The Unapologetic Mathematician

Mathematics for the interested outsider

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<n, 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.

July 30, 2010 Posted by | Analysis, Measure Theory | Leave a comment

   

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