Let $(X,S,\mu)$ and $(Y,T,\nu)$ be two $\sigma$-finite measure spaces. Define $\phi_{A}(x)=\nu(A_{x})$ for $A_{x}=\{y \in Y: (x,y) \in A\}$ and $A \in S \otimes T$ (the $\sigma$-field generated by the measurable rectangles), and similarly $\psi_{A}(y)=\mu(A^{y})$. Show that:
$\displaystyle\int \phi_{A} d\mu=\displaystyle\int \psi_{A} d\nu$
I have show this exercise for both $(X,S,\mu)$ and $(Y,T,\nu)$ being finite measure spaces, but I'm not sure on how to extend this result to $\sigma$-finite spaces. I was thinking that since we can find an increasing sequence of measurable sets of finite measure such that $X = \cup_{i=1}^{\infty}X_{i}$ (similarly for $Y$), then we can set $\mu_{n}(A)=\mu(A \cap X_{n})$ and similarly $\nu_{n}(A)=\nu(A \cap Y_{n})$ and then apply the result for finite measure spaces, but one would need to show:
$\displaystyle\int \phi_{A} d\mu=\lim_{n \to \infty}\int \nu(A_{x} \cap Y_{n}) d\mu_{n}=\lim_{n \to \infty}\int \mu(A^{y} \cap X_{n}) d\nu_{n}=\displaystyle\int \psi_{A} d\nu$
Would this last equality necessarily hold? If it does, how would one be able to show it? I'm not sure on how to proceed.
Thanks for the help.
Your idea is correct.
Note: You're using $A$ both for the set you want to prove it for (in product sigma algebra), as well as a set in either component $\sigma$-algebras when defining $\mu_n,\nu_n$.
Therefore that just set the records straight, we define the finite measures
$$\mu_n (B) = \mu(B \cap X_n),~\nu_n(C) = \nu (B \cap Y_n),$$
for $B\in {\cal S},C\in{\cal T}$.
Now $\mu_n\times nu_n$ is a finite measure space. For $A$ in the product sigma algebra, you have shown that
$$ \int \nu_n (A_x) d \mu_n = \int \mu_n (A^y) d \nu_n .$$
Note however, that the lefthand side is equal to
$$ \int_{X_n}{\bf 1}_{X_n}(x) \nu ( A_x \cap Y_n) d \mu,$$
and similarly, the righthand side is equal to
$$ \int {\bf 1}_{Y_n} (y) \mu (A^y \cap X_n) d \nu.$$
All is left is to apply the monotone convegence theorem to both sides.