# The Unapologetic Mathematician

## Infinite Products, Part 2

After all of yesterday’s definitions, we continue examining the countably infinite product of a sequence $\{(X_n,\mathcal{S}_n,\mu_n)\}$ of totally finite measure spaces such that each $\mu_n(X_n)=1$.

If $m$ and $n$ are positive integers with $m, it may happen that a non-empty subset $E\subseteq X$ is both a $\{1,\dots,m\}$-cylinder and a $\{1,\dots,n\}$-cylinder. By yesterday’s result, we find that we can write both $E=A\times X^{(m)}$ and $E=B\times X^{(n)}$ for some subsets $A\subseteq X_1\times\dots\times X_m$ and $B\subseteq X_1\times\dots\times X_n$. But we can rewrite the first of these equations as $E=(A\times X_{m+1}\times\dots\times X_n)\times X^{(n)}$, and thus we conclude that $B=A\times X_{m+1}\times\dots\times X_n$.

If $E$ is measurable, then both $A$ and $B$ are measurable. We calculate

\displaystyle\begin{aligned}\left[\mu_1\times\dots\times\mu_n\right](B)&=\left[\mu_1\times\dots\times\mu_n\right](A\times X_{m+1}\times\dots\times X_n)\\&=\left[\mu_1\times\dots\times\mu_m\right](A)\mu_{m+1}(X_{m+1})\dots\mu_n(X_n)\\&=\left[\mu_1\times\dots\times\mu_m\right](A)\end{aligned}

using the normalization of all the measures $\mu_i(X_i)=1$. It follows that we can define a set function $\mu$ unambiguously on each measurable $\{1,\dots,n\}$-cylinder $A\times X^{(n)}$ by the equation

$\displaystyle\mu\left(A\times X^{(n)}\right)=\left[\mu_1\times\dots\times\mu_n\right](A)$

Since every measurable rectangle $E$ is the product of a sequence $\{A_i\}$ with all but finitely many $A_i=X_i$, we can choose a large enough $n$ so that $A_i=X_i$ for all $i>n$. Then $E$ is a $\{1,\dots,n\}$-cylinder, and $\mu$ is defined on all measurable rectangles. It’s straightforward to see that $\mu$ is finite, non-negative, and finitely additive where it’s defined.

We define the analogue of $\mu$ in $X^{(n)}$ by $\mu^{(n)}$. If $\mu$ is defined on a cylinder $E$ then $\mu^{(n)}$ will be defined on each section $E(x_1,\dots,x_n)$, and we find that

$\displaystyle\mu(E)=\int\cdots\int\mu^{(n)}\left(E(x_1,\dots,x_n\right)\,d\mu_1(x_1)\cdots\,d\mu_n(x_n)$

So now, if $\{(X_n,\mathcal{S}_n,\mu_n)\}$ is a sequence of totally finite measure spaces with $\mu_n(X_n)=1$ for all $n$, then there exists a unique measure $\mu$ defined on the countably infinite product of the measure spaces $\{(X_n,\mathcal{S}_n)\}$ so that for every measurable $E\subseteq X$ of the form $A\times X^{(n)}$ we have

$\displaystyle\mu(E)=\left[\mu_1\times\dots\times\mu_n\right](A)$

This measure $\mu$ is called the product of the measures $\{\mu_n\}$.

From what we know about continuity, we just have to show that $\mu$ is continuous from above at $\emptyset$ to show that it’s a measure. That is, if $\{E_n\}$ is a decreasing sequence of cylinders on which $\mu$ is defined so that $0<\epsilon\leq\mu(E_j)$ for all $j$, then the countable intersection of the $E_j$ is non-empty.

For each $j$ we define the set

$\displaystyle F_j=\left\{x_1\in X_1\bigg\vert\mu^{(1)}(E_j(x_1))>\frac{\epsilon}{2}\right\}$

We then find that

\displaystyle\begin{aligned}\mu(E_j)&=\int\mu^{(1)}(E_j(x_1))\,d\mu_1(x_1)\\&=\int\limits_{F_j}\mu^{(1)}(E_j(x_1))\,d\mu_1(x_1)+\int\limits_{F_j^c}\mu^{(1)}(E_j(x_1))\,d\mu_1(x_1)\end{aligned}

and so $\mu(E_j)\leq\mu(F_j)+\frac{\epsilon}{2}$. Therefore, $\mu_1(F_j)\geq\frac{\epsilon}{2}$.

Now $F_j$ is a decreasing sequence of measurable subsets of $X_1$, which is bounded in measure away from zero. And so by the continuity of the measure $\mu_1$, we conclude that there is at least one point $\bar{x}_1$ in their intersection. That is, $\mu^{(1)}(E_j(\bar{x}_1))\geq\frac{\epsilon}{2}$ for all $j$.

But now everything we’ve said about $X$ is true as well for $X^{(1)}$. We can thus replace the sequence $\{E_j\}$ with the sequence $\{E_j(\bar{x}_1)\}$, and the bound $\epsilon$ with the bound $\frac{\epsilon}{2}$. We then find a point $\bar{x}_2\in X_2$ so that $\mu^{(2)}(E_j(\bar{x}_1,\bar{x}_2))\geq\frac{\epsilon}{4}$ for all $j$. We can repeat this process to find a sequence $(\bar{x}_1,\bar{x}_2,\dots)$ with $\bar{x}_n\in X_n$, such that

$\displaystyle\mu^{(n)}(E_j(\bar{x}_1,\dots,\bar{x}_n))\geq\frac{\epsilon}{2^n}$

for all $j$.

I say that this sequence belongs to all the $E_j$. Indeed, given any of them, choose the $n$ so that $E_j$ is a $\{1,\dots,n\}$-cylinder. We know that $\mu^{(n)}(E_j(\bar{x}_1,\dots,\bar{x}_n))>0$, and so there must be at least one point $(x_1,x_2,\dots)\in E_j$ with $x_i=\bar{x}_i$ for all $i$ from $1$ to $n$. But then, since $E_j$ is a $\{1,\dots,n\}$-cylinder, it must contain the point $(\bar{x}_1,\bar{x}_2,\dots)$.

Thus, as asserted, the intersection of the $E_j$ is nonempty, and so $\mu$ is continuous from above at $\emptyset$, and is thus a measure.