Prokhorov tightness theorem on polish spaces

The empty family satisfies both definitions vacuously. Otherwise fix a compatible complete metric on S. By F1 there is a homeomorphic embedding e into $H=[0,1{]}^{\mathbb{N}}$. H has complete metric $\rho (x,y)={\sum }_{j=0}^{\mathrm{\infty }}{2}^{-(j+1)}\mathrm{\mid }{x}_j-y_j\mathrm{\mid }$ by F2. It is totally bounded: choose an integer $N\ge 1$ with tail ${\sum }_{j\ge N}{2}^{-(j+1)}<\eta \mathrm{/}2$ and a finite mesh in coordinates $0,\dots ,N-1$ with weighted error below $\eta$/2, putting zero in later coordinates. AC restricted to countable nonempty families gives the countable choice required by F3, which makes H compact.

Assume tightness and take any sequence ${\mu }_n$ in the family. Push it forward by e. F4 gives a subsequence ${\nu }_{n_j}=e_{\ast }{\mu }_{n_j}\Rightarrow \nu$ on H. For every integer $m\ge 1$, choose compact $K_m$ in S with all ${\mu }_n$($K_m$)>1-1/m. Their images are compact and closed in H, so F5 gives $\nu (e(K_m))\ge {\mathrm{lim\hspace{0.17em}sup}}_j{\nu }_{n_j}(e(K_m))\ge 1-1\mathrm{/}m$. Hence the Borel set $E={\bigcup }_{m\ge 1}e(K_m)\subseteq e(S)$ has $\nu$ mass one.

For Borel B in S, e(B) is Borel relative to e(S). Since E is ambient Borel and contained in e(S), $E\cap e(B)$ is Borel in H. Define $\mu (B)=\nu (E\cap e(B))$; disjoint unions are preserved and $\mu$(S)=$\nu$(E)=1. For closed F in S there is a closed Z in H with $Z\cap e(S)=e(F)$, by the relative topology. Then ${\mu }_{n_j}(F)={\nu }_{n_j}(Z)$ and $\nu (Z)=\nu (Z\cap E)=\mu (F)$. The closed bound from F5 gives ${\mathrm{lim\hspace{0.17em}sup}}_j{\mu }_{n_j}(F)\le \mu (F)$, hence weak convergence on S. This proves tightness implies relative sequential compactness without assuming e(S) is Borel.

For the reverse, let $U_i$ be any countable open cover of S. Fix $\epsilon$>0. If no finite initial union works uniformly, AC selects ${\mu }_n$ in the family with ${\mu }_n({\bigcup }_{i\le n}{U}_i)\le 1-\epsilon$. Relative sequential compactness gives a weakly convergent subsequence with probability limit $\mu$. For any fixed r, eventually $n_j$>=r, so for the fixed open set $G_r={\bigcup }_{i\le r}{U}_i$ one has ${\mu }_{n_j}(G_r)\le 1-\epsilon$ eventually. The open bound in F5 yields $\mu (G_r)\le {\mathrm{lim\hspace{0.17em}inf}}_j{\mu }_{n_j}(G_r)\le 1-\epsilon$. F6 as r tends to infinity would give $\mu$(S)<=1-$\epsilon$, a contradiction. Thus a uniform finite initial union exists.

Apply step 1.4 to the dense-center ball cover of radius $2^{-m}$ and loss $\epsilon 2^{-m-1}$ at each $m\ge 1$. Let $C_m$ be the corresponding finite union of closed balls and $K={\bigcap }_m{C}_m$. F7 bounds every $\mu$(S\K) by ${\sum }_m\epsilon 2^{-m-1}<\epsilon$. K is closed in the complete S and is totally bounded: for any $\eta$ choose m with $2^{1-m}<\eta$ and select one point of K from each of the finitely many balls meeting it. These form an $\eta$-net. F3 makes K compact, proving tightness.