If the underlying metric space is separable, then weak convergence is equivalent to convergence in Prokhorov metric

Question

I'm rewriting the proof that weak convergence is equivalent to convergence in Prokhorov metric in separable metric space. Could you verify if my attempt is fine?

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Let $(X, d)$ be a metric space and $\mathcal{M} :=\mathcal{M}(X)$ the set all non-negative finite Borel measures on $X$. Let $d_P$ be the Prokhorov metric on $\mathcal{M}$. The weak convergence on $\mathcal{M}$ is the convergence w.r.t. the space $\mathcal C_b(X)$ of all bounded continuous functionals on $X$.

Theorem: If $X$ is separable, then weak convergence implies convergence in $d_P$.

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I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Answer 1

• Lemma 1: Let $\mu, \mu_1, \mu_2,\ldots \in \mathcal{M}$ and $g \in \mathcal C_b(X)$. If $\mu_i \to \mu$ weakly, then $\mu_i(A) \to \mu(A)$ for all Borel set $A \subseteq X$ with $\mu(\partial A) = 0$.

• Lemma 2: If $X$ is separable and $\mu \in \mathcal{M}$. Then for each $\delta>0$ there are countably many open (or closed) balls $B_{1}, B_{2}, \ldots$ such that $\bigcup_{i=1}^{\infty} B_{i}=X$, the radius of $B_{i}$ is less than $\delta$, and $\mu\left(\partial B_{i}\right)=0$ for all $i$.

Fix $\varepsilon>0$. We want to show that $\exists N, \forall i \geq N: d_{P}\left(\mu_{i}, \mu\right) \leq \varepsilon$, i.e., $\mu_{i}(B) \leq \mu\left(B_{\varepsilon}\right)+\varepsilon$ and $\mu(B) \leq \mu_{i}\left(B_{\varepsilon}\right)+\varepsilon$ for all Borel subset $B$.

Fix $\delta \in (0, \varepsilon/4)$. By Lemma 2, there are countably many open balls $B_{1}, B_{2}, \ldots$ with radius less than $\delta/2$ such that $\bigcup_{i=1}^{\infty} B_{i}=X$ and $\mu\left(\partial B_{i}\right)=0$ for all $i$. Fix $k$ such that $$ \mu\left(\bigcup_{j=1}^{k} B_{j}\right) \ge \mu(X)-\delta. $$

Let $\mathcal A$ be the finite collection of subsets built by combining the balls $B_1, \ldots, B_k$, i.e., $$ \mathcal{A}:=\left\{\bigcup_{j \in I} B_{j} \,\middle\vert\, J \subset \{1, \ldots, k\}\right\}. $$

We will use this collection to approximate any Borel set. For each $A \in \mathcal{A}, \partial A \subset \partial B_{1} \cup \cdots \cup \partial B_{k}$, so $\mu(\partial A) \leq$ $\mu\left(\partial B_{1}\right)+\cdots+\mu\left(\partial B_{k}\right)=0$. By Lemma 1, $\mu_{i}(A) \rightarrow \mu(A)$ for all $A \in \mathcal{A}$. Fix $N$ such that $$ \left|\mu_{i}(A)-\mu(A)\right|<\delta \quad \forall i \geq N, \forall A \in \mathcal{A}. $$

In particular, $$ \mu_i \left(\bigcup_{j=1}^{k} B_{j}\right) \ge \mu \left(\bigcup_{j=1}^{k} B_{j}\right) -\delta \ge \mu(X) - 2 \delta \quad \forall i \ge N. $$

Now we fix a Borel set $B$ and approximate it by $$ A := \bigcup \{B_j \mid j = 1,\ldots,k \text{ such that } B_j \cap B \neq \emptyset\}. $$

Then

• $A \subset B_{\delta} := \{x \mid d(x, B)<\delta\}$ because $\operatorname{diam} B_{j}<\delta$,
• $B=\left[B \cap \bigcup_{j=1}^{k} B_{j}\right] \cup\left[B \cap\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right] \subset \left [ A \cup\left(\bigcup_{j=1}^{k} B_{j}\right)^{c} \right ]$,
• $\left|\mu_{i}(A)-\mu(A)\right|<\delta$ for all $i \geq N$, and
• $\mu\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \leq \delta$ and $\mu_{i}\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \leq \mu_i(X)-\mu(X)+ 2 \delta \le \mu_i(X) + 3\delta$ for all $i \geq N$.

It follows that for every $i \geq N$ : \begin{aligned} \mu(B) & \leq \mu(A)+\mu\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \\ & \leq \mu(A)+\delta \\ & \leq \mu_{i}(A)+2 \delta \\ & \leq \mu_{i}\left(B_{\delta}\right)+2 \delta \\ &\leq \mu_{i}\left(B_{\varepsilon}\right)+\varepsilon \\ \mu_{i}(B) & \leq \mu_{i}(A)+\mu_{i}\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \\ &\leq \mu_{i}(A)+ 3 \delta \\ &\leq \mu(A)+4 \delta \\ & \leq \mu\left(B_{\delta}\right)+4 \delta \\ &\leq \mu\left(B_{\varepsilon}\right)+\varepsilon. \end{aligned}

This is true for every $B \in \mathcal{B}$, so $d_{P}\left(\mu_{i}, \mu\right) \leq \varepsilon$ for all $i \geq N$.