Timeline for Existence of dominating measure for weak*-compact set of measures
Current License: CC BY-SA 3.0
18 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| May 16, 2013 at 14:44 | comment | added | George Lowther | @andy: I can post an answer with proof of my statement later tonight. | |
| May 16, 2013 at 14:23 | comment | added | Davide Giraudo | Not necessarily, but here (I think) we need to show that each totally ordered subset admits a maximal element. | |
| May 16, 2013 at 13:50 | comment | added | andy teich | @Davide : Does a weka*-compact substet necessarily have to be totally ordered? | |
| May 16, 2013 at 9:31 | history | undeleted | Davide Giraudo | ||
| May 16, 2013 at 9:31 | history | edited | Davide Giraudo | CC BY-SA 3.0 |
added 467 characters in body
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| May 15, 2013 at 15:15 | history | deleted | Davide Giraudo | ||
| May 15, 2013 at 12:55 | comment | added | Davide Giraudo | @andy Could you please unaccept my answer. Then I will be able to delete it. | |
| May 15, 2013 at 6:28 | comment | added | andy teich | @George Lowther : do you have a proof for this? So the answer given by Davide Giraudo is not true when the maps $Z$ are bounded? | |
| May 14, 2013 at 21:48 | comment | added | Gerald Edgar | So the set $\{\delta_x : x \in [0,1]\}$ is not compact; it is discrete. | |
| May 14, 2013 at 21:47 | comment | added | George Lowther | @andy: With your modified definition of the weak-* topology, the anwer is now yes, there is a dominating measure. | |
| May 14, 2013 at 19:45 | comment | added | andy teich | @Gerald Edgar: "...the usual way to define it is to use continuous $Z$." Maybe you realized that we are working on measurable spaces and there is no notion of continuity... | |
| May 14, 2013 at 19:43 | comment | added | andy teich | I forgot boundedness! sorry | |
| May 13, 2013 at 18:45 | comment | added | Jochen Wengenroth | Andy does not have a definition at all because his functionals $L_Z$ make sense only for bounded $Z$. Anyway, weak*-compactness could be obtained from the Banach-Alaoglu theorem. | |
| May 13, 2013 at 16:56 | comment | added | Gerald Edgar | @Lutz... you are right. Because andy has the wrong definition for the weak* topology. The usual way to define it is to use only continuous $Z$. If we use all measurable $Z$, as andy does, then you get a much stronger topology. Moreover, andy didn't even say that $Z$ should be bounded. So, let's give him a chance to say whether he wants to correct the definition. | |
| May 13, 2013 at 15:52 | comment | added | Lutz Mattner | Does this really answer the intended Question? With the "weak*-topology" as defined above, the map $x\maspto \delta_x$ is not continuous. | |
| May 13, 2013 at 13:48 | vote | accept | andy teich | ||
| May 15, 2013 at 14:49 | |||||
| May 13, 2013 at 13:46 | history | answered | Davide Giraudo | CC BY-SA 3.0 |