Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why is the weak convergence of probability measures on $\mathbb{R}$ with respect to bounded continuous test functions $C^0_b(\mathbb{R})$ metrizable by the bounded Lipschitz metric $$d(\mu, \nu) = \sup_{f \in \text{Lip}(\mathbb{R})} \Big | \int_{\mathbb{R}} f d \nu - \int_{\mathbb{R}} f d \mu \Big |$$ where $$\text{Lip}(\mathbb{R}) = \Big \{ f \in C_b(\mathbb{R}) : \sup_x |f(x) | \leq 1, \sup_{x \neq y} \frac{| f(x) - f(y) |}{|x-y|} \leq 1 \Big \}?$$ For those who would like a reference, this is invoked in the proof of the truncated version of Wigner's semicircle law in Anderson-Guionnet-Zeitouni's $\textit{Introduction to Random Matrices}$ and is cited in the appendix as part of Theorem C.8, though no proof is given there. If anyone could help me with this fact, I'd greatly appreciate it!

share|cite|improve this question
Are $\mu$ and $\nu$ probability measures? I think that a uniform bound of some kind must be given in order to metrize a weak convergence. – Giuseppe Negro Jan 7 '13 at 2:10
Thanks for pointing this out - $\mu$ and $\nu$ are definitely probability measures! I'll fix this. – StackQs Jan 7 '13 at 17:08
up vote 2 down vote accepted

There is a proof in Section 8.3 of Bogachev's Measure Theory.

share|cite|improve this answer
And another one in the book by Ambrosio, Gigli, and Savaré, Gradient Flows in Metric Spaces and in the Space of Probability Measures. It is Proposition 7.1.5 on pages 154-155. They prove a more general result for $p$-Wasserstein distance, for which convergence is equivalent to weak ("narrow") plus uniform integrability of $p$th moments. – user53153 Jan 7 '13 at 3:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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