@MattN By definition the product $\sigma$-algebra $\mathcal{E} \otimes \mathcal{F}$ is the $\sigma$-algebra generated by the measurable rectangles $E \times F$, that is it is characterized by being the smallest $\sigma$-algebra containing $E \times F$. Now for a measurable rectangle, the $\sigma$-algebra $\mathcal{A} = \bigcap_{x \in E} \{C \subset E \times F\,:\,C(x) \in \mathcal{F}\}$ certainly contains the measurable rectangles, hence $\mathcal{A} \supset \mathcal{E \otimes F}$.

Now by definition $\mathcal{H}$ consists of maps such that $f_C : x \mapsto Q(C(x))$ is measurable and that's the only thing used in Davide's answer.

A few subtleties: It is very rare that the product measure on $\mathcal{E} \otimes \mathcal{F}$ is complete, so one usually works with the completion and this is where things get hairy. It's no longer true that all slices of a set in the completion are measurable (think of a null-set times a non-measurable set, which is a null-set, hence in the completion) but part of Fubini's theorem states that almost all slices are measurable.

Hi @tb

Hi Kannappan

@Matt: Another subtlety is that if $X$ and $Y$ are topological spaces with Borel $\sigma$-algebras $\mathcal{B}(X)$ and $\mathcal{B}(Y)$ then it is not true in general that $\mathcal{B}(X) \otimes \mathcal{B}(Y)$ coincides with $\mathcal{B}(X \times Y)$ when $X \times Y$ carries the product topology. See here for a positive result ($X$ second countable is enough) and here for counterexamples.