# Bounded Lipschitz metric convergence implies weak convergence of probability measures

Given $d(\mu, \nu) = \sup_{f \in \mathcal{L}_b^1} \left| \int f d\mu - \int f d\nu \right|$ for $\mu, \nu \in \mathcal{P}(\mathbb{R})$ the probability measures on $\mathbb{R}$ and where $\mathcal{L}_b^1$ constitutes the 1-Lipschitz functions bounded by $1$. Suppose we have $d(\mu_n, \mu) \to 0$. We want to show that $\mu_n \to \mu$ in distribution, that is, weakly. That is, we want to show that $\int f d\mu_n \to \int f d\mu$ for any bounded continuous function.

My progress so far is as follows. $d(\mu_n, \mu) \to 0$ gives us that $\int f d\mu_n \to \int f d\mu$ for all $f$ Lipschitz. By the Weierstrass Approximation theorem we have that Lipschitz functions are dense in $C_b^0([-M, M])$ for any $M$ real. If we could show that $d(\mu_n, \mu) \to 0$ implies tightness of $\mu_n$, then we could use this density result to extend the limits to all continuous bounded $f$. Is this the right way to go about this or is there a better way? I haven't been able to establish the tightness sought. Could someone help with that?

• What's the actual definition of $d(\mu,\nu)$? It can't be what you said, because that implies $d(\mu,\nu)=\infty$ unless $\mu=\nu$. – David C. Ullrich Mar 7 '18 at 19:14
• @DavidC.Ullrich I amended the definition slightly. It's called the Wasserstein distance if I'm not mistaken. The measures are probability measures so distance infinity can be avoided now. With this in place, is there some way you would recommend proceeding? – user423750 Mar 7 '18 at 20:54

Let $\epsilon>0$. Choose $A$ so $$\mu([-A,A])>1-\epsilon.$$There exists a Lipschitz funtion $f$ with $0\le f\le 1$, $f=1$ on $[-A,A]$, such that $f$ vanishes off $[-A-1,A+1]$. In other words $$\chi_{[-A,A]}\le f\le\chi_{[-A-1,A+1]}.$$If $n$ is large enough then $$\mu_n([-A-1,A+1])\ge\int f\,d\mu_n\ge\int f\,d\mu-\epsilon\ge\mu([-A,A])-\epsilon>1-2\epsilon.$$So, taking the union of $[-A-1,A+1]$ with finitely may compact sets, one for each $\mu_n$ where $n$ is not "large enough", there is a compact set $K$ so $\mu_n(K)>1-2\epsilon$ for every $n$. That's what "tight" means here, right?