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Let $(X,d)$ be a compact metric space and let $(\mathcal K(X),d_H)$ and $(\mathcal P(X),d_W)$ denote its space of nonempty compact subsets with Hausdorff metric $d_H$, and its space of Borel probability measures with 1-Wasserstein metric $d_W$.

Let $\operatorname{supp}:\mathcal P(X)\to \mathcal K(X)$ denote the function which associates to each measure its support. This function is generally not continuous e.g. consider the support of $\mu=(1-\epsilon)\delta_x + \epsilon \delta_y$ with $x\neq y$ as $\epsilon\to 0$. I'm curious about the following:

Does there exist a continuous $\operatorname{dude}:\mathcal K(X)\to \mathcal P(X)$ so that $\operatorname{supp}\circ\operatorname{dude}=\operatorname{id}_{\mathcal K(X)}$?

It seems associating a normalized Hausdorff measure $\mu_K(A) = \frac{1}{\mathcal H^\alpha(K)}\mathcal H^\alpha(A\cap K)$ does not work according to Wikipedia since some fractals have $\mathcal H^\alpha(K)\in\{0,+\infty \}$ for all $\alpha$.

As a follow-up to any answer, it would be nice to know if anything changes if we merely assume $X$ is compact Hausdorff and $\mathcal K(X)$ and $\mathcal P(X)$ are equipped with the Vietoris and weak-* topologies.

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  • $\begingroup$ Hausdorff measure already has a problem with finite sets, since it would be the uniform measure, and this is clearly wrong by considering sets like $\{-1/n, 1/n, 1\}$ in $[-1,1]$. $\endgroup$ – Nate Eldredge Nov 27 '19 at 8:43
  • $\begingroup$ Are you interested in the special case where $X$ is a closed interval? I suspect the s ax newer is yes in that case. $\endgroup$ – Anthony Quas Nov 27 '19 at 15:45
  • $\begingroup$ @Nate That's a good point and I should have noticed that. The empirical measure of a finite set failing to continuously vary with its support was actually the original motivation for this question. $\endgroup$ – Christian Bueno Nov 27 '19 at 22:50
  • $\begingroup$ @Anthony Though I'm mostly interested in greater generality, I'd definitely be interested in seeing if there's any special approaches that could be taken for closed intervals. $\endgroup$ – Christian Bueno Nov 27 '19 at 22:52
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    $\begingroup$ How much of this question was motivated by the goal of writing $\operatorname{supp} \circ \operatorname{dude}$? $\endgroup$ – Geoffrey Irving Dec 1 '19 at 9:54
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If $X$ is a subset of $\mathbb{R}^n$, one way is to take the uniform measure on say a ball containing $X$, and take the pushforward of it by the closest point projection on your compact subset $K$.

The support of that pushforward will be $K$, and it can be shown that this operation is continuous for the Hausdorff distance vs $1$-Wasserstein distance (actually it is even $1/2$-Holder).

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  • $\begingroup$ If $K$ consists of two concentric spheres, won't this put all the mass on the outer sphere? So the support won't be all of $K$. Or likewise, if $K$ is a ball, all the mass will go on the surface. $\endgroup$ – Nate Eldredge Nov 27 '19 at 19:25
  • $\begingroup$ Or maybe you mean a sphere in $\mathbb{R}^{n+1}$? $\endgroup$ – Nate Eldredge Nov 27 '19 at 19:31
  • $\begingroup$ meant a ball, sorry $\endgroup$ – alesia Nov 27 '19 at 19:32
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    $\begingroup$ It is included in the set of non differentiability points of the distance function, which is Lipschitz $\endgroup$ – alesia Nov 27 '19 at 20:04
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    $\begingroup$ link.springer.com/article/10.1007/s10208-009-9056-2. there are free versions online $\endgroup$ – alesia Nov 27 '19 at 22:42
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The affirmative answer to this question is given by the following theorem, proved by the technique of continuous selections.

Theorem. For any compact metrizable space $X$ there exists a continuous map $\phi:\mathcal K(X)\to\mathcal P(X)$ assigning to each nonempty compact set $K\subset X$ a probability measure $\phi(K)\in\mathcal P(X)$ such that $\mathrm{supp}(\phi(K))=K$.

