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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?

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  • $\begingroup$ May be; take a a union of a Borel set and a subset of a measure 0 $G_\delta$ set ?? $\endgroup$
    – Bumblebee
    Oct 31, 2018 at 20:31
  • $\begingroup$ @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. $\endgroup$ Oct 31, 2018 at 20:36
  • $\begingroup$ Oh, I see. This might be helpful. $\endgroup$
    – Bumblebee
    Oct 31, 2018 at 20:40
  • $\begingroup$ @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. $\endgroup$ Oct 31, 2018 at 22:28
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    $\begingroup$ @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. $\endgroup$ May 30, 2019 at 3:04

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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.

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