Timeline for Completeness of Borel measure
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2013 at 23:39 | answer | added | George Lowther | timeline score: 1 | |
| Sep 9, 2013 at 9:17 | review | Close votes | |||
| Sep 9, 2013 at 11:09 | |||||
| Sep 8, 2013 at 16:23 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced deprecated tag 'analysis' (currently cleaning up tag 'analysis')
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| Mar 29, 2013 at 20:42 | comment | added | orlandoweber | @Pietro: I understand that it follows from [ams.org/journals/proc/1957-008-01/S0002-9939-1957-0085475-7/…, Theorem 5] that $X$ must contain a perfect set since it's assumed to be atomless. But it's not obvious to me why we could assume such a perfect set to have measure zero. Is this a well-known result? Could you please give me a hint or a reference? Thanks! | |
| Feb 11, 2012 at 0:05 | comment | added | Bill Johnson | @Ramiro: I also do not see that this would be enough. | |
| Feb 10, 2012 at 12:56 | comment | added | Ramiro de la Vega | @Bill: I don´t quite understand why that would be enough. Can you explain a bit more how would you finish the argument? | |
| Feb 9, 2012 at 16:58 | comment | added | Bill Johnson | Wouldn't it be enough to have a continuous function from a closed subset $C$ of $X$ onto the Cantor set $\Delta$? You have a subset $D$ f $\Delta$ s.t. neither $D$ nor its complement contains an uncountable closed subset. And can't you build such a function by doing the usual tree construction of a Cantor set in $X$ and identifying to points the branches in the tree? (Or maybe this is a suggestion I should post under the nom de plume unknown (google).) | |
| Feb 9, 2012 at 15:13 | comment | added | Ramiro de la Vega | @Pietro: Is it always (i.e. for any $T_2$ space) true that there are as many Borel sets as open sets? Making a rough estimate, apparently one needs that $|\tau|^{\aleph_0}=|\tau|$ (which is true for compact $X$ by a non-trivial result of Shelah, but not true in general). | |
| Feb 9, 2012 at 13:23 | comment | added | Gerald Edgar | Suppose you have such a measure. The problem then would be to construct a compact set $K$ of measure zero, with a subset that is not a Borel set. Certainly this can be done in compact metric space, but what about in exotic compact Hausdorff spaces? | |
| Feb 9, 2012 at 13:22 | comment | added | Pietro Majer | I understand the question as, whether such a measure can be complete on the $\sigma$ algebra of Borel sets. If $X$ is metrizable, the answer is no for cardinality reasons exactly as for the Lebesgue measure -there is a $\mu$-null perfect set, thus with more subsets than there are Borel sets in $X$. The case of a non-metrizable $X$ is the point of the question. If we want to use the same cardinality argument in the latter case, the question is: can we still find a $\mu$-null Borel set $C\subset X$ of cardinality $\tau(X)$? (here $\tau(X)$ = card. of the open sets = card. of Borel sets) | |
| Feb 9, 2012 at 12:56 | comment | added | Matthew Daws | @Ramiro: This is exactly my point of confusion... | |
| Feb 9, 2012 at 12:54 | comment | added | Ramiro de la Vega | I´ve frequently seen complete regular atomless Borel measures on compact spaces in the literature, but they usually refer to the completion of a Borel measure, which is not a Borel measure strictly speaking. | |
| Feb 9, 2012 at 12:47 | comment | added | Ramiro de la Vega | @Matthew: One could say that the $\sigma$-algebra is determined by the measure; after all, one is the domain of the other. | |
| Feb 9, 2012 at 12:36 | comment | added | Matthew Daws | I don't quite understand: "complete" is a property of the measure and the $\sigma$-algebra. So is it the case that Lebesgue measure on $[0,1]$ wouldn't be an example, as this is not "Borel" because the $\sigma$-algebra is larger than just the Borel sets? | |
| Feb 9, 2012 at 11:45 | history | asked | arc | CC BY-SA 3.0 |