# Lebesgue Measurable Set which is not a union of a Borel set and a subset of a null $F_\sigma$ set?

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 $$0$$ $$F_\sigma$$ set?

Or does no such example exist?

• May be; take a a union of a Borel set and a subset of a measure 0 $G_\delta$ set ?? Oct 31, 2018 at 20:31
• @Bumblebee You can't just choose any union of a Borel set and a subset of a measure $0$ $G_\delta$ set, because it's possible that that set can also be written as a union of a Borel set and a subset of a measure $0$ $F_\sigma$ set. Oct 31, 2018 at 20:36
• Oh, I see. This might be helpful. Oct 31, 2018 at 20:40
• @Bumblebee It's easy to find an example of a Borel set that is neither $F_\sigma$ nor $G_\delta$, but how would that help us? Regardless of what Borel set of measure $0$ you take, the union of a given subset of it with a Borel set might still be able to be written as a union of a Borel set and a subset of a measure 0 $F_\sigma$ set. Oct 31, 2018 at 22:28
• @Bach Every Lebesgue measurable measure zero set is a subset of a measure 0 Borel set. In any case, the fact that every Lebesgue measurable set can be written as a union of a Borel set and a subset of a measure 0 Borel set is a standard result, usually stated as “the Lebesgue sigma algebra is the completion of the Borel algebra”. A more general result which implies this result is proven in Theorem 1.6.6 in this chapter of Junghenn’s Measure Theory book: gdurl.com/dw43 If it’s too difficult to understand I can find a simpler proof for you. May 30, 2019 at 3:04

I found an example in this journal paper. Let $$\beta$$ be a Bernstein set, i.e. a subset of $$\mathbb{R}$$ such that both it and its complement intersects every uncountable closed subset of $$\mathbb{R}$$. (This post describes how to construct such a set using the axiom of choice.) And let $$\gamma$$ be a dense measure-$$0$$ $$G_\delta$$ subset of the fat Cantor set. (This answer describes how to construct such a set.)
Then $$\beta\cap\gamma$$ is a Lebesgue measurable set which cannot be written as a union of a Borel set and a subset of a measure $$0$$ $$F_\sigma$$ set. This is proven in the linked paper.