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Dec 17, 2014 at 20:26 vote accept user38200
Dec 17, 2014 at 20:13 comment added Joel David Hamkins I edited to give a little more explanation of the complexity. Basically, it is $\Sigma_2$ because you can verify it in any sufficiently large $V_\theta$, and indeed, you don't have to go very high. Statements that are not $\Sigma_2$ must involve set-theoretic properties that stretch up arbitrarily high in the set-theoretic universe. Lebesgue measurability is not like that, since once you have the reals and all the sets of reals, then all the issues about measurability will be determined.
Dec 17, 2014 at 20:11 history edited Joel David Hamkins CC BY-SA 3.0
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Dec 17, 2014 at 20:03 comment added user38200 Thanks, I suspected so but couldn't figure that the complexity of the definition of the set on nonmeasurable sets is $\Sigma_2$.
Dec 17, 2014 at 19:59 history answered Joel David Hamkins CC BY-SA 3.0