# 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 $[-1,1]$. – Nate Eldredge Nov 27 '19 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. – Anthony Quas Nov 27 '19 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. – Christian Bueno Nov 27 '19 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. – Christian Bueno Nov 27 '19 at 22:52
• How much of this question was motivated by the goal of writing $\operatorname{supp} \circ \operatorname{dude}$? – Geoffrey Irving Dec 1 '19 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. – Nate Eldredge Nov 27 '19 at 19:25
• Or maybe you mean a sphere in $\mathbb{R}^{n+1}$? – Nate Eldredge Nov 27 '19 at 19:31
• meant a ball, sorry – alesia Nov 27 '19 at 19:32
• It is included in the set of non differentiability points of the distance function, which is Lipschitz – alesia Nov 27 '19 at 20:04
• link.springer.com/article/10.1007/s10208-009-9056-2. there are free versions online – alesia Nov 27 '19 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.