I:A_i\ne {\mathrm{\Omega }}_i\}$ is finite. Then $J={\bigcup }_{n=1}^{\mathrm{\infty }}{I}_n$ is countable. Let ${\mathrm{\Omega }}_J={\prod }_{i\in J}{\mathrm{\Omega }}_i$, let ${𝒞}_J$ be the collection of cylinder sets corresponding to the probability spaces $\{({\mathrm{\Omega }}_i,{ℱ}_i,P_i):i\in J\}$, and define $P_J:{𝒞}_J\to [0,1]$ by

$$P_J(B)=\prod _{i\in J_B}{P}_i(B_i)=\prod _{i\in J}{P}_i(B_i),$$

for $B\in {𝒞}_J$ and with $J_B=\{i\in J:B_i\ne {\mathrm{\Omega }}_i\}$, which is finite. $P_J$ satisfies

$$P_J(B)=P\left(B\times \prod _{i\in I\setminus J}{\mathrm{\Omega }}_i\right),B\in {𝒞}_J.$$

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 ${𝒞}_J$ with pairwise disjoint elements whose union is equal to ${\mathrm{\Omega }}_J$, and applying Lemma 3 we get that

$$\sum _{n=1}^{\mathrm{\infty }}{P}_J(B_n)=1,$$

and therefore

$$\sum _{n=1}^{\mathrm{\infty }}P(A_n)=1.$$

∎

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

$$P(A)=\prod _{i\in I}P(A_i)$$

for $A={\prod }_{i\in I}{A}_i\in 𝒞$, 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

$$\underset{i\in I}{\otimes }{ℱ}_i=\sigma (𝒞)$$

and

$$\prod _{i\in I}{P}_i=\mu .$$