product of power sets For a set $X$, let $\mathcal P(X)$ denote its power set and let $\mathcal P(X)\otimes\mathcal P(X)$ denote the product $\sigma$-algebra in $X^2$.  When $|X|\leq\aleph_0$ then $\mathcal P(X)\otimes\mathcal P(X)=\mathcal P(X^2)$ but when $|X|>2^{\aleph_0}$ this equality is known to fail.  What happens when $\aleph_0<|X|\leq 2^{\aleph_0}$?
 A: The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here.
It is consistent (assuming large cardinals) that there is an extension of Lebesgue measure defined on all sets of reals. Here, $X=\mathbb R$ and $\Sigma=\mathcal P(\mathbb R)$. Since $\nu$ extends Lebesgue measure, the space satisfies the assumptions of the result just stated, and $\mathcal P(\mathbb R)\otimes\mathcal P(\mathbb R)$ is not $\mathcal P(\mathbb R^2)$.

By the way, there is a recent article in the Monthly dealing precisely with this problem and discussing how $\mathsf{CH}$ implies that $\mathcal P(\mathbb R)\otimes\mathcal P(\mathbb R)$ is $\mathcal P(\mathbb R^2)$ while the existence of extensions of Lebesgue measure gives a negative answer:

MR3626256 Avilés, Antonio; Plebanek, Grzegorz. A little ado about rectangles. Amer. Math. Monthly 124 (2017), no. 4, 345–350. 

