# Is a subset of a Borel set of measure zero always Borel?

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.
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.