Concerning Borel measurability, it was already pointed out that there is an explicit formula s(x) such that ZFC proves "The set { x in R: s(x) } is not Borel". (This is not true for ZF, as was pointed out elsewhere.)
Concerning Lebesgue measurability, ZFC neither proves nor refutes the following:
There is an OD-definition (or: OD(R)-definition) of a subset of the reals which is non-measurable.
There is a slight fuzziness here, because there are many non-equivalent notions of definability; OD(R)-definability is perhaps the most prominent and useful.
But an explicit formula can be given: There is a formula phi(x) in the language of set theory (without parameters) such that ZFC neither proves nor refutes
"The set { x in R : phi(x) } is non-measurable".
In fact, phi(x) can be of a rather simple form ($\Delta^1_2$, as you remarked above). Here is an abbreviated version of phi: For each real number x, let $x_1$ be the number obtained from x by deleting all even decimal places, $x_2$ by deleting all odd decimal places (do what you want for the countably many reals where this is not well-defined). This defines a measure-preserving Borel map from $\mathbb R$ to $\mathbb R\times \mathbb R$. Now consider the set M of all reals x for which there is some $\alpha$ such that $x_1\in L_\alpha$, but $x_2\notin L_\alpha$. ZFC does neither prove nor refute that M is Lebesgue-measurable.
(I think that the fact that ZFC does not prove that M is measurable is already due to Gödel.)
