A tight collection of probability measures on space of probability measures

Question

Let $E$ be a Polish space, and let $\mathcal M_1(E)$ denote the space of probability measures on $E$. I want to show the following:

A collection $\mathcal K \subset \mathcal M_1\left(\mathcal M_1(E)\right)$ is tight if and only if, for any $\epsilon > 0$, there exists a compact set $K \subset E$ for which $$ \tilde \mu\left(\left\{ \mu \in \mathcal M_1(E) : \mu\left(K^c\right)>\epsilon\right\}\right)< \epsilon \quad \forall \tilde \mu \in \mathcal K. $$

I'm having trouble in both directions. My thinking for "only if" is to show that given any $\epsilon > 0$, every compact subset $L \subset \mathcal M_1(E)$ (in the weak* topology) is contained in some subset of the form $$ \tilde K_\epsilon := \left\{\mu \in \mathcal M_1(E) : \mu(K^c) \leq \epsilon\right\} . $$ Since $E$ is Polish, every probability measure is inner regular, so every $\mu \in \mathcal M_1(E)$ lies in some subset $\tilde U_{K,\epsilon} = \left\{\mu \in \mathcal M_1(E) : \mu(K^c) < \epsilon\right\}$. If these subsets were open, then by compactness of $L$, we could take a finite union of such sets $\tilde U_{K,\epsilon}$ and we'd be done with "only if". But these subsets aren't open in general (see Counterexample 1 below).

For "if", this would be immediate if $\tilde K_\epsilon$ were compact. It's easy to show $\tilde K_\epsilon$ is closed: if $\left(\mu_n\right)\subset\tilde K_\epsilon$ and $\mu = \displaystyle \mathop{\textrm{w-lim}}_{n\to\infty} \mu_n$, by the Portemanteau theorem, since $K^c$ is open, $$ \mu(K^c) \leq \liminf_{n \to \infty} \mu_n(K^c) \leq \epsilon, $$ so $\mu \in \tilde K_\epsilon$. To show compactness, by Prohorov's theorem, we need only show that $\tilde K_\epsilon$ is tight. But this also may not be true (see Counterexample 2 below).

Any suggestions for other strategies I might consider?

Counterexample 1: In general, $\tilde U_{K,\epsilon} \subset \mathcal M_1(E)$ need not be open in the weak* topology. Take $\mu, \mu_n \in \mathcal M_1(\mathbb R)$ to be $$ \mu_n = \frac 1 2 \left(\delta_{-1/n} + \lambda|_{\left( -\frac 1 n, 1-\frac 1 n\right]}\right), \quad \mu = \frac 1 2 \left(\delta_0 + \lambda|_{\left(0,1\right]}\right) $$ where $\lambda$ is Lebesgue measure and $\delta_x$ are Dirac measures. In this case, $\mu([0,1/2]^c) = 1/4$, but $\mu_n([0,1/2]^c) > 1/2$, so using $K = [0,1/2]$ and $\epsilon = 1/2$, $\tilde U_{K,\epsilon}$ isn't sequentially open.

Counterexample 2: In general, $\tilde K_\epsilon \subset \mathcal M_1(E)$ need not be tight. Let $\beta\in (0,\epsilon/2)$, and define $\mu_n$ by $$ \mu_n = \left(1-\frac\epsilon 2\right)\lambda|_{[0,1)} + \frac \epsilon 2 \delta_{n+1} $$ Then $\mu_n\left([0,1]^c\right) = \frac\epsilon 2 < \epsilon$ for all $n$, so $(\mu_n) \subset \tilde K_\epsilon$ for $K = [0,1]$. However, for any $a > 0$, there is an $n \in \mathbb N$ for which $\mu_n\left([-a,a]^c\right) = \epsilon/2 > \beta$. So $\tilde K_\epsilon$ cannot be tight.

Answer 1

$(\Longrightarrow)$ Assume $\mathcal K \subseteq \mathcal M_1(\mathcal M_1(E))$ is tight. Let $\epsilon > 0$. Then there is a set $\tilde L \subseteq \mathcal M_1(E)$ that is compact in the weak* topology for which $\sup_{\tilde \mu \in \mathcal K}\tilde \mu(\tilde L^c)<\epsilon$. Since $\tilde L$ is weak*-compact and $E$ is Polish, by Prohorov's Theorem, $\tilde L \subseteq \mathcal M_1(E)$ is tight. So there is a compact $K \subseteq E$ for which $\mu(K^c) < \epsilon$ for every $\mu \in \tilde L$. In particular, $\tilde L \subseteq \left\{\mu \in \mathcal M_1(E) : \mu(K^c) < \epsilon\right\}$. So, for $\tilde \mu \in \mathcal K$, $$ \tilde\mu\left(\left\{ \mu \in \mathcal M_1(E) : \mu(K^c) > \epsilon\right\}\right) \leq \tilde \mu\left(\tilde L^c\right) < \epsilon. $$

$(\Longleftarrow)$ Let $\epsilon > 0$. For $n \in \mathbb N$, by hypothesis, there is a compact $K_n \subseteq E$ for which $$ \tilde\mu\left(\left\{\mu \in \mathcal M_1(E) : \mu(K_n^c) > \frac{\epsilon}{2^n}\right\}\right) < \frac{\epsilon}{2^n} \quad \forall \tilde\mu \in \mathcal K. $$ Define $\tilde K_n = \left\{\mu \in \mathcal M_1(E) : \mu(K_n^c) \leq \frac{\epsilon}{2^n}\right\}$. Suppose $(\mu_k)_{k \geq 1}$ is a sequence in $\tilde K_n$ that converges (in the weak* topology) to a measure $\mu \in \mathcal M_1(E)$. By the Portemanteau Theorem, since $K^c_n \subseteq E$ is open, $$ \mu(K^c_n) \leq \limsup_{k \to \infty} \mu_k(K^c_n) \leq \frac{\epsilon}{2^n}, $$ so $\mu \in \tilde K_n$, so $\tilde K_n$ is closed. Thus, the set $$ \tilde K := \mathop{\bigcap}_{n=1}^\infty \tilde K_n $$ is also closed. Let $\delta > 0$, and choose $n \in \mathbb N$ so that $\frac{\epsilon}{2^n} < \delta$. Then $K_n$ is compact, and for every $\mu \in \tilde K$, we have that $\mu(K_n^c) \leq \frac{\epsilon}{2^n} < \delta$, so $\tilde K$ is tight. By Prohorov's Theorem, $\tilde K$ is relatively compact in the weak* topology, and since $\tilde K$ is closed, in fact $\tilde K$ is compact. Thus, for $\tilde \mu \in \mathcal K$, $$ \tilde\mu \left(\tilde K^c\right) = \tilde\mu\left(\mathop{\bigcup}_{n=1}^\infty \tilde K^c_n\right) \leq \sum_{n=1}^\infty \tilde\mu\left(\tilde K_n^c\right) < \epsilon. $$ It follows that $\mathcal K$ is tight.