The Unapologetic Mathematician

Mathematics for the interested outsider

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


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



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