3
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Please change your title to something like "Does Dudley's theorem hold for nonseparable metric spaces?" because the current title makes it sound like you have what you think is one. $\endgroup$
    – user5810
    Oct 24, 2010 at 20:11
  • $\begingroup$ @Ricky Demer: Good point, I have changed it to your suggestion. $\endgroup$
    – JT_NL
    Oct 24, 2010 at 20:18
  • $\begingroup$ Could you give (or link) the definitions of "converges narrowly" and "the bounded Lipschitz metric"? $\endgroup$ Oct 24, 2010 at 20:41
  • $\begingroup$ @Nate: Okay, done! $\endgroup$
    – JT_NL
    Oct 24, 2010 at 21:03
  • 2
    $\begingroup$ If all Borel measures are tight (or just have a separable support set), then your whole sequence has common separable support, so apply the theorem quoted. Now maybe when you say $\mu_i$ you mean not a sequence but a net. For non-tight measures (as least if the metric space is complete) you need to have a (real-valued) measurable cardinal. $\endgroup$ Oct 24, 2010 at 21:40

1 Answer 1

5
$\begingroup$

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 $\int f d\mu$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.