0

In proving the reverse direction of Prokhorov theorem, I have to prove this auxiliary result. Could you verify if my attempt is fine?

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

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

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

Analyst
  • 5,637

1 Answers1

0

Notice that $x \mapsto \delta_{x}$ is a homeomorphism from $X$ onto $\left\{\delta_{x} \mid x \in X\right\}$ which is closed in $\mathcal{P}$. If $\mathcal{P}$ is compact, then so is $\left\{\delta_{x} \mid x \in X\right\}$ and thus is $X$. Let's prove the other direction.

Let $E:= \mathcal C_b(X)$. Then $(E, \| \cdot\|_\infty)$ is a Banach space. Let $E'$ be its dual and $$ \begin{align} F := &\{\varphi \in E' \mid \|\varphi\| = 1, \varphi \text{ positive}\} \\ = &\{\varphi \in E' \mid \|\varphi\| \le 1, \varphi \text{ positive}, \varphi(1) = 1\}. \end{align} $$

Then $F$ is closed in weak* topology of $E'$ and is a subset of the unit ball. By Banach-Alaoglu theorem, $F$ is compact in weak* topology. Because $X$ is compact, $E$ coincides with the space $\mathcal C_0 (X)$ of functions vanishing at infinity. By Riesz-Markov-Kakutani representation theorem, the map $$ T: \mathcal P \to F, \mu \mapsto \left ( \varphi_\mu: f \mapsto \int_X f \mathrm d \mu \right ). $$ is a bijection.

  • Let $\mu, \mu_1,\mu_2, \ldots \in \mathcal P$ such that $\mu_n \to \mu$ in $d_P$. Then $\mu_n \to \mu$ weakly, so $\varphi_{\mu_n} \to \varphi_\mu$ in weak* topology. This implies $T(\mu_n) \to T(\mu)$ in weak* topology.

  • Let $\varphi_\mu, \varphi_{\mu_1}, \varphi_{\mu_2}, \ldots \in \mathcal P$ such that $\varphi_{\mu_n} \to \varphi_\mu$ in weak* topology. Then $\varphi_{\mu_n} (f) \to \varphi_\mu (f)$ for all $f\in E$, so ${\mu_n} \to \mu$ weakly. Notice that compact metric space is separable. In separable metric space, weak convergence is equivalent to convergence in $d_P$. This implies $\mu_n \to \mu$ in $d_P$.

As such, $T$ is a sequential homeomorphism. This completes the proof.

Analyst
  • 5,637