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For the purposes of this discussion, let a Vitali Set be any subset $V\subseteq{}[0,1)$ such that for $V_q:=\{x+q\;|\;x<1-q,\;x\in{}V\}\cup\{x+q-1\;|\;x\geq{}1-q,\;x\in{}V\}$ there is a countable subset $I\subset[0,1)$ such that

  1. $[0,1)=\bigcup{}_{q\in{}I}V_q$
  2. For $r,q\in{}I$ distinct, $V_r\cap{}V_q=\emptyset$

Can such a $V$ be constructed without AC?

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up vote 4 down vote accepted

No Vitali set in your sense can be measurable. I am assuming this is the reason for defining a Vitali set in this way. But Solovay has shown (assuming the consistency of a certain large cardinal, namely an inaccessible) that there is a model of ZF in which all sets of reals are Lebesgue measurable. In particular, there is no Vitali set in Solovay's model. Hence you need some fragment of AC to construct one.

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I actually just cited this article in response to mathoverflow.net/questions/72904/… but apparently countable additivity of Lebesgue Measure is not a theorem of ZF. So, the standard proof of the nonmeasurability of $V$ would not work in ZF alone. Do you know of a proof of the nonmeasurability of $V$ in ZF alone? –  user17100 Aug 15 '11 at 7:12
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Solovay's model also satisfies DC, so Lebesgue measure is countably additive in that model. –  François G. Dorais Aug 15 '11 at 7:26
    
I don't know a proof of the nonmeasurability of $V$ is ZF alone, but the countable additivity of Lebesgue measure can certainly be proved using the axiom of dependent choice (DC), which holds in Solovay's model. DC says that given a relation $R$ on a set such that every element has an upper bound with respect to $R$, then you can choose an $R$-increasing sequence of ordertype $\omega$. This is stronger than AC for countable families. –  Stefan Geschke Aug 15 '11 at 7:27
    
Oh, Francois beat me in his comment. –  Stefan Geschke Aug 15 '11 at 7:28
    
Thanks, I just saw this after posting to the other question. Thank you! –  user17100 Aug 15 '11 at 8:54
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