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?
Best Answer
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.