Timeline for Measure of the support of a Borel probability on a metric space
Current License: CC BY-SA 2.5
9 events
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| Nov 1, 2010 at 15:34 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| Nov 1, 2010 at 15:27 | comment | added | Joel David Hamkins | That is, if there is a countably complete non-principal ultrafilter on a set, then there is a measurable cardinal, and your new answer is the same as my answer. | |
| Nov 1, 2010 at 15:23 | comment | added | Joel David Hamkins | About your revised answer: the filter should be countably closed to have an additive measure, and it needs to measure all the Borel sets, which in the discrete case means all sets. So it has to be an ultrafilter. And this takes a measurable cardinal. | |
| Nov 1, 2010 at 15:18 | history | undeleted | Pietro Majer | ||
| Nov 1, 2010 at 15:18 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| Nov 1, 2010 at 15:11 | history | deleted | Pietro Majer | ||
| Nov 1, 2010 at 15:10 | comment | added | Pietro Majer | ops I was initially thinking of the restricted sigma algebra indeed | |
| Nov 1, 2010 at 15:02 | comment | added | Joel David Hamkins | Since there are disjoint uncountable sets, this isn't a measure (it's not additive). It is a measure on the sigma-algebra of countable/co-countable sets, but that algebra does not contain all the Borel sets in your space. | |
| Nov 1, 2010 at 14:33 | history | answered | Pietro Majer | CC BY-SA 2.5 |