# A question on models of set theory and Lebesgue measure

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 $\Delta_2$.

• Thanks, I suspected so but couldn't figure that the complexity of the definition of the set on nonmeasurable sets is $\Sigma_2$. Commented Dec 17, 2014 at 20:03
• 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. Commented Dec 17, 2014 at 20:13