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Let $S$ be a separable metric space, and $\mu,\mu_1,\mu_2,\ldots$ be Borel probability measures on $S$. We know that $\mu_n \to \mu$ weakly if and only if $\pi(\mu_n,\mu) \to 0$ where $\pi$ is the Prokhorov metric. However, this (a priori) doesn't imply that the metric topology induced by $\pi$ is the same as the topology of weak convergence of probability measures (which is the smallest topology such that $\mu \mapsto \int f \,d\mu$ is continuous for all continuous bounded $f:S \to \mathbb{R}$), since the topology of weak convergence of probability measures might not be first-countable. So my question is: is the topology of weak convergence of probability measures on $S$ first-countable? Thank you in advance.

Davide Giraudo
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Wooyoung Chin
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  • See https://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric – Kavi Rama Murthy Jul 15 '18 at 11:55
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    @KaviRamaMurthy That Wikipedia articles references the book by Billingsley. The proof that the Prokhorov metric metrizes weak convergence in that book is incomplete exactly because it does not show the weak topology is first countable. – Michael Greinecker Jul 15 '18 at 12:00
  • Billingsley's book clearly states that if $S$ is separable then the space $Z$ of Borel probability measures on it with the topology of weak convergence is metrizable (and separable). This is in line14 of page 239. If you show this fact it follows as a consequence that weak convergence topology is first countable. – Kavi Rama Murthy Jul 15 '18 at 12:15
  • @KaviRamaMurthy I'm not sure which book you are talking about. The second edition of "Convergence of Probability Measures" has nothing like that on page 239. The metrizability of weak convergence is discussed when discussing the Prohorov metric on page 72. Billingsley shows that the Prohorov metric gives the same convergent sequences as the weak topology. But in order to show that this implies that the Prohorov metric metrizes weak convergence, one needs to show this either for general nets or prove that the topology of weak convergence is first countable. – Michael Greinecker Jul 15 '18 at 12:24
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    I probably have the first edition. The statement I am quoting appears exactly four lines above the heading "Prohorov's Theorem" – Kavi Rama Murthy Jul 15 '18 at 12:27
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    @KaviRamaMurthy I just looked at the first edition and you are right. This seems to be one of the cases where earlier editions are better. – Michael Greinecker Jul 15 '18 at 12:38

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