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S Aug 13, 2013 at 1:20 history suggested Michael Albanese CC BY-SA 3.0
Replaced \\, by \,.
Aug 13, 2013 at 1:11 review Suggested edits
S Aug 13, 2013 at 1:20
Oct 25, 2010 at 6:45 vote accept CommunityBot moved from User.Id=5295 by developer User.Id=481663
Oct 25, 2010 at 2:06 answer added Gerald Edgar timeline score: 5
Oct 24, 2010 at 22:43 history edited Jonas T CC BY-SA 2.5
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Oct 24, 2010 at 22:38 comment added Thierry Zell Could you also define $\|f\|_{BL}$?
Oct 24, 2010 at 22:27 comment added Jonas T @Gerald: Okay, but that happens for instance if our space is separable and complete (that all finite Borel measures are tight) and in this question we don't assume separability. Am I missing something?
Oct 24, 2010 at 21:40 comment added Gerald Edgar 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.
Oct 24, 2010 at 21:09 history edited Jonas T CC BY-SA 2.5
added 2 characters in body; added 4 characters in body
Oct 24, 2010 at 21:03 comment added Jonas T @Nate: Okay, done!
Oct 24, 2010 at 21:02 history edited Jonas T CC BY-SA 2.5
added 574 characters in body
Oct 24, 2010 at 20:41 comment added Nate Eldredge Could you give (or link) the definitions of "converges narrowly" and "the bounded Lipschitz metric"?
Oct 24, 2010 at 20:18 comment added Jonas T @Ricky Demer: Good point, I have changed it to your suggestion.
Oct 24, 2010 at 20:17 history edited Jonas T CC BY-SA 2.5
edited title
Oct 24, 2010 at 20:11 comment added user5810 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.
Oct 24, 2010 at 20:01 history asked Jonas T CC BY-SA 2.5