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Let $I=[a,b]$ with $a<b\in\mathbb{R}$ and denote by $\mathcal{M}(I)$ the set of Borel probability measures on $I$ equipped with the topology induced by the weak convergence of measures.

Does there exist a correspondence $\phi:\mathcal{M}(I) \to 2^I\setminus\{\emptyset\}$ which fulfills $\mathrm{supp}(\sigma)\nsubseteq\phi(\sigma)$ (equivalently $\sigma(\phi(\sigma))< 1$) for all $\sigma\in \mathcal{M}(I)$ while simultaneously having a closed graph: $\{(\sigma,t)\in \mathcal{M}(I)\times I \mid t \in \phi(\sigma)\}$ closed (wrt. the product topology) in $\mathcal{M}(I)\times I$.

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    $\begingroup$ Now that the question has been edited/clarified, my answer no longer is an answer. So I have deleted it. $\endgroup$
    – user95282
    Commented Mar 15, 2023 at 11:58
  • $\begingroup$ (i) In general, $\mathrm{supp}(\sigma)\nsubseteq\phi(\sigma)$ is not equivalent to $\sigma(\phi(\sigma))< 1$. E.g., take $\sigma$ to be uniform on $[0,1]$ and $\phi(\sigma)=(0,1]$. (ii) The values of $\phi$ are, not in $I$, but in $2^I\setminus\{\emptyset\}$. So, you need to specify a topology on $2^I\setminus\{\emptyset\}$. $\endgroup$
    – Iosif Pinelis
    Commented Mar 16, 2023 at 17:04
  • $\begingroup$ @Iosif Pinelis (i) It is immediately verified that it is equivalent for the case of $\phi(\sigma)$ being closed in $I$. The latter is of course implied by the closed graph property I stated in the question. (ii) The closed graph property $\phi$ shall satisfy is completely independent of the topology on $2^I$. So a topology on $2^I$ is not required. However, you could equip $2^I$ with the topology induced by the upper Hausdorff hemimetric . Then the required closed graph property should be equivalent to $\phi$ being continuous and $\phi(\sigma)$ being closed in $I$. $\endgroup$
    – Julian
    Commented Mar 17, 2023 at 11:56

1 Answer 1

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$\newcommand\M{\mathcal M}\newcommand\F{\mathcal F}\newcommand\si{\sigma}\newcommand\th{\theta}$I think the answer is no. Here is the idea of how to show this:

Suppose that such a map $\phi$ exists. In view of this answer or this answer, there is a continuous mapping $\psi$ from the set $\F$ of all nonempty closed subsets of $I$ endowed with the Hausdorff metric to $\M:=\M(I)$ such that each $F\in\F$ is the support $S_\mu$ of the measure $\mu:=\psi(F)$. Then the map $\psi\circ\phi\colon\M(I)\to\M(I)$ will be (semi-)continuous in some sense. So, by an appropriate version of the fixed-point theorem, $\mu=(\psi\circ\phi)(\mu)$ for some $\mu\in\M$, whence $S_\mu=\phi(\mu)$, a contradiction.

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  • $\begingroup$ If this answer works, it will have to use the fact that the underlying space is an interval and not for example a circle. Perhaps that will be used in the "appropriate version of the fixed-point theorem". $\endgroup$
    – user95282
    Commented Mar 21, 2023 at 11:23
  • $\begingroup$ @user7427029 : What would be wrong with a circle? $\endgroup$
    – Iosif Pinelis
    Commented Mar 21, 2023 at 13:34
  • $\begingroup$ \@IosifPinelis You are right, it's not clear why it wouldn't work for the circle. $\endgroup$
    – user95282
    Commented Mar 21, 2023 at 15:12

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