Product $\sigma$-algebra of power sets Let $X,Y$ be any two sets. In general, what is the product $\sigma$-algebra $\mathcal{P}(X) \times \mathcal{P}(Y)$? In the context (using Fubini's Theorem to prove that one can reverse the order of summation for absolutely convergent double series) the author makes an implicit assumption that, at least in the case $X=Y=\mathbb{N}$, we have $\mathcal{P}(X) \times \mathcal{P}(Y)=\mathcal{P}(X \times Y)$. But I do not see why is this true or in what cases it holds. Here it seems that every subset of $\mathbb{N}^2$ must somehow be a union (possibly with complements) of Cartesian products of subsets of $\mathbb{N}$.
 A: It's about $\sigma$-algebras.
Take any $R\subset \mathcal P(\mathbb N\times\mathbb N)$. Define 
$$
R^x=\{y\in\mathbb N:\ (x,y)\in R\}.
$$
Then 
$$
R=\bigcup_{x\in\mathbb N} \{x\}\times R^x.
$$
As each of the sets $\{x\}\times R^x\in\mathcal P(\mathbb N)\times \mathcal P(\mathbb N)$, their (countable) union is in the $\sigma$-algebra they generate. Thus the $\sigma$-algebra generated by $ \mathcal P(\mathbb N)\times \mathcal P(\mathbb N)$ is $ \mathcal P(\mathbb N\times\mathbb N)$.
A: You cannot show that $\mathcal{P}(X \times Y) = \mathcal{P}(X) \times \mathcal{P}(Y)$, and this is not what your author claims.
For countable $X$ and $Y$, the sigma-algebra generated by all sets $A \times B$, where $A \in \mathcal{P}(X)$ and $B \in \mathcal{P}(Y)$ contains all singletons $\{(x,y)\} = \{x\} \times \{y\}$ and thus all subsets of the countable set $X \times Y$, so the generated sigma-algebra for the product is indeed $\mathcal{P}(X \times Y)$. 
This no longer holds for uncountable sets though. In that case it's a classic fact that the diagonal $D= \{(x,x): x \in X\}$ is not in the generated sigma-algebra by $\mathcal{P}(X) \times \mathcal{P}(X)$ when $|X| > \mathfrak{c}$ (see Nedoma's pathology), so we generate a proper sub-sigma-algebra of $\mathcal{P}(X \times Y)$. For $|X| =\aleph_1$ we do get the desired result. (due to V. Rao.) Daniel Schepler (in the comments below) shows that this also extends to sets of size $\mathfrak{c}$, so we have a division of uncountable sizes here: small uncountable (where we do get $\mathbb{P}(X \times Y) = \mathcal{P}(X) \otimes \mathcal{P}(Y)$) and large uncountable where this fails to be the case. The countable case is straightforward, as we saw.
