Timeline for Ultrafilter theorem and translation invariant measures
Current License: CC BY-SA 3.0
12 events
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| Jul 7, 2017 at 5:30 | comment | added | Taras Banakh | By the way, the Vitali construction shows that there are no translation-invariant $\sigma$-additive measures defined on all subsets of $\mathbb R$ because a translation invariant finitely additive measure does exist by the amenability of the group $\mathbb R$. | |
| Jul 6, 2017 at 12:27 | answer | added | Taras Banakh | timeline score: 3 | |
| Oct 14, 2013 at 22:05 | comment | added | Ashutosh | I am only speculating, but I meant looking into models of the form $L(R)[U]$, where $U$ is $P(\omega)/Fin$-generic over $L(R)$. See, for example, Andres Caicedo's posting: mathoverflow.net/questions/69615/… | |
| Oct 14, 2013 at 20:48 | comment | added | Carlos | Maybe I don't understand what you are suggesting. A negative answer would require a model of ZF + UT in which there is a translation invariant measure. A model with an ultrafilter but no Vitali set would not allow us to conclude. And if we could prove that there is no such model, this would mean that UT implies the existence of a Vitali set. I have no idea about how this could be proved. I think that a positive answer would consist of generalizing some proof of the existence of a non Lebesgue measurable set from UT in the same way than Vitali's argument can be generalizec. | |
| Oct 14, 2013 at 17:43 | comment | added | Ashutosh | Do you know any models with an ultrafilter but no Vitali set? | |
| Oct 14, 2013 at 11:41 | comment | added | Carlos | I have added the $\sigma$-finite hypothesis to the question. | |
| Oct 14, 2013 at 11:39 | history | edited | Carlos | CC BY-SA 3.0 |
added 17 characters in body
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| Oct 14, 2013 at 11:35 | comment | added | Carlos | Yes, I am thinking of $\sigma$-finite measures. A translation invariant $\sigma$-finite measure defined on $\mathcal P\mathbb R$ for which $[0,1]$ has measure $1$ would extend Lebesgue measure. Assuming the consistency of a measurable cardinal, it is consistent with AC the existence of an extension of Lebesgue measure to $\mathcal P\mathbb R$. AC implies that such extension cannot be translation invariant (by the Vitali argument). I am asking whether the ultrafilter theorem suffices. | |
| Oct 14, 2013 at 2:41 | comment | added | Joel David Hamkins | Could you say a little more about what kind of measures you are considering? After all, I could define $\mu(X)$ to be $\infty$, if $X$ is uncountable and otherwise $0$. This measures every set, with no mass at points and it is translation-invariant. So I guess you want $\sigma$-finite measures. But if the unit interval gets finite measure, then this determines the measure of any interval, since we can split the interval into two halves, or thirds, etc. which must get equal measure, and so on. So if an interval has finite measure, then one is back to the arguments to which you refer. | |
| Oct 14, 2013 at 2:36 | comment | added | Joel David Hamkins | Related: mathoverflow.net/a/57108/1946. | |
| Oct 13, 2013 at 22:29 | review | First posts | |||
| Oct 13, 2013 at 23:08 | |||||
| Oct 13, 2013 at 22:11 | history | asked | Carlos | CC BY-SA 3.0 |