Levy prokhorov metric metrizes weak convergence

The decreasing closed enlargements have finite mass, so F5 applies. If $\pi$ tends to zero, for any $\eta$>0 it is eventually less than $\eta$, so $\eta$ is admissible by the upward-closed admissibility set. Hence for closed F, ${\mathrm{lim\hspace{0.17em}sup}}_n{\mu }_n(F)\le \mu (F^{[\eta ]})+\eta$. Decreasing $\eta$ to zero makes the right-hand side tend to $\mu$(F), by finite measure continuity; for F empty the inequality is immediate. F1 proves weak convergence.

Conversely suppose weak convergence. Fix $\epsilon$>0 and choose $\delta$>0 with 3delta<$\epsilon$. By F2, select a partition with atom diameters less than $\epsilon$. Finitely many atoms $A_1,\dots ,A_m$ cover $\mu$ mass greater than 1-$\delta$, by F3. F1 gives convergence of each atom mass. Thus eventually ${\sum }_{i\le m}\mathrm{\mid }{\mu }_n(A_i)-\mu (A_i)\mathrm{\mid }<\delta$, and the complement of their union has ${\mu }_n$ mass below 2delta.

For any Borel B, let V be the union of those selected atoms meeting B. Then $V\subseteq B^{\epsilon }$ and B is contained in V together with the uncovered complement. Step 1.2 gives ${\mu }_n(B)\le {\mu }_n(V)+2\delta \le \mu (V)+3\delta \le \mu (B^{\epsilon })+\epsilon$. Similarly $\mu (B)\le \mu (V)+\delta \le {\mu }_n(V)+2\delta \le {\mu }_n(B^{\epsilon })+\epsilon$. These bounds hold simultaneously for every B, so F4 gives $\pi$(${\mu }_n$,$\mu$)<=$\epsilon$ eventually. Since $\epsilon$ is arbitrary, $\pi$ tends to zero.