Let $A$ be a subset of $\mathbb{R}^2$ which intersects every straight line in exactly two points. Is there a such set which is Lebesgue measurable or Borel? A well-known fact is that there exists such set which is not Lebesgue measurable.
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1$\begingroup$ For Borel: mathoverflow.net/a/21864/30186 $\endgroup$– WojowuCommented Nov 7, 2021 at 13:08
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There is such a set which is Lebesgue measurable, and indeed of Lebesgue measure zero. To see this, start with a subset $S$ of $\mathbb R^2$ such that every line intersects it in continuum many points, for instance $C\times\mathbb R\cup\mathbb R\times C$, where $C$ is the Cantor set. Now repeat your favorite transfinite recursive construction of a set $A$ intersecting each line at exactly two points, modifying it in such a way that we all the points picked belong to $S$. Since $A$ is a subset of a measure zero set $S$, it itself is Lebesgue measurable of measure zero.
It appears that existence of such a set which is Borel is an open problem, see this MO post.
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$\begingroup$ Does this construction depend on Choice? $\endgroup$ Commented Nov 8, 2021 at 20:51
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$\begingroup$ @Acccumulation Yes, more specifically on a well-ordering of $\mathbb R$. I'm not aware of a choiceless construction. $\endgroup$– WojowuCommented Nov 8, 2021 at 20:56
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$\begingroup$ @Accumulation Since it is consistent with $\mathsf{ZF}$ that all subsets of $\Bbb R$ are Borel and the existence of such a Borel set is open, it must be open whether $\mathsf{ZF}$ proves that such a set exists as well $\endgroup$ Commented Nov 9, 2021 at 18:01