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I'm trying to wrap my head around the "Measurable maximum theorem", Thm 14.91 in "A hitchhiker's guide to infinite dimensional analysis" by Aliprantis & Border. I wonder if I can use it in a case when the underlying measurable space is a probability space. (I will explain in which sense below).

The statement of the theorem goes like this:

Let $X$ be a polish space and $(S,\Sigma)$ a measurable space. Let $\varphi: S \twoheadrightarrow X$ (My clarification: $\varphi$ is set valued/a correspondance) be a weakly measurable correspondance with nonempty compact values, and suppose that $f: S \times X \to \mathbb{R}$ is a Caratheodory function (measurable in $s$ and continuous in $x$). Define the value function $m$ by: $$m(s) = \max_{x \in \varphi(s)} f(s,x),$$ and the correspondence $$ \mu(s) = \{ x \in \varphi(s): f(s,x) = m(s) \}.$$ Then:

  • The value function $m$ is measurable.
  • The argmax correspondance $\mu$ is measurable, has nonempty compact values and admits a measurable selector.

Clarification: By weakly measurable correspondence is meant that $\{s \in S: \varphi(s) \cap K \neq \emptyset \}$ belongs to $\Sigma$ for each compact set $K$.

My question is the following; suppose that I have a probability space $\left( \Omega, \mathcal{A}, \mathbb{P} \right)$ and a Caratheodory function $f: \Omega \times \mathbb{R}^d \to \mathbb{R}$. I would like to check that the argmax correspondence $$\text{argmax}_{x \in \mathbb{R}} = f(\omega,x)$$ is a measurable correspondence and admits a measurable selector. Is it possible to use the theorem above to do this? If yes, how do I define $\varphi$ and check that it's weakly measurable in the above sense?

kodlu
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user202542
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1 Answers1

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$\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\argmax}{\mathrm{argmax}}$Since you are maximizing $x \in \RR$ you are going to run into trouble defining $\varphi$ canonically. Strategies to overcome this are limited, but include these tricks:

  • If actually you are content with $\argmax_{x \in \kappa(\omega)} f(\omega, x)$, where $\kappa : \Omega \twoheadrightarrow \RR$ is upper semicontinuous and compact valued. (For example, if $\kappa$ is compact and constant.)
  • You can argue that $f(\omega, x)$ is sup-compact, meaning $$ \{(\omega, x) : f(\omega, x) \geq \lambda \} $$ is a compact subset of $\Omega \times \RR$ for every $\lambda \in \RR$. Then you can perform a more delicate analysis by introducing the multifunctions $\varphi_\lambda(\omega) = \{x \in \RR : f(\omega, x) \geq \lambda\}$, which are upper semicontinuous, compact-valued, and nonempty for carefully chosen $\lambda$.

The above works because upper semicontinuous multifunctions are weakly measurable: since $\varphi^{u}(G)$ is open whenever $G \subset \RR$ is open, it follows immediately that $\varphi$ is weakly measurable.

I'm not aware of other general techniques for getting around measurable parametric optimization over non-compact sets.

David Kraemer
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  • Thanks for your reply. Can I always choose $\kappa$ such that it's upper semicontinuous and compact valued, as in your first point? How do you define semicontinuity in this case? – user202542 Oct 31 '22 at 12:05
  • Always? Definitely not, because it depends on the problem you're trying to study. For example, if you're maximizing $f : \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y) = 1 - \exp(-y^2)$, then $\mathrm{arg,max}_{y \in \mathbb{R}} f(x,y)$ is empty. But if you actually only care about $y \in [-1,1]$, much more can be said. You need to examine the problem in question. – David Kraemer Nov 01 '22 at 18:07