Measurability of supremum over measurable set

Consider a finite-valued function $f : \mathbb{R}^n \rightarrow \mathbb{R}$; and a closed-valued, measurable, set-valued mapping $S: \mathbb{R}^m \rightrightarrows \mathbb{R}^n$ .

Measurability is intended with respect to a finite measure $m: \mathcal{B}(\mathbb{R}^m) \rightarrow [0,1]$, where $\mathcal{B}(\mathbb{R}^m)$ are the Borel sets.

I am wondering if the following mapping is measurable as well. $$ x \mapsto \sup_{y \in S(x)} f(y) $$

What I thought is to define the mapping "$M: C \mapsto \sup_{y \in C} f(y)$", which takes a closed set as argument, and look at the measurability of $x \mapsto M(S(x))$. However, I am not clear what properties of $M$ should be exploited for the claim.

Solution 1:

If $S$ is compact-valued, this follows from the measurable maximum theorem, Theorem 18.19 in Aliprantis & Border 2006. Here is the statement:

Let $X$ be a separable metrizable space and $(S,\Sigma)$ a measurable space. Let $\phi:S\rightrightarrows X$ be a weakly measurable correspondence with nonempty compact values, and suppose $f:S\times X\to\mathbb{R}$ is a Carathéodory function (measurable in the first argument and continuous in the second). Define the value function $M:S\to\mathbb{R}$ by $$m(s)=\max_{x\in\phi(s)}f(s,x),$$ and the correspondence $\mu:S\rightrightarrows X$ of maximizers by $$\mu(s)=\{x\in\phi(s):f(s,x)=m(s)\}.$$

Then:

1. The value function $m$ is measurable.

2. The argmax correspondence $\mu$ has nonempty and compact values.

3. The argmax correspondence is measurable and admits a measurable selector.

If $S$ is not compact, similar results exist in which $m$ is only measurable with respect to every complete measure which make use of the theory of analytic sets. See the paper Some Measurability Results for Extrema of Random Functions over Random Sets by Stinchcombe and White.