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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.

user693
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1 Answers1

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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.

Michael Greinecker
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  • 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? – user693 Sep 12 '13 at 13:30
  • I mean that I would agree with your answer if $f$ is a continuous function. – user693 Sep 12 '13 at 13:41
  • @Adam, you are right. The Stinchcombe-White paper should cover your problem though. – Michael Greinecker Sep 12 '13 at 19:53
  • Thanks again for the tip. Do you mean because, in view of Theorem 2.17 of the Stinchcombe-White paper, $f$ is constant (wrt $x$) and hence measurable (Assumption iii)? – user693 Sep 13 '13 at 06:36
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    @Adam, yes. And measurability is enough there. And an analytic function is automatically measurable with respect to every complete measure. – Michael Greinecker Sep 13 '13 at 09:56
  • Is this result stated elsewhere? The journal "Review of economic studies" does not seem mathematically appealing... – user693 Oct 25 '13 at 14:32
  • @Adam, well you can just follow the proof. The article is mainly an exposition for how one can use the theory of analytic sets as developed by Dellarcherie and Meyer to these kind of problems. RES is a fairly good journal though. – Michael Greinecker Oct 25 '13 at 17:36
  • Is there another reference for this result? – user693 Oct 25 '13 at 18:08
  • But analytic sets regard universally measurable sets, not Borel measurable set, right? – user693 Oct 25 '13 at 18:24
  • @Adam What exact result do you want to have a reference for? – Michael Greinecker Oct 25 '13 at 18:52
  • So I took a look in the book [Dudley - Real Analysis and Probability], Section 13.2 on Analytic Sets. The main result is that "any analytic set is universally measurable", which is not the same of Borel measurability. Now, do you think that Borel measurability follows because in my case the measure is finite? – user693 Oct 25 '13 at 19:32
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    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. – Michael Greinecker Oct 26 '13 at 13:15
  • So when in the paper you mention it comes out that my function $x \mapsto \sup_{ y \in S(x) } f(y)$ is "analytic", what kind of measurability do we get? – user693 Oct 26 '13 at 16:54
  • It's definition 2.10 of the paper on page 499. – Michael Greinecker Oct 27 '13 at 03:10
  • Is the map then continuous with respect to the Hausdorff space (of non-empty compacts) on $X$? –  Apr 28 '20 at 07:50
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    @AnnieTheKatsu If the function to be maximized is continuous, yes (from the BErge maximum theorem). But that is a completely different issue. – Michael Greinecker Apr 28 '20 at 08:03