In a question and an answer at MO, Joel David Hamkins showed that (if ZFC is consistent) there are models of ZFC in which $V\neq HOD$ and every $\Sigma_2$-definable set has a definable member.

Let $\mathfrak{M}$ be such a model. My question is: Can such a model $\mathfrak{M}$ satisfy further the following:

(*) Every ordinal definable set of reals is Baire (or Lebesgue) measurable?

I suspect very much that the answer is negative, but I would like the confirmation of an expert.


This is impossible; there is no model of ZFC like that. The reason is that the set of non-measurable sets of reals (or non-Baire sets, respectively) is definable, and moreover $\Sigma_2$ definable; so under the first part of your conditions, it would have a definable member, which would violate the second part of your requirements.

The question of whether a set of reals $A$ is measurable or not is something that can be checked in a comparatively small rank-initial segment of the universe, in $V_{\omega+3}$ or so, and for this reason, it is a local property, which therefore has complexity at worst $\Delta_2$.

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  • $\begingroup$ Thanks, I suspected so but couldn't figure that the complexity of the definition of the set on nonmeasurable sets is $\Sigma_2$. $\endgroup$ – user38200 Dec 17 '14 at 20:03
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    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Dec 17 '14 at 20:13

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