Dudley's theorem (1966) states that if $(X, d)$ is a metric space and if $X$ is separable and $\mu$, $\mu_i$ are Borel probability measures then $\mu_i \to \mu$ narrowly iff $d_{\text{BL}}(\mu_i, \mu) \to 0$ where $d_{\text{BL}}$ is the bounded Lipschitz metric.

**Definitions:**
$(\mu_i)$ converges narrowly to $\mu$ (where all measures are Borel probability measures) if

$$\int f \, d\mu_i \to \int f \, d\mu \text{ for all $f$ bounded and continuous on $X$}$$

The bounded Lipschitz metric is a metric on the space $\text{BL}(X,d) := \{f : X \to \mathbb{R} : f \text{ is bounded and Lipschitz} \}$. Then define

$$d_\text{BL}(\mu, \nu) := \sup \left \{ \left | \int f \, d\mu - \int f \, d\nu\, \right | : f \in \text{BL}(X,d), \|f\|_\text{BL} \leq 1 \right \}$$

where $\|f\|_{\text{BL}}$ is the sum of the Lipschitz-norm and the $\infty$-norm.

The proof uses Arzela-Ascoli, but I wonder what would be a counterexample if $X$ isn't separable? From right-to-left still works.

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