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Jul 29, 2023 at 1:05 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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Apr 4, 2022 at 23:11 comment added Julian Newman To be precise: if $B$ is any separable metric space, then the Borel $\sigma$-algebra of the topology of weak convergence on $\mathcal{P}_B$ is precisely the $\sigma$-algebra generated by the set of evaluation mappings $\{\mu \mapsto \mu(A) : A \in \Sigma_B\}$. [Any proof where the result is stated for "Polish spaces" should work for more general separable metric spaces - i.e. being completely metrisable shouldn't play a role.]
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Mar 5, 2022 at 21:10 answer added Steve timeline score: 3
Jan 19, 2022 at 18:20 comment added ECL Thanks! I'll have a look!
Jan 19, 2022 at 18:16 comment added Michael Greinecker The answer is yes. I don't have them at hand, but you can find a proof in the discrete time optimal control book by Bertsekas and Shreve, or (an appendix of) the Bayesian nonparametrics book by Ghosal and van der Vaart.
Jan 19, 2022 at 16:59 history edited ECL CC BY-SA 4.0
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S Jan 19, 2022 at 15:40 review First questions
Jan 19, 2022 at 15:52
S Jan 19, 2022 at 15:40 history asked ECL CC BY-SA 4.0