Supremum of a product measurable function... This is an interesting question that has stumped the entirety of my measure theory class, including the professor:
Prove or disprove:
Let $(X,\mathcal A)$, $(Y,\mathcal B)$ be measure spaces.
Let $f$ be an $\mathcal A$ $\times$ $\mathcal B$ measurable nonnegative function, and let $g(x)$ $=$ $sup_{y \in Y}f(x,y)$, with $g(x)<\infty$ for all $x$. Is $g(x)$ necessarily an $\mathcal A$-measurable function?
We all feel the answer is no, given that slices are measurable, and sups of measurable functions are only guaranteed to be measurable over a countable index. We think the correct answer is to start with a nonmeasurable set $S$ in $X$, and to try to build a set $T$ in $Y$ that makes $S \times T$ measurable in $\mathcal A \times \mathcal B$, but we suspect this is quite difficult with no further guidance. 
Any ideas?
 A: Here is an example in which the resulting function $g$ is nonmeasurable, but only slightly so:
We let $X=Y=[0,1]$, endowed with the Borel $\sigma$-algebra. We let $A\subseteq [0,1]^2$ be a Borel set such that the projection $\pi_X(A)$ to the first coordinate is not Borel (and hence only analytic) and $f=1_A$ be the indicator function of $A$. Then $g$ is the indicator function of $\pi_X(A)$ and hence not Borel measurable. However, $g$ will be measurable with respect to every complete probability measure on $X$. 
This problem is actually a typical example for why one uses analytic sets in optimization problems. For a nice introduction, see Some Measurability Results for Extrema of Random Functions over Random Sets by Stinchcombe and White. 
A: The following is Corollary 2.13 from Crauel, Random Probability Measures on Polish Spaces:
Suppose that $f\colon X\times\Omega\to \mathbb{R}$ is product measurable, where $(\Omega,\mathcal{F})$ is a measurable space and $X$ is a Polish space equipped with its Borel $\sigma$-algebra. Let $C\colon\Omega\to 2^X$, $\omega\mapsto C(\omega)$, be any set valued mapping such that $\textrm{graph}(C)$ is product measurable, then $\omega\mapsto\sup_{x\in C(\omega)}f(x,\omega)$ is measurable with respect to the universal completion of $\mathcal{F}$.
In particular, one may then take $C(\omega)\equiv X$.
