Tonelli's theorem holds for an arbitray $(Y,\mathcal{Y},\nu)$ measurable space in this case My question is the exercise 10.M) of Bartle's book.
I want to prove that Tonelli's theorem holds for an arbitrary $(Y,\mathcal{Y},\nu)$ measurable space if $(X,\mathcal{X},\mu)$ is the measurable space with $X=\Bbb{N},\mathcal{X}=\mathcal{P}(X)$ is the $\sigma-$algebra of all subsets of $X$, and $\mu$ is the counting measure in $\mathcal{X}.$
The Tonelli's theorem suppose that $(X,\mathcal{X},\mu),(Y,\mathcal{Y},\nu)$ are $\sigma-$ finite measurable spaces. The demonstration follows by:
(i) First proving the case where $F$ is the characteristic function of a measurable set,  
(ii) then proving the case where $F$ is a simple function by linearity, and, finally, 
(iii) proving the case where $F$ is a nonnegative measurable arbitrary function. It defines increasing sequences of simples functions converging to each integral $f(x)=\int_{Y}F_{x}d\nu,g(y)=\int_{X}F^{y}d\mu.$ Then, applies the Monotone Convergence Theorem to prove that $\int_{X}fd\mu=\int_{Y}gd\nu=\int_{Z}Fd\pi,$ where $Z=X\times Y$ and $\pi$ is the product measure (unique by $\sigma-$finiteness).
In my question, $(X,\mathcal{X},\mu)$ is a $\sigma-$finite space, but $(Y,\mathcal{Y},\nu)$ doesn't needs to be it. I found two problems:
Proving that (i)$\implies$(ii) is easy, since the assumption of $\sigma-$finiteness is not used. But I have two problems:
1) How can I prove (i) if $(Y,\mathcal{Y},\nu)$ is not a $\sigma-$finite in this case? Is there something about $(\Bbb{N},\mathcal{P}(\Bbb{N}),\mu)$ that make it true?
2) How can I deal with the problem that $\pi$ is not the unique product measure in $Z=X\times Y$? 
 A: For your first question:
We have that
$$
\nu \times \mu (A):=\int \chi _A \,\mathrm d (\nu ,\mu )\tag1
$$
Now note that $A=\bigcup_{n\in \Bbb N }\{n\}\times [A]_n$, where $[A]_n:=\{y\in Y:(n,y)\in A \}$ and that $(\{n\}\times [A]_n)\cap (\{m\}\times [A]_m)=\emptyset $ if $n\neq m$, therefore
$$
\nu \times \mu (A)=\sum_{n\in \Bbb N }\nu \times \mu (\{n\}\times [A]_n)
=\sum_{n\in \Bbb N }\int \chi _{\{n\}\times [A]_n}\,\mathrm d (\nu,\mu )\tag2
$$
Now from the definition of product measure we knows that
$$
\nu\times \mu (\{n\}\times [A]_n)=\mu (\{n\})\cdot \nu ([A]_n)=\nu([A]_n)\tag3
$$
Thus
$$
\nu \times \mu (\{n\}\times [A]_n)=\iint \chi _{\{n\}\times [A]_n}\,\mathrm d \nu \,\mathrm d \mu =\iint \chi _{\{n\}\times [A]_n}\,\mathrm d \mu \,\mathrm d \nu\tag4
$$
where we used the fact that $\chi _{\{n\}\times [A]_n}=\chi _{\{n\}}\cdot \chi _{[A]_n}$. Then by the monotone convergence theorem we find that
$$
\sum_{n\in \Bbb N }\iint \chi _{\{n\}\times [A]_n}\,\mathrm d \mu \,\mathrm d \nu=\iint \sum_{n\in \Bbb N }\chi _{\{n\}\times [A]_n}\,\mathrm d \mu \,\mathrm d \nu=\iint \chi_A \,\mathrm d \mu \,\mathrm d \nu\tag5
$$
that in view's of $\rm(2)$ and $\rm(4)$ shows that Tonelli's theorem holds.

For your second question:
In view of the above it seems that when $X$ is countable and the $\sigma$-algebra on $X$ is it power set then the product measure is unique.
