# Can a Vitali set be Lebesgue measurable? (ZF)

Here is the definition of Lebesgue measure.

The standard proof that Vitali sets are not Lebesgue measurable uses countable additivity of Lebesgue measure, which is not a theorem of ZF. (In particular, it is consistent that the real line is a countable union of countable sets, and thus a countable union of measure zero sets.) Since ZF does prove that Lebesgue measure is super-additive, that proof can be easily adapted to show in ZF that if a Vitali set is measurable, then its measure is zero. By the Caratheodory construction, this is equivalent to having outer measure zero.

Does ZF prove that all Vitali sets have positive outer measure?

If no, does ZF prove "if there exists a Vitali set, then there exists a Vitali set with positive outer measure"?

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People seem to be answering a different question than what was asked... – Harry Altman Aug 15 '11 at 11:56
Now when you say measurable, in ZF+$\lnot$Countable choice for $\Bbb R$; do you mean measures by Borel codes, or the usual Borel measure without the sigma-additivity? Also, what is the outer measure in this case? – Asaf Karagila Mar 3 '13 at 20:25
I mean Caratheodory measurable wrt Lebesgue outer measure. $\:$ – Ricky Demer Mar 3 '13 at 20:59
Yes, but suppose that the real numbers are a countable union of countable sets; requiring countable subadditivity of the outer measure immediately makes all sets measure zero. – Asaf Karagila Mar 4 '13 at 16:24

The Vitali set can never be "made measurable", primarily due to translation invariance. As you recall, the Vitali construction yield a countable collection of sets whose union is [0,1). However, through simple translations of these same sets, they can be redistributed so that their union is now [0,2), or [0,n), or any rational length. Since the union of these sets can have differing measures, the measure of these sets must therefore remain undefined.

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This only applies if the measure is sigma-additive which cannot be proven in the complete absence of any form of choice. You need at least countable choice or something like that I think. – Johannes Hahn Mar 3 '13 at 17:02
@Johannes: $\:$ Do you mean, "which you think cannot be proven in ..."? $\:$ If you have a proof $\hspace{.4 in}$ that that cannot be proven, then I'd like a link or sketch. $\;\;$ – Ricky Demer Mar 3 '13 at 21:02
There is a model of ZF (without choice) in which the real line is the union of countably many countable sets. This is Theorem 10.6 in Jech's book "The Axiom of Choice"; the result is due to Feferman and Lévy, very shortly after Cohen's proof of the independence of AC from ZF. Of course in such a model Lebesgue measure cannot be countably additive. – Andreas Blass Mar 3 '13 at 21:43
Oh yeah, I'd completely forgotten that one. $\:$ – Ricky Demer Mar 3 '13 at 22:08

Sigma additivity is not the crucial issue issue here. In R^3 for example, the Banach-Tarski decomposition can divide the sphere up into five pieces (four non-measurable parts, and one point), and via translations and rotations only, create two spheres the same size as the original. This instance involves only finite additivity. Sigma additivity is required for R^1 and R^2, but that is not the essential issue of non-measurability.

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It is an essential issue of defining the measure, thus an essential issue of non-measurability. – Asaf Karagila Mar 4 '13 at 20:04
Are you saying that your argument only uses a finite redistribution? – François G. Dorais Mar 4 '13 at 21:05