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?
Unless I'm missing something this is fairly straightforward, because we're given that the limit is actually a probability measure.
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?