One simple thing that can go wrong is purely related to the size of the space (polish spaces are all size $\leq 2^{\aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology: Let $X$ be a measure space with $|X| > 2^{\aleph_0}$. The diagonal in $X^2$ is not measurable.
We'll prove this by way of a theorem:
Let $U \subseteq X^2$ be measurable. $X$ can be written as a union of at most $2^{\aleph_0}$ spaces of the form $A \times B$.
Proof: We can find some countable $A_i$ such that $U \subseteq \sigma(A_i \times A_j)$.
For $x \in \{0, 1\}^\mathbb{N}$ define $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i \}$.
Consider all sets which can be written as a (possibly uncountable) union of $B_x \times B_y$ for some $y$. This is a sigma algebra and obviously contains all the $A_i \times A_j$, so contains $A$.
But now we're done. There are at most $2^{\aleph_0}$ of the $B_x$, and each is certainly measurable in $X$, so $U$ can be written as a union of $2^{\aleph_0}$ sets of the form $A \times B$.
QED
Corollary: The diagonal is not measurable.
Evidently the diagonal cannot be written as a union of at most $2^{\aleph_0}$ rectangles, as they would all have to be single points, and the diagonal has size $|X| > 2^{\aleph_0}$.
