Combinations of Lebesgue measurability, the property of Baire and the perfect set property

Lebesgue measurability (LM), the property of Baire (BP) and the perfect set property (PSP) are probably the most prominent among all the regularity properties of sets of reals. Such a set can either satisfy any one of these properties or not and so this leaves us with 8 possible combinations of LM, BP and PSP.

Just for the heck of it, I recently tried to construct examples for all these 8 combinations. I was able to do so for 7 of them. This leads me to my question:

Is there, provably in ZFC, a set of reals which is Lebesgue measurable, has the property of Baire, but not the perfect set property?

I know that the existence of such a set is consistent both with CH and $$\neg$$CH. On the one side, in $$L$$ there is a $$\Pi_1^1$$-set which does not have the PSP and it being $$\Pi_1^1$$ certainly makes it satisfy LM and BP. On the other hand, in a model of $$\neg$$CH where $$\operatorname{non}(\mathcal N)=\operatorname{non}(\mathcal M)=2^{\omega}$$, any set of reals of size $$\omega_1$$ would be an example. Here, $$\operatorname{non}(\mathcal N)$$ and $$\operatorname{non}(\mathcal M)$$ are the smallest cardinalty of a set of reals which does is not LM, respectively does not have BP.

The standard examples for sets without the PSP are the Bernstein sets, i.e. sets $$B\subseteq\mathbb R$$ for which both $$B$$ and $$\mathbb R\setminus B$$ meet any perfect set. However, these are are not LM and do not have the BP as well.

• Is there a problem with constructing a Bernstein set on a closed null and nowhere dense set? Or do you actually mean universally BP/LM? Mar 20, 2019 at 12:43
• No, the only problem is that i didnt think about that. Thanks! Mar 20, 2019 at 13:00
• @Jonathan please turn this comment into an answer! Mar 21, 2019 at 10:00
• Of possible interest is Relations among some classes of subsets of $R$ by Harvey Diamond and Gregory Gelles (1984), which gives a nice Venn diagram for these kinds of collections of sets. Regarding Bernstein sets, I posted a lot of information about them in this 6 February 2006 sci.math post (correction here) archived at Math Forum. Mar 21, 2019 at 16:54

Fix an uncountable perfect subset $$C$$ of $$\mathbb{R}$$ that is null and nowhere dense (for example the classical Cantor set works).
Then construct a Bernstein set $$B$$ on the space $$C$$ in the same was as it is constructed on $$\mathbb{R}$$. $$B$$ is uncountable, has no perfect subset, is null and meager (as a subset of $$\mathbb{R}$$). Thus it does not have the perfect set property, it is Lebesgue measurable and it has the Baire property.