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.

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    $\begingroup$ 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? $\endgroup$ Commented Mar 20, 2019 at 12:43
  • $\begingroup$ No, the only problem is that i didnt think about that. Thanks! $\endgroup$ Commented Mar 20, 2019 at 13:00
  • $\begingroup$ @Jonathan please turn this comment into an answer! $\endgroup$ Commented Mar 21, 2019 at 10:00
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    $\begingroup$ 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. $\endgroup$ Commented Mar 21, 2019 at 16:54

1 Answer 1


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.


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