Metrizability of the set of probability measures with the Kantorovich–Rubinstein metric

147 Views Asked by At

Let $X$ be a compact metric space and let $\mathcal{M}(X)$ be the set of probability measures. Then the weak topology on $\mathcal{M}(X)$ is metrizable, for example with the Wasserstein-metric and with the Prokhorov-metric, which seem to be the most famous ones. But both metrics seem kind of unhandy to me. Now I read something about a third metric, the Kantorovich–Rubinstein metric, which should be compatible to the weak topology, too. It is defined by

$$\mathrm{d}(\mu,\nu)=\sup\left\{\left|\int f\mathrm{d}\mu-\int f\mathrm{d}\nu\right|,~f\in\mathrm{Lip}_1(X,\mathbb{R}),~\sup_{x\in X}f(x)\leq 1\right\},$$

and I'm wondering if there is an easy way to show that it indeed metrizes the weak topology. Especially I can't figure out, why an open ball wrt. this metric, which should be defined as

$$B_\epsilon(\bar{\mu})=\left\{\mu\in\mathcal{M}(X):\left|\int f\mathrm{d}\mu-\int f\mathrm{d}\bar{\mu}\right|<\epsilon~\forall f\in\mathrm{Lip}_1(X,\mathbb{R}),~\sup_{x\in X}f(x)\leq 1\right\},$$

is indeed open in the weak topology. $\mathrm{Lip}_1(X,\mathbb{R})$ should be separabel but I don't know how this can help. Do you have any hints for me how one can prove that?