I'm pretty sure that a subset of a Borel set (in $\mathbb{R}$) of measure zero may not be Borel, but I don't know how to show it.

I do know that there are more Lebesgue-measurable sets than Borel sets.

I also know that every subset of a Borel set of measure zero is Lebesgue-measurable.

I would prefer a general argument (for example, using cardinality) over a counterexample.


If $A$ is a subset of $\Bbb R$ of measure $0$, then all its subsets, $2^{|A|}$ many of them, are measure $0$ (and so Lebesgue measurable, Borel or not). Any subspace of a separable metric space is itself separable metrisable and so has at most $\mathfrak{c} = |\Bbb R|$ many Borel subsets.

So whenever $2^{|A|} > \mathfrak{c}$, $A$ has a Lebesgue measurable subset that is not Borel in $A$. This is the case for the standard Cantor middle third set e.g.


This answer is inspired by Henno Brandsma's answer, and perhaps easier:

Any set of measure zero is contained in a Borel set of measure zero (since any Lebesgue-measurable set can be approximated from outside by a Borel set), so it suffices to show that there is a set of measure zero which is not Borel.

The Cantor set has measure zero and cardinality $\mathfrak{c}$, so it has $2^{\mathfrak{c}}$ subsets of measure zero. There are $\mathfrak{c}$ Borel sets, so there must be some set of measure zero which is not Borel.


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