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.
