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.

  • $\begingroup$ How do you prove that every uncountable Borel set contains an uncountable closed set? $\endgroup$ – Keshav Srinivasan Nov 11 '18 at 15:27
  • $\begingroup$ 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. $\endgroup$ – bof Nov 11 '18 at 22:25

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