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I'm trying to prove below result. Could you verify if my attempt is fine?

Let $(X, d)$ be a metric space and $\mathcal{P} :=\mathcal{P}(X)$ the space of all Borel probability measures on $X$. Let $d_P$ be the Prokhorov metric on $\mathcal{P}$.

Theorem: Assume that $X$ is separable. If $X$ is complete, then so is $\mathcal P$.

I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.

Analyst
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2 Answers2

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Lemma 1: Assume $D$ is countable and dense in $X$. Then $$ \mathcal D := \{a_1 \delta_{x_1} + \cdots + a_n \delta_{x_n} \mid n \in \mathbb N_{>0} \text{ and } x_1, \ldots, x_n \in D \text{ and } a_1, \ldots, a_n \in \mathbb Q_{\ge 0} \text{ and } a_1 + \cdots + a_n=1\} $$ is countable and dense in $\mathcal P$.

Lemma 2: Suppose $X$ has a countable dense subset $A$ such that every Cauchy sequence in $A$ converges to a point in $X$. Then $X$ is complete.

Lemma 3: Let $X$ be complete and $\Gamma \subset \mathcal{P}$. If $$ \forall \varepsilon, \delta>0, \exists a_{1}, \ldots, a_{n} \in X, \forall \mu \in \Gamma: \mu\left(\bigcup_{i=1}^{n} B\left(a_{i}, \delta\right)\right) \geq 1-\varepsilon, $$ then $\Gamma$ is uniformly tight.

Let $D$ be a countable dense subset of $X$. By Lemma 1, let $\mathcal D$ be a countable dense subset of $\mathcal P$ that is induced by $D$. Let $(\mu_n)$ with $\mu_n := a_{n,1} \delta_{x_{n,1}} + \cdots + a_{n, \lambda(n)} \delta_{x_{n, \lambda(n)}}$ be a Cauchy sequence in $\mathcal D$. By Lemma 2, it suffices to show that there is $\mu \in \mathcal P$ such that $\mu_n \to \mu$ in $d_P$. By Prokhorov theorem, it suffices to show that $(\mu_n)$ is uniformly tight. As such, we will prove that $(\mu_n)$ satisfies the condition of Lemma 3.

Fix $\varepsilon, \delta>0$. Let's $\varepsilon' := \frac{1}{2} \min\{\varepsilon, \delta\}$. There is $N$ such that $d_P(\mu_n, \mu_N) < \varepsilon'$ for all $n \ge N$, i.e., $$ \begin{cases} \mu_N(A) \leq \mu_n \left(A_{\varepsilon'}\right)+\varepsilon' \\ \mu_n(A) \leq \mu_N \left(A_{\varepsilon'}\right)+\varepsilon' \end{cases} \quad \forall n \ge N, A \in \mathcal B(X). $$

Let $$ A := \bigcup_{i=1}^{\lambda(N)} B (a_{N,i}, \varepsilon' ). $$

It follows that $\mu_n \left(A_{\varepsilon'}\right) \ge 1-\varepsilon'$ for all $n \ge N$. Notice that $$ A_{\varepsilon'} = \{x\in X \mid d(x, A) < \varepsilon'\} \subset \bigcup_{i=1}^{\lambda(N)} B (a_{N,i}, 2\varepsilon' ) \subset \bigcup_{i=1}^{\lambda(N)} B (a_{N,i}, \delta). $$

It follows that $$ \mu_n \left( \bigcup_{i=1}^{\lambda(N)} B (a_{N,i}, \delta) \right ) \ge \mu_n(A_{\varepsilon'}) \ge 1- \varepsilon' \ge 1-\varepsilon \quad \forall n \ge N. $$

It follows that the collection of centers $\{a_{n,i} \mid n \le N, i\le \lambda(n)\}$ satisfies the condition.

Analyst
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I have found another approach that does not need above Lemmas 1 and 2.

Lemma: Let $X$ be complete and $\Gamma \subset \mathcal{P}$. If $$ \forall \varepsilon, \delta>0, \exists a_{1}, \ldots, a_{n} \in X, \forall \mu \in \Gamma: \mu\left(\bigcup_{i=1}^{n} B\left(a_{i}, \delta\right)\right) \geq 1-\varepsilon, $$ then $\Gamma$ is uniformly tight.

Let $(\mu_n)$ be a Cauchy sequence in $\mathcal P$. By Prokhorov theorem, it suffices to show that $(\mu_n)$ is uniformly tight. As such, we will prove that $(\mu_n)$ satisfies the condition of Lemma.

Fix $\varepsilon, \delta>0$. Let's $\varepsilon' := \frac{1}{2} \min\{\varepsilon, \delta\}$. There is $N$ such that $d_P(\mu_n, \mu_N) < \varepsilon'$ for all $n \ge N$, i.e., $$ \begin{cases} \mu_N(A) \leq \mu_n \left(A_{\varepsilon'}\right)+\varepsilon' \\ \mu_n(A) \leq \mu_N \left(A_{\varepsilon'}\right)+\varepsilon' \end{cases} \quad \forall n \ge N, A \in \mathcal B(X). $$

Let $D := \{a_1, a_2, \ldots\}$ be countable and dense in $X$. Notice that $\bigcup_{i=1}^{\infty} B (a_i, \varepsilon' ) = X$, so there exist $a_1, a_2, \ldots, a_M \in D$ such that $$ \mu_n (A) \ge 1- \varepsilon' \quad \forall n \le N, $$ where $$ A := \bigcup_{i=1}^{M} B (a_i, \varepsilon' ). $$

It follows that $\mu_n \left(A_{\varepsilon'}\right) \ge \mu_N(A)-\varepsilon' \ge1-2\varepsilon'$ for all $n \ge N$. Notice that $$ A_{\varepsilon'} = \{x\in X \mid d(x, A) < \varepsilon'\} \subset \bigcup_{i=1}^{M} B (a_{i}, 2\varepsilon' ) \subset \bigcup_{i=1}^{M} B (a_{N,i}, \delta). $$

It follows that $$ \mu_n \left( \bigcup_{i=1}^{M} B (a_{i}, \delta) \right ) \ge \mu_n(A_{\varepsilon'}) \ge 1- 2\varepsilon' \ge 1-\varepsilon \quad \forall n \ge 1. $$

It follows that the collection of centers $\{a_1, a_2, \ldots, a_M\}$ satisfies the condition.

Analyst
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