Timeline for Continuous Strictly Positive Measures on Countable Boolean Algebras
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jun 27, 2012 at 8:46 | comment | added | Joel David Hamkins | François, you are right. I was thinking that one could get a measurable cardinal out of it, but your example shows that this version of countable additivity does not suffice for that. | |
| Jun 27, 2012 at 7:00 | vote | accept | provocateur | ||
| Jun 27, 2012 at 5:21 | answer | added | Joseph Van Name | timeline score: 7 | |
| Jun 27, 2012 at 1:46 | comment | added | François G. Dorais | Joel, I'm not sure none of them are. For example, consider the ultrafilter $\mathcal{U} = \lbrace a \in B : \sqrt2 \in a \rbrace,$ where $B$ is the Boolean algebra generated by the intervals $[p,q)$ where $p,q \in \mathbb{Q}$. It seems that the measure associated with $\mathcal{U}$ is actually countably additive, in the sense described in the question. | |
| Jun 27, 2012 at 0:52 | comment | added | provocateur | Yes, that's the problem, I don't see how that construction guarantees countable additivity - in particular, the ultrafilters mentioned in the construction there need not be closed under countable limits. Hence my question. | |
| Jun 27, 2012 at 0:18 | comment | added | Joel David Hamkins | Jason, it appears that Francois's answer provides only a finitely additive measure, rather than a countably additive one. For example, few of the functions $m_n$ in that answer are countably additive, and in the atomless case, none of them are. | |
| Jun 26, 2012 at 23:16 | comment | added | Jason Rute | Doesn't the answer you linked to work? | |
| Jun 26, 2012 at 22:31 | history | edited | Asaf Karagila♦ |
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| Jun 26, 2012 at 22:24 | history | asked | provocateur | CC BY-SA 3.0 |