Variant of relative compactness implying variant of tightness Let $\mu_{n, \theta}$ be a family of probability measures on a Polish space for $n \in \mathbb{N}$ and $\theta \in \Theta$. Assume that for every $\theta_n \subseteq \Theta$, the sequence $\mu_{n, \theta_n}$ has a weakly convergent sub-sequence.
I'm trying to determine whether this assumption is sufficient to imply that for all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ and a compact set $K$ such that
$$
\sup_\theta \sup_{n \geq N} \mu_{n, \theta}(K^c) < \varepsilon.
$$
This seems very reminiscent of the "indirect" half of Prohorov's theorem but I struggle to repeat the same argument here as is given in Billingsley. Any ideas? I'm happy to use the ordinary Prohorov's theorem as part of the argument.
 A: Your assertion is correct on $\mathbb R^d$ (or any hemicompact space).  I am not sure what happens on more general spaces.
Let $B_n$ be an (inclusion-wise) increasing sequence of compacta such that any compact $K$ is contained in $B_n$ for some $n$ - on $\mathbb R^d$ you can take $B_n=\{x:\|x\|\le n\}$.  Suppose that the assertion is wrong.  Then there exists $\epsilon>0$ such that for all $N\in\mathbb N$ there are $k_N\ge N$ and $\rho_N\in\Theta$ such that $\mu_{k_N,\rho_N}(B_N)\le 1-\epsilon$.  Since $k_N\ge N$ we can extract a strictly increasing subsequence $N_l$ such that $k_{N_l}$ is strictly increasing.  Define a new sequence $\theta_n$ by $\theta_n=\rho_{N_l}$ if $k_{N_l}=n$ and $\theta_n=\rho_1$ if no such $l$ exists.  The sequence $\mu_{n,\theta_n}$ is relatively compact by assumption, and so is its subsequence $\mu_{k_{N_l},\theta_{k_{N_l}}}$.  Therefore, it must be tight.  But any compact $K$ is included in $B_{{N_l}}$ for $l$ large,  so $$\limsup_{l\to\infty}\mu_{k_{N_l},\theta_{k_{N_l}}}(K)\le \limsup_{l\to\infty}\mu_{k_{N_l},\theta_{k_{N_l}}}(B_{{N_l}})= \limsup_{l\to\infty}\mu_{k_{N_l},\rho_{N_l}}(B_{{N_l}})\le 1-\epsilon,$$
which contradicts tightness.
