# Lebesgue Measurable Set which is not a union of a Borel set and a subset of a null meager set?

This is a follow-up to my question here. The Lebesgue Sigma algebra is the completion of the Borel Sigma algebra under the Lebesgue measure, which means that every Lebesgue measurable set can be written as a union of a Borel set and a subset of a measure $$0$$ Borel set. But my question is, what is an example of a Lebesgue measurable set which cannot be written as a union of a Borel set and a subset of a measure $$0$$ meager set?

Or does no such example exist?

Example of a set $$S\subseteq\mathbb R$$ of Lebesgue measure zero which is not the union of a Borel set and a meager set.

Let $$G$$ be a dense $$G_\delta$$ set of measure zero.

Let $$B$$ be a Bernstein set, i.e., $$B$$ meets every uncountable closed set but contains no uncountable closed set. (Such sets exists assuming the axiom of choice.)

Let $$S=B\cap G$$.

$$S$$ has measure zero, since $$S\subseteq G$$.

Assume for a contradiction that $$S=A\cup M$$ where $$A$$ is a Borel set and $$M$$ is meager.

Since $$B$$ is a Bernstein set, $$B$$ contains no uncountable closed set; since every uncountable Borel set contains an uncountable closed set, $$B$$ contains no uncountable Borel set; since $$A\subseteq S\subseteq B$$ and $$A$$ is a Borel set, $$A$$ is countable.

Since $$S=A\cup M$$, where $$A$$ is countable and $$M$$ is meager, $$S$$ is meager.

Since $$S$$ is meager, there is a dense $$G_\delta$$ set $$H$$ such that $$S\cap H=\emptyset$$.

Since $$G$$ and $$H$$ are dense $$G_\delta$$ sets, $$G\cap H$$ is a dense $$G_\delta$$ set.

Since $$G\cap H$$ is a dense $$G_\delta$$ set, there is an uncountable closed set $$F\subseteq G\cap H$$.

Now $$B\cap F\subseteq B\cap G\cap H=S\cap H=\emptyset$$, so $$B\cap F=\emptyset$$. Since $$F$$ is an uncountable closed set, this contradicts the fact that $$B$$ is a Bernstein set.

• How do you prove that every uncountable Borel set contains an uncountable closed set? – Keshav Srinivasan Nov 11 '18 at 15:27
• It's a classical result, the Perfect Set Theorem for Borel sets: every uncountable Borel set contains a nonempty perfect set. Probably the easiest approach is to show that Borel sets are analytic, and every uncountable analytic set contains a perfect set. Too complicated to post here. – bof Nov 11 '18 at 22:25