Let $(E, d)$ be a metric space and $\mathcal C$ the set of all non-empty compact subsets of $E$. We define the Hausdorff metric $d_H$ on $\mathcal C$ by $$ d_H (A, B) := \max \left \{ \max_{x \in B} d(x, A), \max_{y \in A} d (y, B) \right \}\quad \forall A,B \in \mathcal C, $$ with $d(x, A) := \inf_{y \in A} d(x, y)$. I'm trying to prove a property from page 37 of Santambrogio's Optimal Transport for Applied Mathematicians: Calculus of Variations, PDEs, and Modeling.
Theorem: Let $A, A_n \in \mathcal C$ such that $A_n \to A$ in $d_H$. Let $\mu, \mu_n$ be non-negative finite Borel measures such that $\operatorname{supp} \mu_n \subset A_n$ and that $\mu_n \overset{\ast}{\rightharpoonup} \mu$. Then $\operatorname{supp} \mu \subset A$.
In below attempt, I can only prove the thereom in the case $\mu, \mu_n$ are probability measures.
How can we prove this theorem in its full generality?
My attempt: First, we have the following equivalent formula of $d_H$, i.e., $$ d_H (A, B) = \inf \{ \varepsilon >0 \mid A \subset B_{\varepsilon}, B \subset A_{\varepsilon} \} \quad \forall A,B \in \mathcal C. $$
Here $A_{\varepsilon} := \{x \in E \mid d(x, A) \le \varepsilon\}$ and $B_{\varepsilon} := \{x \in E \mid d(x, B) \le \varepsilon\}$ being the $\varepsilon$-neighborhoods of $A, B$ respectively.
Fix $\varepsilon > 0$. First, we'll prove that $A_{\varepsilon}$ is closed. Let $(x_n) \subset A_{\varepsilon}$ such that $x_n \to x \in E$. We have $$ d(x, a) \le d(x, x_n) + d(x_n, a) \quad \forall a \in A, \forall n. $$
Then for all $n$, $$ \begin{align} d(x, A) &= \inf_{a \in A} d(x,a) \\ &\le d(x, x_n) + \inf_{a \in A} d(x_n,a) \\ &= d(x, x_n) + d(x_n, A) \\ &\le d(x,x_n) + \varepsilon. \end{align} $$
Taking the limit $n \to \infty$, we get $d(x,A) \le \varepsilon$ and thus $x \in A_{\varepsilon}$.
Because $A_n \to A$ in $d_H$, there is $n_{\varepsilon }$ such that $A_n \subset A_{\varepsilon }$ for all $n \ge n_{\varepsilon }$. This implies $$ \mu_n (A_{\varepsilon}) \ge \mu_n (A_n) = 1 \quad \forall n \ge n_{\varepsilon}. $$
By Portmanteau theorem, we have $$ \mu (A_{\varepsilon }) \ge \limsup_n \mu_n (A_{\varepsilon }) \ge 1. $$
Notice that $A$ is closed, so $A = \bigcap_{\varepsilon >0} A_{\varepsilon}$ and thus $A_{\varepsilon} \searrow A$ as $\varepsilon \searrow 0$. It follows that $$ \mu (A) = \lim_{\varepsilon \searrow 0} \mu (A_{\varepsilon}) = 1. $$
It follows that $\operatorname{supp} \mu \subset A$.
Update: I have found another proof as follows.
We define $f_n (x) := d(x, A_n)$ and $f (x) := d(x, A)$. Notice that $$ f_n (x) \le d(x, y) + d (y, A_n) = d(x, y) + f_n (y) \quad \forall x,y \in E. $$
By symmetry, we have $|f_n (x)-f_n (y)| \le d(x,y)$. Thus $f_n$ is $1$-Lipschitz. Moreover, $f_n$ is compactly supported. It follows that $f_n \in C_b (E)$, i.e., $f_n$ is continuous bounded. Similarly, $f \in C_b (E)$. We have the following equivalent formula of $d_H$, i.e., $$ d_H (A, B) = \max_{x \in E} |d(x, A) - d(x, B)|. $$
It follows that $|f_n (x)-f(x)| = |d(x, A_n) - d(x, A)| \le d_H(A_n, A)$. It follows that $f_n \to f$ uniformly. By this convergence result, we get $$ 0 = \int_E f_n \mathrm d \mu_n \to \int_E f \mathrm d \mu. $$
Hence $\mu$ is supported on $A$. This completes the proof.
