The Fubini-Tonelli theorem is valid when $(X,\mathcal{M},\mu)$ is an arbitrary measure space and $Y$ is a countable set $\mathcal{N}=\mathcal{P}(Y)$, and $\nu$ is counting measure on $Y$.
This is an exercise from Folland. I tried to prove this as in the proof of the original theorem, but as you can see below, the proof of the Tonelli theorem uses Theorem 2.36, which in turn requires the $\sigma-$finiteness of both measures. So how can I prove this? I would greatly appreciate any help.
Edit: I came up with a solution below but need it verified.
My Attempt
For convenience, let $Y=\mathbb{N}$, and let $Y$ be the union of $Y_n=\{1,\dots, n\}$. Let $f\in L^{+}(X \times Y)$. Let $f_n=f\chi_{X\times Y_n}$. Then clearly $f_n$ is an increasing sequence converging to $f$, so we can use the Monotone Convergence Theorem later. Now we have:
$\int f_n d(\mu \times \nu)=\sum_{y=1}^n\int_{X\times y} f d(\mu \times \nu)=\sum_{y=1}^n\int_X f_n (x,y) d\mu(x)$.
$g_n(x)=\int f{_n}(x,y)d\nu(y)=\sum_{y=1}^n f_n(x,y)$.
$h_n(y)=\int f_n(x,y)d\mu(x)$.
$\int g_n(x) d\mu(x)=\int \sum_{y=1}^n f_n(x,y)d\mu=\sum_{y=1}^n \int_X f_n(x,y) d\mu(x)$.
$\int h_n(y) d\nu(y)=\sum_{y=1}^n \int_X f_n(x,y) d\mu(x)$.
Also, $\lim_n g_n(x)=\lim_n \sum_{y=1}^n f_n(x,y)=\lim_n \sum_{y=1}^n f(x,y)\cdot \chi_{X\times Y_n}=\sum_{y=1}^\infty f(x,y)$.
and $\lim_n h_n(y)=\lim \int_X f_n(x,y) d\mu(x)=\int_X \lim f_n(x,y) d\mu(x)=\int_X f(x,y) d\mu(x)$ by the Monotone Convergence Theorem.
So the three integrals are equal, and now we can apply the Monotone Convergence Theorem and the rest of the proof is identical to the one given below.
Best Answer
There is only a need to prove the case where $f$ is a characteristic function. Following which, the rest of proof for theorem 2.37 applies.
To this end let $f=\chi_E$ where $E \subseteq X\times Y$. Writing $Y=\{1,2,3,\dots\}$, we may write $E$ as a disjoint union of countably many rectangles $E=\coprod\limits_{n\in \mathbb{Z}^+} (E_n \times\{n\})$. Going back to the definition of $\mu \times\nu$ as an outer measure that extends the pre-measure on the algebra of "finite disjoint rectangles", we see that
$$(\mu\times\nu)(E)=\sum\limits_{n=1}^{\infty}\mu(E_n)$$
That is,
$$\int \chi_E d(\mu\times\nu) = \sum\limits_{n=1}^{\infty} \left[\int \chi_{E_n} d\mu(x)\right] =\int \left[\int \chi_{E_n} d\mu(x)\right] d\nu(y)$$
We also know in general for $f_1,f_2,\dots\in L^{+}$ that $\int \sum f_n = \sum \int f_n$, so both of equations (2.38) are proven for $f$ being a characteristic function.