I am studying out of Folland and I think I understand the proof of Theorem 2.36, but I am not sure why we can't just look at disjoint unions of rectangles.
Here is the Theorem:
Let $(X,\mathcal{M}, \mu)$, $(Y,\mathcal{N}, v)$ be $\sigma$-finite measure spaces. Then for any $E\in\mathcal{M}\bigotimes\mathcal{N}$, $x\to v(E_x)$ and $y\to \mu(E^y)$ are measurable (on $X$ and $Y$ respectively) and
$$\mu\times v (E)=\int \mu(E^y)\ dv=\int v(E_x)\ dx.$$
It is obvious for rectangles ($\chi_{E_x}=\chi_{A}v(B)$ if $E=A\times B$) and for finite disjoint unions of rectangles, so why can we take an countable disjoint collection of rectangles (Theorem 2.15)? Wouldn't this just be taking the limit of a finite disjoint union of rectangles?
If so, then the claim follows. So why use the Monotone Class Lemma? I know I have to be missing something.
Not every element of $\mathcal{M}\otimes\mathcal{N}$ is a countable union of rectangles. So it does not suffice to just prove the theorem when $E$ is a countable union of rectangles.
For instance, if $\mathcal{M}$ and $\mathcal{N}$ are the Borel algebra on $\mathbb{R}$, note that every open subset of $\mathbb{R}^2$ can be written as a countable union of open rectangles, and so every open set is in $\mathcal{M}\otimes\mathcal{N}$. But then so is every closed set (since the complement of any element of $\mathcal{M}\otimes\mathcal{N}$ is in $\mathcal{M}\otimes\mathcal{N}$), including sets like $\{(x,x):x\in\mathbb{R}\}$ which cannot be written as a countable union of rectangles. (As an exercise, try and prove for yourself that this set cannot be written as a countable union of rectangles!)