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Could someone help me with the following problem?

Let $S$ be a compact metric space, $\mu$ be a Borel measure on a Souslin space, $f:S\times X\to\mathbb{R}$ be a function continuous in the first argument, and Borel measurable in the second argument. Show that there is a Borel measurable function $F:X\to S$ such that $F(x)$ is a point of maximum of $f(x,s)$ over $s\in S$.

I was given two hints:

  1. We are looking for the function $F$ such $f(F(x),x)=\max\limits_{s\in S}f(x,s)$
  2. We may use the fact that a Borel measurable mapping between Souslin spaces has a Borel measurable right inverse.

This question might seems as duplicate of the question Measurability of supremum over measurable set, but the answer looks like an overkill. I'm looking for a more down to Earth solution.

Norbert
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  • Do you really mean the function is $F:X\rightarrow S$, instead of $F:S\rightarrow X$? – Michael Nov 30 '21 at 06:26
  • @Michael, thank you! This is a typo. Now fixed. – Norbert Nov 30 '21 at 14:17
  • I wonder if the axiom of choice is needed here, or if you can use the metric to uniquely break ties when there are many $s \in S$ that maximize $f(x,s)$ for a given $x$. For example you could try to break ties by picking a point $v \in S$ and choosing the maximizer that is closest to $v$, but you are not guaranteed to have a unique closest one. – Michael Nov 30 '21 at 19:12

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