# Topological proof that a Vitali set is not Borel

This question is purely out of curiosity, and well outside my field — apologies if there is a trivial answer. Recall that a Vitali set is a subset $$V$$ of $$[0,1]$$ such that the restriction to $$V$$ of the quotient map $$\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Q}$$ is bijective. It follows easily from the definition that $$V$$ is not Lebesgue measurable. Now suppose you don't know the Lebesgue measure (which, after all, is not that easy to construct). Is there a topological proof that $$V$$ is not a Borel subset of $$\mathbb{R}$$?

• The family of sets satisfying the property of Baire is a sigma-algebra, in particular all Borel sets have this property. On the other hand, Vitali set does not have this property. This is a nice instance of Baire property being a convenient substitute of Lebesgue measurability in topological contexts. May 25, 2020 at 9:33
• Very good, thanks! If you feel like writing this as an answer I'll accept it.
– abx
May 25, 2020 at 9:38
• You may be interested in Oxtoby's book Measure and Category that goes through several parallel examples of theorems about measure and Baire category. Wojowu's answer explains the "category analogue" of Vitali's theorem. May 25, 2020 at 23:40

Sometimes a convenient substitute for Lebesgue measurability is the property of Baire. Just like Lebesgue measurability, the class of sets with this property is a $$\sigma$$-algebra containing the open subsets - indeed, open sets clearly have property of Baire, this class is closed under countable unions since meager sets are closed under countable unions, and it's closed under complements essentially since the boundary of an open set is nowhere dense.

It of course follows that every Borel set has the property of Baire. It remains to show the Vitali set doesn't have a property of Baire. Indeed, assume it was, then either $$V$$ is meager, or it is comeager in some open interval. The former cannot hold, since $$[0,1]$$ is contained in countable union of translates of $$V$$, so would be meager itself. In the other hand, if it were comeager in some open interval $$I$$, then for any rational $$q$$ the intersection $$I\cap(V+q)$$, since it's contained in $$I\setminus V$$, is meager, from which we would conclude $$V$$ is meager.

This might be a slightly overly detailed answer, but I did try to make it reasonably self-contained - it indeed is quite easier than if we had to build up the theory of Lebesgue measure from scratch!

• Just for completeness: a subset $Y$ of a topological space $X$ has the Baire property, or is almost open in $X$ if it differs from an open subset by a meager subset, i.e., can be written as $U\triangle M$ with $U$ open and $M$ meager. And meager means, contained in a countable union of closed subsets with empty interior.
– YCor
May 25, 2020 at 10:48
• Also I guess that "$V$ is meager in some open interval" should be understood as "there exists a nonempty open interval $I$ such that $V\cap I$ is comeager in $I$". (It took me a while to unravel so I'm adding this as a comment.)
– YCor
May 25, 2020 at 10:53
• @YCor Thank you for these remarks! Yes, this is precisely what I meant with "comeager in an open interval". May 25, 2020 at 11:01
• (yes, and sorry for the typo you corrected: I should have quoted "$V$ is co-meager in some open interval)
– YCor
May 25, 2020 at 11:03
• This might be a slightly overly detailed answer, Not at all. I would expect all answers to be detailed as this. Most of the answers on MathOverflow lack details making them basically useless. Thank you. I just added OP's question to the homework in my measure theory course and I included hints that follow your answer. Sep 25, 2020 at 21:14