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.



Does Dudley's theorem hold for nonseparable metric spaces?
Let $X$ be a set with $2$-valued measurable cardinal. (Real-valued measurable can also be done, but with some more complications, so I do not do that now.) Give it the discrete metric. Let $\mu$ be a countably-additive measure on the Borel sets (i.e., the power set) with values $0$ and $1$ such that each singleton has measure $0$ but $\mu(X) = 1$. There is a net $\mu\_i$ of point-masses converging to $\mu$ narrowly, but not in the BL metric. 


A **point-mass** is a measure that assigns measure $1$ to a certain singleton, and measure $0$ to the complement. As long as our net of point-masses is eventually outside each set of measure $0$, we have convergence to $\mu$ in the narrow topology. But any point-mass $\mu\_i$ at the point $a\_i$ is far away from $\mu$ in the BL topology, since the indicator function of the singleton $a\_i$ is a BL function with norm $2$. 


Another note. For any bounded function $f \colon X \to \mathbb{R}$, there is a set $F\subseteq X$ with $\mu(F)=1$ and $f$ is constant on $F$; the constant value there is the integral $\int f d\mu$.


