# Continuous section of support - Is it possible to map compact sets to measures supported on them?

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.

• 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 $[-1,1]$. Nov 27, 2019 at 8:43
• Are you interested in the special case where $X$ is a closed interval? I suspect the s ax newer is yes in that case. Nov 27, 2019 at 15:45
• @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. Nov 27, 2019 at 22:50
• @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. Nov 27, 2019 at 22:52
• How much of this question was motivated by the goal of writing $\operatorname{supp} \circ \operatorname{dude}$? Dec 1, 2019 at 9:54

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).

• 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. Nov 27, 2019 at 19:25
• Or maybe you mean a sphere in $\mathbb{R}^{n+1}$? Nov 27, 2019 at 19:31
• meant a ball, sorry Nov 27, 2019 at 19:32
• It is included in the set of non differentiability points of the distance function, which is Lipschitz Nov 27, 2019 at 20:04
• link.springer.com/article/10.1007/s10208-009-9056-2. there are free versions online Nov 27, 2019 at 22:42

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.