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We know that the subset $N$ of $[0,1) \subset \mathbb R$ that contains exactly one member of the equivalence classes of $\sim$ defined on $[0,1)$, where $x \sim y$ iff $x - y$ is rational, is an example of a Lebesgue set that is not a Borel set. Is $N$ a subset of a measure zero Borel set? This seems to be true since intuitively it seems like we can shrink the size of $N$ arbitarily small, but I cannot prove it at the moment. If so, how can we find such a Borel superset?

More generally, is it true that any uncountable Borel set $E$ contains a subset that is a Lebesgue measurable set but not a Borel set?

Squirrel-Power
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  • Isn't this actually an example of a non Lebesgue measurable set? And as such, it can't be contained in a set of measure zero. (since the Lebesgue measure is complete) – Mark Jan 25 '24 at 22:01

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is it true that any uncountable Borel set $E$ contains a subset that is a Lebesgue measurable set but not a Borel set?

Yes. An uncountable Borel set $E$ contains a subset $F$ homeomorphic to the Cantor set. We may assume $F$ has Lebesgue measure zero. Argue by cardinality: There are $\mathfrak c = 2^{\aleph_0}$ Borel subsets of $F$, but there are $2^{\mathfrak c}$ subsets (and all subsets are Lebesgue measurable null sets). Thus, there is a Lebesgue measurable subset that is not Borel.

Is this a duplicate question? For example, HERE

GEdgar
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