If $F\subset X\times Y$ is a measurable, is it true that $\chi_F$, seen as a function from $X$ to $L^\infty(Y)$ is measurable?

Question

Let $(X,\mathcal F_{X})$ and $(X,\mathcal F_{Y})$ measurable spaces and $F\in \mathcal F_{X}\otimes\mathcal F_{Y}$. Let $L^\infty(Y)$ be the space of bounded measurable functions from $(Y,\mathcal{F}_Y)$ to $\mathbb{R}$, equipped with the sup metric, and let $\mathcal{F}$ its Borel $\sigma$-algebra.

Consider the function $$f :X\to L^\infty(Y), x\mapsto(y\mapsto\chi_F(x,y))$$ where $\chi_F:X\times Y\to\mathbb{R}$ is the indicator function of $F$.

Is it true that $f$ is measurable from $(X,\mathcal{F}_X)$ to $(L^\infty(Y),\mathcal{F})$?

Answer 1

Yes. Here are some hints: If $F=F_1 \times F_2$ with $F_1$ measurable in $X$ and $F_2$ measurable in $Y$ then it is easy to prove that the result is true. [The inverse image of any open set belongs to $\{\emptyset, F_1,F_1^{c}, X\}$]. Now consider the class of all measurable sets $F$ in $X \times Y$ such that $\chi_F$ is measurable. It is trivial to see the this class is closed under complementation. It is closed under countable intersections: $\chi_{\cap F_n} =\prod_n \chi_{F_n}$. Hence this class is a sigma algebra. This completes the proof.