Infinite Products, Part 2
After all of yesterday’s definitions, we continue examining the countably infinite product of a sequence of totally finite measure spaces such that each .
If and are positive integers with , it may happen that a non-empty subset is both a -cylinder and a -cylinder. By yesterday’s result, we find that we can write both and for some subsets and . But we can rewrite the first of these equations as , and thus we conclude that .
If is measurable, then both and are measurable. We calculate
using the normalization of all the measures . It follows that we can define a set function unambiguously on each measurable -cylinder by the equation
Since every measurable rectangle is the product of a sequence with all but finitely many , we can choose a large enough so that for all . Then is a -cylinder, and is defined on all measurable rectangles. It’s straightforward to see that is finite, non-negative, and finitely additive where it’s defined.
We define the analogue of in by . If is defined on a cylinder then will be defined on each section , and we find that
So now, if is a sequence of totally finite measure spaces with for all , then there exists a unique measure defined on the countably infinite product of the measure spaces so that for every measurable of the form we have
This measure is called the product of the measures .
From what we know about continuity, we just have to show that is continuous from above at to show that it’s a measure. That is, if is a decreasing sequence of cylinders on which is defined so that for all , then the countable intersection of the is non-empty.
For each we define the set
We then find that
and so . Therefore, .
Now is a decreasing sequence of measurable subsets of , which is bounded in measure away from zero. And so by the continuity of the measure , we conclude that there is at least one point in their intersection. That is, for all .
But now everything we’ve said about is true as well for . We can thus replace the sequence with the sequence , and the bound with the bound . We then find a point so that for all . We can repeat this process to find a sequence with , such that
for all .
I say that this sequence belongs to all the . Indeed, given any of them, choose the so that is a -cylinder. We know that , and so there must be at least one point with for all from to . But then, since is a -cylinder, it must contain the point .
Thus, as asserted, the intersection of the is nonempty, and so is continuous from above at , and is thus a measure.