Proposition 1.6.3.label Let $\seq{\mu_n}\subset \mathbf{M}_{1}(\Omega)$, then $\seq{\mu_n}$ is weakly precompact if and only if the following holds:
- (1)
$\seq{(\pi_0)_*\mu_n}$ is tight.
- (2)
For any $N \in \nat$ and $\eta > 0$,
\[\lim_{\delta \to 0}\sup_{n \in \nat}\mu_{n}(\bracs{V_{\delta, N} > \eta}) = 0\]
Proof. Using Prokhorov’s theorem, it is sufficiently to show that $\seq{\mu_n}$ is tight.
To this end, let $\eps > 0$, then there exists $K \subset \real$ compact such that
\[\sup_{n \in \nat}\mu_{n}(\bracs{\pi_0 \not\in K}) \le \eps\]
Fix $N \in \nat$, then for each $k \in \nat$, there exists $\delta(N, k) > 0$ such that
\[\sup_{n \in \nat}\mu_{n}\paren{\bracs{V_{\delta(N, k), N} > \frac{1}{k}}}\le 2^{-N-k}\cdot \eps\]
In which case,
\[\sup_{n \in \nat}\mu_{n}\paren{\bracs{\omega \in \Omega \bigg | \forall k \in \nat, V_{\delta(N, k), N}(\omega) \le \frac{1}{k}}^c}\ge 1 - 2^{-N}\eps\]
Let $K'$ be the intersection of $\bracs{\pi_0 \in K}$ and the above compact set, then $K'$ is precompact by the Arzela-Ascoli theorem, and $\sup_{n \in \nat}\mu_{n}(\Omega \setminus K') \le 2\eps$.$\square$