TITLE: Is Lebesgue/Borel non-measurability actually caused by non-uniqueness?
QUESTION [7 upvotes]: In ZFC, every construction of a Lebesgue or Borel non-measurable set uses the axiom of choice. None of them that I've seen use choice to define a unique set, even though it's entirely possible to do so (e.g. under the AoC, if $\kappa = |A|$ is the cardinality of set $A$, then $\kappa$ is unique). So I've been wondering lately whether the strength of the AoC is enough by itself to construct a non-measurable set.
Here's an attempt at a specific question. In the language of set theory (first-order logic with equality extended with the ZFC axioms), is there a formula without parameters that identifies a unique, Lebesgue/Borel non-measurable set?

REPLY [4 votes]: 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.)