Let $(X=\Bbb{N},\mathcal{X}=\mathcal{P}(\Bbb{N}),\mu)$ a measurable function, where $\mu$ is the counting measure. Let $(Y,\mathcal{Y},\nu)$ an arbitrary measurable space.
First, I already proved that $E\in\mathcal{Z}=\mathcal{X}\times\mathcal{Y}\iff E_{n}\in\mathcal{Y}$ for all $n\in\Bbb{N},$ where $E_{n}=\{(y\in Y:(n,y)\in E\}$ is the $n$-section of $E$.
I also proved that, if $E\in\mathcal{Z}$, so the unique product measure is defined by $$\pi(E)=\sum_{n\in\Bbb{N}}\nu(E_{n}). $$
Also, I already proved that a function $f:Z=X\times Y\rightarrow \Bbb{R}$ is measurable $\iff$ each $f_{n}$ is $\mathcal{Y}$-measurable.
Now, I want to prove that $f$ is integrable with respect to $\pi \iff\sum_{n\in\Bbb{N}}\int_{Y}|f_{n}|d\nu$ converges, and in this case
$$\int_{Z}fd\pi=\sum_{n\in\Bbb{N}}\left[\int_{Y}f_{n}d\nu\right]=\int_{Y}\left[\sum_{n\in\Bbb{N}}f_{n}\right]d\nu $$
My problem is that, for me, this result is so "trivial", that I don't know how to prove it from the basic definitions. I can't use Tonelli's Theorem or Fubini's Theorem, because $(Y,\mathcal{Y},\nu)$ needs to be $\sigma$-finite. What can I do?