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.


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 $$\mu(s)=\{x\in\phi(s):f(s,x)=m(s)\}.$$


  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.

  • $\begingroup$ Thanks a lot for the answer. However, I am missing why the $f$ in the question is a Carathéodory mapping. Also, I guess that to be a normal integrand, since we take the $\sup$, it should be upper semicontinuous. Can you please explain? $\endgroup$ – user693 Sep 12 '13 at 13:30
  • $\begingroup$ I mean that I would agree with your answer if $f$ is a continuous function. $\endgroup$ – user693 Sep 12 '13 at 13:41
  • $\begingroup$ @Adam, you are right. The Stinchcombe-White paper should cover your problem though. $\endgroup$ – Michael Greinecker Sep 12 '13 at 19:53
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    $\begingroup$ @Adam, yes. And measurability is enough there. And an analytic function is automatically measurable with respect to every complete measure. $\endgroup$ – Michael Greinecker Sep 13 '13 at 9:56
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    $\begingroup$ The notion of analytic sets used by Dellarcherie and Meyer is more general than what you find in Dudley. In general, you will not get Borel measurability without further assumptions, which is why mentioned the completion. $\endgroup$ – Michael Greinecker Oct 26 '13 at 13:15

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