After considering the product of measurable spaces, let’s consider what happens if our spaces are actually equipped with measures. That is, let and be -finite measure spaces, and consider the product space .
Now, if is any measurable subset of , then we can define two functions — and — by the formulæ and , where and are the –section determined by and -section determined by , respectively. I say that both and are non-negative measurable functions, and that
where we take the first integral over and the second over .
Let be the collection of subsets of for which the assertions we’ve made are true. It’s straightforward to see that is closed under countable disjoint unions. Indeed, if is a sequence of disjoint sets for which the assertions hold — call the functions and respectively — then if is the disjoint union of the we can calculate
Since each of the are measurable, so is , and is similarly measurable. The equality of the integrals should be clear.
Now, since and are -finite, every measurable subset of can be covered by a countable disjoint union of measurable rectangles, each side of each of which has finite measure. Thus we just have to verify that the result holds for such measurable rectangles with finite-measure sides, and it will hold for all measurable subsets of , and that is a monotone class. That is a monotone class follows from the dominated convergence theorem and the monotone convergence theorem, and so we have only to show that the assertions hold for measurable rectangles with finite-measure sides.
Okay, so let be a measurable rectangle with and . We can simplify our functions to write
These are clearly measurable, and we find that
so the assertions hold for measurable rectangles with finite-measure sides, and thus for all measurable subsets of .