Portmanteau theorem

(i) implies (ii) because a uniformly continuous function is continuous. Suppose (ii), and let F be nonempty and closed. By F1, $f_m(x)=\max (0,1-md(x,F))$ is bounded and uniformly continuous. Moreover $f_m\downarrow {\mathbf{1}}_F$. Thus ${\mathrm{lim\hspace{0.17em}sup}}_n{\mu }_n(F)\le \int f_{m}d\mu$ for every m. F2 with majorant one gives (iii) as m tends to infinity. For F empty the inequality is zero<=zero.

For G open, apply (iii) to its closed complement and use ${\mu }_n(G)=1-{\mu }_n(S\setminus G)$ to obtain (iv). Conversely the same complement calculation obtains (iii) from (iv). If A is a Borel continuity set, $A^{\circ }\subseteq A\subseteq \bar{A}$ and $\mu (A^{\circ })=\mu (A)=\mu (\bar{A})$. The open lower bound and closed upper bound therefore squeeze ${\mu }_n(A)$ to $\mu (A)$, proving (v).

Assume (v), and fix a bounded continuous real f and $\eta$>0. The disjoint level sets with $\mu (f=t)\ge 1\mathrm{/}r$ number at most r for each positive integer r. Their union over r contains all positive-mass levels and is countable (each finite subset of the real line can be listed in increasing order). Choose finitely many increasing levels $t_0<\cdots <t_m$ outside this countable set, with $t_0<-\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, $t_m>\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and mesh below $\eta$. Such levels exist in every open interval, since an interval is uncountable.

For $A_j=\{t_{j-1}\le f<t_j\}$, continuity of f gives $\mathrm{\partial }{A}_j\subseteq \{f=t_{j-1}\}\cup \{f=t_j\}$, so (v) applies. The simple function $s={\sum }_j{t}_{j-1}{\mathbf{1}}_{A_j}$ satisfies $\mathrm{\mid }f-s\mathrm{\mid }\le \eta$ everywhere. Therefore $\mathrm{\mid }\int fd{\mu }_n-\int fd\mu \mathrm{\mid }\le 2\eta +\mathrm{\mid }{\sum }_j{t}_{j-1}({\mu }_n(A_j)-\mu (A_j))\mathrm{\mid }$. The finite sum tends to zero, by step 1.3 and (v). Letting $\eta$ tend to zero proves (i), closing all equivalences.