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Jan 4, 2016 at 7:36 comment added Guillaume Dehaene @PaulSiegel: thank you for the answer ! However, there is a simple metric topology which I think works (though, since I've forgotten my topology, I'm not completely sure what "making the map continuous" is exactly). That's the Wasserstein-1 metric. More generally, the Wasserstein-k metric (I think) works for all statistics with at most polynomial growth
Dec 21, 2015 at 1:28 comment added Paul Siegel @YemonChoi Arg, continuity of that map is essentially a tautology. After recovering from flashbacks to my qualifying exams in graduate school, I amended the answer.
Dec 21, 2015 at 1:26 history edited Paul Siegel CC BY-SA 3.0
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Dec 21, 2015 at 0:45 comment added Yemon Choi I sort of agree with the basic point, but the choice of example in your first paragraph is unfortunate, if I am not mistaken; the map $f\to \int f$ is continuous as a linear functional on the normed space $L^1({\bf R})$
Dec 20, 2015 at 23:33 history edited Paul Siegel CC BY-SA 3.0
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Dec 20, 2015 at 23:27 history edited Paul Siegel CC BY-SA 3.0
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Dec 20, 2015 at 23:25 comment added Paul Siegel Uniform integrability reminds me a little of equicontinuity. I wonder if the theorem above is a special case of a suitable generalization of the Arzela-Ascoli theorem.
Dec 20, 2015 at 23:18 history answered Paul Siegel CC BY-SA 3.0