# The Unapologetic Mathematician

## Infinite Products, Part 1

Because we know that product spaces are product objects in the category of measurable spaces — at least for totally measurable spaces — we know that the product functor is monoidal. That is, we can define $n$-ary products unambiguously as iterated binary products. But things start to get more complicated as we pass to infinite products.

If $\{X_n\}_{n=1}^\infty$ is a countably infinite collection of sets, the product is the collection of all sequences $(x_1,x_2,\dots)$ with $x_n\in X_n$ for all $n$. If each $X_n$ is equipped with a $\sigma$-ring $\mathcal{S}_n$ and a measure $\mu_n$, it’s not immediately clear how we should equip the product space with a $\sigma$-ring and a measure. However, we can give meaning to these concepts if we specialize. First we shall insist that each $\mathcal{S}_n$ be a $\sigma$-algebra, and second we shall insist that $\mu_n$ be totally finite. Not just totally finite, though; we will normalize each measure so that $\mu_n(X_n)=1$.

This normalization is, incidentally, always possible for a totally finite measure space. Indeed, if $\mu$ is a totally finite measure on a space $X$, we can define $\displaystyle\hat{\mu}(E)=\frac{\mu(E)}{\mu(X)}$

It is easily verified that $\hat{\mu}$ is another totally finite measure, and $\hat{\mu}(X)=1$.

Now we will define a rectangle as a product of the form $\displaystyle\prod\limits_{n=1}^\infty A_n$

where $A_n\subseteq X_n$ for all $n$, and where $A_n=X_n$ for all but finitely many $n$. We define a measurable rectangle to be one for which each $A_n$ is measurable as a subset of $X_n$. We then define a subset of the countably infinite product $X$ to be measurable if it’s in the $\sigma$-algebra $\mathcal{S}$ generated by the measurable rectangles. This defines the product of a countably infinite number of measurable spaces.

Now if $J\subseteq\mathbb{Z}^+$ is any set of positive integers, we say that two sequences $(x_1,x_2,\dots)$ and $(y_1,y_2,\dots)$ agree on $J$ if $x_j=y_j$ for all $j\in J$. A set $E\subseteq X$ is called a $J$-cylinder if any two points which agree on $J$ are either both in or both out of $E$. That is, membership of a sequence $x=(x_n)$ in $E$ is determined only by the coordinates $x_j$ for $j\in J$.

We also define the sets $\displaystyle X^{(n)}=\prod\limits_{i=n+1}^\infty X_i$

so that we always have $X=(X_1\times\dots\times X_n)\times X^{(n)}$ Each $X^{(n)}$ is itself a countably infinite product space. For every set $E\subseteq X$ and each point $(x_1,\dots,x_n)\in X_1\times\dots\times X_n$, we define the subset $E(x_1,\dots,x_n)\subseteq X^{(n)}$ as the section of $E$ determined by $(x_1,\dots,x_n)$. It should be clear that every section of a measurable rectangle in $X$ is a measurable rectangle in $X^{(n)}$

Now if $J=\{1,\dots,n\}$, and if E\subseteq X\$ is a (measurable) $J$-cylinder, then $E=A\times X^{(n)}$, where $A$ is a (measurable) subset of $X_1\times\dots X_n$. Indeed, let $(\bar{x}_{n+1},\bar{x}_{n+2},\dots)$ be an arbitrary point of $X^{(n)}$, and let $A\subseteq X_1\times\dots\times X_n$ be the $X^{(n)}$-section of $E$ determined by this point. The sets $E$ (by assumption) and $A\times X^{(n)}$ (by construction) are both $J$-cylinders, so if $(x_1,x_2,\dots)$ belongs to either of them, then so does the point $(x_1,\dots,x_n,\bar{x}_{n+1},\bar{x}_{n+2},\dots)$.

It should now be clear that if such a point $(x_1,\dots,x_n,\bar{x}_{n+1},\bar{x}_{n+2},\dots)$ belongs to either $E$ or $A\times X^{(n)}$, then it belongs to the other as well. Again using the fact that both $E$ and $A\times X^{(n)}$ are $J$-cylinders, if $(x_1,\dots,x_n,\bar{x}_{n+1},\bar{x}_{n+2},\dots)$ belongs to either $E$ or $A\times X^{(n)}$ then so does the point $(x_1,\dots,x_n,x_{n+1},x_{n+2},\dots)$. We can conclude that $E$ and $A\times X^{(n)}$ consist of the same points. The parenthetical assertion on measurability follows from the fact that every section of a measurable set is measurable.

July 29, 2010 Posted by | Analysis, Measure Theory | 1 Comment