In general, my problem can be formulated as follows: Let $X$ be a random variable with value in $\mathbb R^2$, and let $G:\mathbb R^2 \times \mathbb R\rightarrow \mathbb R$ be a function which is continuous in the first argument and measurable in the second(i.e., a Caratheodory function). Assume the partial maximization $x\mapsto \sup\_yG(x,y)$ is measurable. I am considering the values given by


\begin{equation}
V\_1=\sup\{\mathbb E(G(X,y(X))| y:\mathbb R^2\rightarrow \mathbb R \: measurable\} \\
V\_2=\mathbb E(\sup\_{y\in\mathbb R}G(X,y))
\end{equation}
I would like to know if these two values are the same or not, assuming problem $V\_1$ admits a maximizer. 




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In my original problem, the function $G$ is given by $G(x,y)=(x\_1-f(y))(y-x\_2)$, where $f:\mathbb R \rightarrow \mathbb R$ is a Borel measurable function. It seems that continuity of $G$ with respect to $y$ is quite crucial for the argmax correspondence $\Phi(x)=\{y:G(x,y)=sup\_yG(x,y)\}$ to have a measurable selector. Unfortunately, this is something I do not have.
I would pretty much like to have a positive answer(i.e.,$V\_1=V\_2$), but a counter-example will be equally appreciated! Thanks!



measurable selection and values of optimization problem
Undery our assumptions, the argmax-correspondence need not even be nonempty-valued. But one can find almost-selections of an almost-argmax-correspondence, and that is enough.


Let $\epsilon>0$. Define the function $h$ by $h(x)=\sup\_yG(x,y)$. By assumption, $h$ is measurable. Therefore, the set 
$$B=\big\{(x,y)\mid G(x,y)>h(x)-\epsilon\big\}$$
is a Borel set such that no $x$-section is empty. By the von-Neumann selection theorem, there exist an analytically measurable selection $f$ of the crrespondece with graph $B$. Now an analytically measurable function is universally measurable, so $f$ is measurable with respect to the completion of the distribution of $X$. But then there must be a Borel measurable function $f'$ such that $f$ and $f'$ agree on a set of measure $1$ under the distribution of $X$.


