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I have the following question:

Let $E$ be a polish space (that is, a topological space, which is separable and metrizable, such that $E$ would be complete if equipped with this metric).

Consider the space of probability measures on $E$, denoted by $\mathcal M (E)$ equipped with the topology of weak convergence.

The claim I am struggling with is the following:

For every $\ell\in\Bbb N$ let $K_\ell$ be a compact set of $E$. Then the set $$\bigcap_{\ell\in\Bbb N} \{ m \in \mathcal M (E): m(K^{\operatorname{c}}_\ell) \leq \frac 1 \ell \}$$ is compact.

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I hope I understand the full meaning of Prokhorov in the right way:

In this setting (i.e. $E$ polish) the set $\mathcal M (E)$ is metrizable with a complete metric, say $d$.

The set $M := \cap_{\ell\in\Bbb N} \{ m : m(K_{\ell}^{\operatorname{c}}) \leq \frac 1 \ell \} $ is tight (For any $\varepsilon > 0$ just choose $\ell$ such that $\frac 1 \ell \leq \varepsilon$ and take $K_\ell$). By Prokhorov this means that the closure of $M$ with respect to $d$ is compact.

Further, we have that for $\eta >0$ and $K\subset E$ compact the sets of the form $\{ m : m(K^{\operatorname{c}}) \leq \eta \}$ are closed with respect to $d$. This can be seen by the following: Take a convergent sequence sequence $(m_n)_{n\geq 1}$ with $m_n (K^{\operatorname{c}}) \leq \eta$ for every $n$. Define $m := \lim_{n\to\infty} m_n$. Since the topology induced by $d$ is equivalent to the topology of weak convergence we have the portmanteau theorem, such that: $$m(K^{\operatorname{c}}) \leq \liminf_{n\to\infty} m_n (K^{\operatorname{c}}) \leq \eta$$ All in all this means $M$ is compact.

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