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I'm trying to prove the reverse of Prokhorov theorem. Could you verify if my attempt is fine?

Let $(X, d)$ be a metric space and $\mathcal{P}(X)$ the set all Borel probability measures on $X$. Let $d_P$ be the Prokhorov metric on $\mathcal{P}(X)$. The set $\Gamma \subseteq \mathcal P(X)$ is called uniformly tight if for all $\varepsilon>0$ there is a compact subset $K$ of $X$ such that $$ \mu(K^c) := \mu(X\setminus K) \le \varepsilon \quad \forall \mu \in \Gamma. $$

Theorem: If $X$ is separable, then [$\Gamma$ is uniformly tight] implies [$\Gamma$ is relatively compact].

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

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1 Answers1

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Lemma 1: If $X$ is separable, then there exists a compact metric space $(Y, d')$ and a map $T: X \to Y$ such that $T$ is a homeomorphism from $X$ onto $T(X)$.

Lemma 2: $(X, d)$ is compact if and only if $(\mathcal P, d_P)$ is compact.

By Lemma 1, there is a compact metric space $(Y, d')$ and a map $T:X \to Y$ such that $T$ is a homeomorphism from $X$ onto $T(X)$.

Let $(\mu)_n$ be a sequence in $\overline \Gamma$. Let $\nu_n := T_{\sharp} \mu_n \in \mathcal P(Y)$ for all $n$. Because $(Y, d')$ is compact, so is $(\mathcal P(Y), d'_{P})$ by Lemma 2. Hence there exist a subsequence $\lambda \in \mathbb N^\mathbb N$ and $\nu \in \mathcal P(Y)$ such that $\nu_{\lambda(n)} \to \nu$.

Let $Z := T(X)$. First, we prove that $\nu$ is concentrated on $Z$. Because $\Gamma$ is tight, so is $\overline \Gamma$. For each $m\ge 1$ there is a compact subset $K_m$ of $X$ such that $$ \mu_n(K_m) \ge 1-1/m \quad \forall n. $$

Clearly, $C_m :=T(K_m)$ is compact and thus closed in $Y$. So $C_m \in \mathcal B(Y)$. Because $T$ is injective, $$ \nu_n (C_m) = \mu_n(K_m) \ge 1-1/m \quad \forall n. $$

Also, $$ \nu(C_m) \ge \limsup_n \nu_n (C_m) \ge 1-1/m. $$

Then $$ \nu \left ( \bigcup_{m=1}^\infty C_m \right ) \ge 1-1/m \quad \forall m. $$

So $$ \nu \left ( \bigcup_{m=1}^\infty C_m \right ) =1. $$

We define a Borel probability measure $\mu$ on $X$ as follows. $$ \mu (B) := \nu \left ( \bigcup_{m=1}^\infty \left [ T (B )\cap C_m \right ] \right ) \quad \forall B \in \mathcal B(X). $$

  • Notice that homeomorphism sends Borel set to Borel set. Also, $\mathcal B(Z) = \{ C \cap Z \mid C \in \mathcal B(Y)\}$.
  • We have $T(B) \in \mathcal B(Z)$, so $\exists C \in \mathcal B(Y)$ such that $T(B) = C \cap Z$. Hence $T (B )\cap C_m = C \cap Z \cap C_m = C \cap C_m \in \mathcal B(Y)$.
  • It follows that $\bigcup_{m=1}^\infty \left [ T (B )\cap C_m \right ] \in \mathcal B(Y)$ and thus $\mu$ is well-defined.

Next we prove that $\mu_n \to \mu$ in $d_P$. Because $X$ is separable, it suffices to show that $\mu_n \to \mu$ weakly. Let $B$ is closed in $X$. We have $$ \begin{align} \mu(B) &= \nu \left ( \bigcup_{m=1}^\infty \left [ T (B )\cap C_m \right ] \right ) \\ &\ge \limsup_n \nu_n \left ( \bigcup_{m=1}^\infty \left [ T (B )\cap C_m \right ] \right ) \\ &= \limsup_n \mu_n \left ( T^{-1} \left ( \bigcup_{m=1}^\infty \left [ T (B )\cap C_m \right ] \right ) \right )\\ &= \limsup_n \mu_n \left ( B \cap \bigcup_{m=1}^\infty K_m \right ) \\ &=\limsup_n \mu_n(B). \end{align} $$ This completes the proof.

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