# Can a Vitali Set be constructed without AC?

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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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
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