The bounded-Lipschitz metric characterizes weak convergence on Polish spaces

Nonnegativity, symmetry and the triangle inequality for ${\beta }_{BL}$ follow by applying these properties to the difference of the two test integrals and then taking the supremum. If ${\beta }_{BL}(\nu ,\rho )=0$, fix a nonempty closed $F$ and put $g_m=(1-md(\cdot ,F){)}_{+}$. The function $g_m\mathrm{/}m$ is an admissible test, so $\int g_{m}d\nu =\int g_{m}d\rho$. By [F4], these integrals decrease to $\nu (F)$ and $\rho (F)$, respectively. The two probabilities therefore agree on every closed set, including the empty set. Closed sets form a pi-system generating the Borel sigma-algebra, and both total masses are one, so [F6] gives $\nu =\rho$. Thus ${\beta }_{BL}$ is a metric; this argument uses no choice.

Assume ${\beta }_{BL}({\mu }_n,\mu )\to 0$, and fix a nonempty closed set $F\subseteq S$. For every integer $m\ge 1$, the function $g_m(x)=(1-md(x,F){)}_{+}$ satisfies $0\le g_m\le 1$ and ${\mathrm{Lip}}_d(g_m)\le m$. Hence $g_m\mathrm{/}m$ is admissible in the definition of ${\beta }_{BL}$, so $\int g_{m}d{\mu }_n\to \int g_{m}d\mu$. Since $1_F\le g_m$, it follows that ${\mathrm{lim\hspace{0.17em}sup}}_n{\mu }_n(F)\le \int g_{m}d\mu$ for every $m$.

Now assume ${\mu }_n\Rightarrow \mu$, and put $\mathcal{A}=\{\mu \}\cup \{{\mu }_n:n\ge 1\}$. Any sequence chosen from $\mathcal{A}$ either has a member repeated infinitely often, yielding a constant convergent subsequence, or has a subsequence of the form $({\mu }_{n_j})$ with $n_j\to \mathrm{\infty }$, which converges weakly to $\mu$. Therefore $\mathcal{A}$ is relatively sequentially compact for weak convergence.

The functions $g_m$ decrease pointwise to $1_F$ and are bounded by the integrable constant function $1$ under the probability $\mu$. By [F4], $\int g_{m}d\mu \to \mu (F)$, so the closed-set bound holds; it is trivial for $F=\mathrm{\varnothing }$. The Portmanteau theorem [F2] gives weak convergence in the sense of [F1].

By [F5], $S$ is Polish; therefore [F3] applies under [A1] and $\mathcal{A}$ is tight. Thus for every $\epsilon >0$ there is a compact $K\subseteq S$ such that ${\sup }_{\nu \in \mathcal{A}}\nu (K^c)<\epsilon$.

Fix $\eta >0$. Compactness gives a finite $\eta \mathrm{/}8$-net $x_1,\dots ,x_q$ of $K$. Quantize each value $g(x_i)\in [-1,1]$ into finitely many bins of width at most $\eta \mathrm{/}2$. There are finitely many resulting value patterns for the bounded 1-Lipschitz test class; choose one representative $h_r$ from each nonempty pattern. If $g$ and $h_r$ share a pattern, then for $x\in K$ a net point $x_i$ gives $\mathrm{\mid }g(x)-h_r(x)\mathrm{\mid }\le 2d(x,x_i)+\mathrm{\mid }g(x_i)-h_r(x_i)\mathrm{\mid }<\eta$.

For any admissible $g$, choose the representative $h_r$ from its pattern. The integral difference is bounded by $\mid \int gd{\mu }_n-\int gd\mu \mid \le {\max }_r\mid \int h_{r}d{\mu }_n-\int h_{r}d\mu \mid +2\eta +2{\mu }_n(K^c)+2\mu (K^c)\mathrm{.}$ The maximum tends to zero by [F1], because there are finitely many bounded continuous representatives, and the last two terms are at most $4\epsilon$ by step 2.2. Taking the supremum over $g$, then the limit superior in $n$, gives ${\mathrm{lim\hspace{0.17em}sup}}_n{\beta }_{BL}({\mu }_n,\mu )\le 2\eta +4\epsilon$. Letting $\eta ,\epsilon \downarrow 0$ proves the reverse implication.