Does Dudley's theorem hold for nonseparable metric spaces?

2010-10-24T20:01:06Z

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.

Answer by Gerald Edgar for Does Dudley's theorem hold for nonseparable metric spaces?

2010-10-25T02:06:49Z

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 $\int f d\mu$.

Does Dudley's theorem hold for nonseparable metric spaces? - MathOverflow

Does Dudley's theorem hold for nonseparable metric spaces?

Answer by Gerald Edgar for Does Dudley's theorem hold for nonseparable metric spaces?