Timeline for Is the space of Radon measures a Polish space or at least separable?
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19 events
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| Dec 15, 2023 at 10:20 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Typo fixed
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| Jul 7, 2023 at 7:38 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Added a **Reference** section
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| S Jul 7, 2023 at 6:42 | history | suggested | ViktorStein | CC BY-SA 4.0 |
fixed tex formatting
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| Jul 6, 2023 at 19:36 | review | Suggested edits | |||
| S Jul 7, 2023 at 6:42 | |||||
| Oct 8, 2019 at 9:34 | vote | accept | Mark | ||
| Sep 12, 2019 at 0:33 | answer | added | Mikhail Ostrovskii | timeline score: 2 | |
| Sep 10, 2019 at 7:57 | comment | added | Mark | @user131781: Thanks for the comment. I am new to all of this so of course I didn't new those results.I am especially interested in the last sentence of yours: If the underlying space is Polish, than the space of Radon probability measures under the weak topology is Polish. | |
| Sep 10, 2019 at 7:53 | comment | added | Mark | @NateEldredge: Thanks for the comments. Besides weak* topology I am sure that I won't use all signed Radon measures in the problem I am facing. I will probably use finite or positive or probability Radon measures etc. Not sure precisely which one for my problem, but I hope I will figure it out in a few days. And it is very interesting to me how different subsets of all Radon measures behave differently. | |
| Sep 10, 2019 at 7:45 | comment | added | Mark | @JochenWengenroth: Thanks for the comment. I am also pretty sure that I need to use some weak* topology. I need just to figure it how. | |
| Sep 10, 2019 at 0:47 | history | became hot network question | |||
| Sep 9, 2019 at 18:35 | answer | added | Robert Furber | timeline score: 3 | |
| Sep 9, 2019 at 18:11 | comment | added | Nate Eldredge | @RobertFurber: Right, that's what I mean. The space $M(\Omega)$ defined in the question is the space of all signed Radon measures, i.e. the dual of $C(\overline{\Omega})$ (or something like that), and this is not weak-* metrizable (nor, as you say, even first countable). | |
| Sep 9, 2019 at 17:57 | comment | added | Robert Furber | @NateEldredge Do you mean as opposed to Radon probability measures? The set of Radon probability measures on a Polish space is Polish as a subspace of the weak-* topology, even though the weak-* topology isn't first-countable on the whole space $C_b(X)^*$. | |
| Sep 9, 2019 at 17:55 | comment | added | user131781 | It is a long established fact that the space of Radon measures on a completely regular space is naturally identifiable with the dual of the space $C^b(S)$ of bounded, continuous functions thereon (with the so-called strict topology—-the finest locally convex topology which coincides with compact convergence on the unit ball). This was shown by Buck in the locally compact case in the 50’s and extended to the general case in the 60‘s. If the underlying space is polish, then so is that of the Radon probability measures (under the weak topology induced by $C^b(S)$)—-again a celebrated result. | |
| Sep 9, 2019 at 17:32 | comment | added | Nate Eldredge | @JochenWengenroth: On the other hand, the space of all Radon measures with the weak-* topology is not metrizable, though it is separable. | |
| Sep 9, 2019 at 17:32 | answer | added | Yuval Peres | timeline score: 9 | |
| Sep 9, 2019 at 17:28 | comment | added | Jochen Wengenroth | You have better chances with the weak $^*$ topology. The dual unit ball of a separable Banach space is weak$^*$ compact and metrizable. | |
| Sep 9, 2019 at 17:10 | answer | added | Michael Greinecker | timeline score: 8 | |
| Sep 9, 2019 at 16:35 | history | asked | Mark | CC BY-SA 4.0 |