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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.


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!

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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 $X$.

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  • $\begingroup$ Thanks! Could you provide me with some reference on von-Neumann selection theorem? $\endgroup$
    – Ryan
    Oct 23, 2017 at 0:51
  • $\begingroup$ @Ryan You can find it in the books "A Course on Borel Sets" by Srivastava and "Classical Descriptive Set Theory" by Kechris. The result is sometimes referred to as a "uniformization theorem" instead of a "selection theorem". $\endgroup$ Oct 23, 2017 at 6:25

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