Timeline for Is "There exists an unbounded non-measurable set but no bounded non-measurable set" consistent with $\mathsf{ZF}$?
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Jun 1, 2021 at 11:39 | vote | accept | Clement Yung | ||
| May 24, 2021 at 5:01 | comment | added | Elliot Glazer | This answer has been significantly overhauled. The above discussion refers to a previous argument which has been generalized to the (3) $\rightarrow$ (4) in the current answer. | |
| May 24, 2021 at 4:59 | history | edited | Elliot Glazer | CC BY-SA 4.0 |
This is a significant overhaul, proving a theorem which immediately (and unambiguously) resolves both of the questions.
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| May 19, 2021 at 17:37 | history | edited | Elliot Glazer | CC BY-SA 4.0 |
added 191 characters in body
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| May 19, 2021 at 8:17 | comment | added | Elliot Glazer | @FrançoisG.Dorais I think your earlier suggestion is reasonable. That if $\mathbb{R}$ is a countable union of countable sets, then there is no total isometry-invariant finitely additive probability measure on $\mathcal{P}([0,1])$ which has the Lebesgue density property. | |
| May 19, 2021 at 8:11 | comment | added | François G. Dorais | I think the theorem is that if $\mathbb{R}$ is a countable union of countable sets and if every set is measurable (in pretty much any decent interpretation of that) then every set of reals has measure zero. So the question is what is "pretty much any decent interpretation of that". | |
| May 19, 2021 at 7:51 | comment | added | François G. Dorais | But that's a fixed definition of Lebesgue measure... I'm bummed since I wanted your argument to be more general! That was the genius part in the first place! | |
| May 19, 2021 at 7:50 | comment | added | Elliot Glazer | @FrançoisG.Dorais In ZF, "Lebesgue measure" is a finitely additive measure with the definition given above. The claim is that if $\mathbb{R}$ is a countable union of countable sets, there is a set which is not Lebesgue measurable, giving an affirmative answer to question 1. | |
| May 19, 2021 at 7:47 | comment | added | François G. Dorais | Nope. I messed up again. What is the claim here? | |
| May 19, 2021 at 7:16 | comment | added | François G. Dorais | That paper fixes a definition of $\lambda^\ast$, using sums of lengths of interval covers, but that's fine since we're using "Lebesgue measure" (whatever that ought to mean). But I think we finally found our point of agreement: if $[0,1]$ is a countable union of countable sets then no countably additive extension of the length-of-interval measure on $\mathcal{P}([0,1])$ is total. And countably additive could be replaced by finitely additive and Lebesgue Densitarian? | |
| May 19, 2021 at 7:09 | comment | added | Elliot Glazer | @AsafKaragila If it makes you feel better we can call it "Lebesgue finitely additive measure." Doesn't have the same ring though. | |
| May 19, 2021 at 7:07 | comment | added | Asaf Karagila♦ | Again with this finitely additive nonsense? | |
| May 19, 2021 at 6:52 | comment | added | Elliot Glazer | Yes, by compactness of the unit interval, which is also choiceless. In any case, see Foreman and Wehrung's "The Hahn-Banach Theorem implies the existence of a non-Lebesgue measurable set" for a full discussion on choiceless Lebesgue measure. | |
| May 19, 2021 at 6:49 | comment | added | François G. Dorais | Can you show that $\varnothing$ is measurable in this sense? | |
| May 19, 2021 at 6:45 | comment | added | Elliot Glazer | Lebesgue measure makes sense in choiceless contexts, but it is only finitely additive. | |
| May 19, 2021 at 6:45 | comment | added | Elliot Glazer | In this context, $X \subset [0, 1]$ is measurable if $\lambda^*(X)+\lambda^*([0,1] \setminus X) = 1.$ If $\lambda(X)=m>0,$ then some open cover of $X$ has measure less than $m(1+\epsilon),$ so there has to be some interval in the open cover on which $X$ has $1/(1+\epsilon)$ of the interval's measure. | |
| May 19, 2021 at 6:44 | comment | added | François G. Dorais | Nope, I didn't quite get it. The real issue is that Lebesgue measure doesn't make sense without DC. Your argument just shows that, in this model, if every set is measurable then every set has measure 0. | |
| May 19, 2021 at 6:40 | comment | added | François G. Dorais | Ah, I think I got it! Lebesgue Density requires you to fix a notion of measurable? | |
| May 19, 2021 at 6:38 | comment | added | Elliot Glazer | @FrançoisG.Dorais No, Lebesgue Density is a choiceless theorem. The standard proof doesn't use any choice. | |
| May 19, 2021 at 6:37 | comment | added | François G. Dorais | You're right! So Lebesgue Density must use CC or DC? | |
| May 19, 2021 at 6:34 | comment | added | Elliot Glazer | @FrançoisG.Dorais A non-principal ultrafilter is a nonmeasurable set because it violates Lebesgue density. I think Raisonnier's Theorem is that an $\omega_1$-sequence of reals implies there is a nonmeasurable set, which I didn't use. | |
| May 19, 2021 at 6:31 | comment | added | François G. Dorais | Aha! This is great! However, I think the last paragraph uses DC... (Raisonnier's Theorem) | |
| May 19, 2021 at 5:01 | history | answered | Elliot Glazer | CC BY-SA 4.0 |