Proof. It can be shown that the multi-valued map $\Phi:\mathcal H(X)\multimap \mathcal P(X)$ assigning to each compact set $K\in\mathcal H(X)$ the compact convex set $\Phi(K)=\{\mu\in\mathcal P(X):\mu(K)=1\}$ is lower-semicontinuous (which means that for any open set $U\subset X$ and any $a\in[0,1]$ the set $\{K\in\mathcal H(X):\exists \mu\in\Phi(K),\;\mu(U)>a\}$ is open in $\mathcal H(X)$).

Fix a countable base $(U_n)_{n\in\omega}$ of the topology of $X$ consisting of non-empty open sets in $X$. For every $n\in\omega$, consider the open set $\;\mathcal U_n=\{K\in\mathcal H(X):K\cap U_n\ne\emptyset\}$ in $\mathcal H(X)$ and the open convex subset set $\mathcal W_n=\{\mu\in P(X):\mu(U_n)>0\}$ in $\mathcal P(X)$.

Since the space $\mathcal H(X)$ is metrizable (and hence perfectly normal), for every $n\in\omega$ we can fix a continuous function $\lambda_n:\mathcal H(X)\to[0,\frac1{2^n}]$ such that $\mathcal U_n=\{K\in\mathcal H(X):\lambda_n(K)>0\}$. It follows that the function $$\lambda:\mathcal H(X)\to [0,2],\;\lambda:K\mapsto\sum_{n=0}^\infty\lambda_n(K),$$is continuous and strictly positive.

For every $n\in\omega$, consider the multi-valued map $\Phi_n:\mathcal U_n\multimap\mathcal W_n$ assinging to each compact set $K\in\mathcal U_n$ the closed convex subset $\Phi_n(K)=\mathcal W_n\cap\Phi(K)$ of $\mathcal W_n$. By Theorem 0.47 in the book [RS], the lower semi-continuity of the multi-valued map $\Phi$ implies the lower semi-continuity of the multi-valued map $\Phi_n$.

By the Compact-Valued Selection Theorem 4.1 in the book [RS], the multi-valued map $\Phi_n$ admits a compact-valued lower semicontinuous selection $\Psi_n:\mathcal U_n\multimap\mathcal W_n$. Let $\overline{\mathrm{co}}\Psi_n:\mathcal U_n\multimap \mathcal W_n$ be the multi-valued map assigining to each compact set $K\in\mathcal U_n$ the closed convex hull $\overline{\mathrm{co}}\Psi_n(K)\subset \Phi_n(K)$ of the compact set $\Phi_n(K)$ in $\mathcal W_n$. By Theorems 0.45 and 0.46 in [RS], the multi-valued map $\overline{\mathrm{co}}\Psi_n$ is lower semicontinuous. It can be shown that the closed convex hull of any compact set in $\mathcal W_n$ is compact. Consequently, for every $K\in\mathcal U_n$ the convex set $\overline{\mathrm{co}}\Psi_n(K)$ is compact.

By the Michael's Convex-Valued Selection Theorem 1.2 in [RS], the multivalued map $\overline{\mathrm{co}}\Psi_n$ admits a continuous selection $\psi_n:\mathcal U_n\to \mathcal P(X)$, which is a continuous map such that for every $K\in\mathcal U_n$ the measure $\psi_n(K)$ belongs to the compact convex set $\overline{\mathrm{co}}\Psi_n(K)\subset\Phi_n(K)$, and hence $\psi_n(K)(U_n)>0$ and $\psi_n(K)(K)=1$.

Finally, consider the continuous map $\phi:\mathcal H(X)\to\mathcal P(X)$ assigning to each compact set $K\in\mathcal H(X)$ the probability measure $$\phi(K)=\frac1{\lambda(K)}\sum_{n\in N(K)}\lambda_n(K){\cdot}\psi_n(K)$$where $N(K)=\{n\in\omega:K\in\mathcal U_n\}$. It can be shown that $\phi$ is a required continuous map assigning to each compact set $K\in\mathcal H(X)$ a probability measure $\phi(K)$ with $\mathrm{supp}(\phi(K))=K$.

Reference.

RS: D.Repovs, P.Semenov, Continuous Selections of multivalued mappings, Springer, 1998.

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