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?



Is Lebesgue/Borel non-measurability actually caused by non-uniqueness?
**The answer below has been edited in light of other answers and comments.**


There are all sorts of models of $ZFC$ in which *every* set is definable without parameters, including nonmeasurable sets; indeed a [recent paper](https://arxiv.org/abs/1105.4597) of Hamkins, Linetsky, and Reitz is devoted to such "pointwise definable" models.


Also, as pointed out in Theo Buehler's comment to the question, there certainly exist definable subsets of reals that are $ZFC$-provably not *Borel*.



> 
> However, the situation is completely different for measurability. The classical work of Solovay [using an inacessible] shows that there is a model of $ZFC$ in which every subset of reals in $OD(\Bbb{R})$ is Lebesgue measurable. Recall that $X$ is in $OD(\Bbb{R})$ if $X$ is definable with parameters from $Ord \cup \Bbb{R}$.
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As pointed out in Demer's answer, Krivine [without an inaccessible] provided a model of $ZFC$ in which every ordinal definable subset of reals is measurable. Moreover, as shown by Harvey Friedman, [here](http://cms.math.ca/cjm/v32/cjm1980v32.0653-0656.pdf), there is a model of $ZFC$ [which is a generic extension of Solovay's model] in which the following property holds:



> 
> **(\*)** Every equivalence class of sets of reals modulo null sets that is in $OD(\Bbb{R})$
> consists of Lebesgue measurable sets.
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Note that **(\*)** implies that no non-measurable subset of reals in definable, since if $X$ is any definable subset of reals that is not measurable, then the equivalence class $\[X\]$ of $X$ modulo null sets satisfies the following two properties:


**(1)** $\[X\]$ definable,


**(2)** No member of $\[X\]$ is measurable.



> 
> So, to sum-up, the answer to the question for Lebesgue measurability is negative, i.e., there is no formula $\phi(x)$ in the language of set theory for which $ZFC$ proves "there is a unique nonmeasurable subset of reals satisfying $\phi$".
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> However, if $ZFC$ is strengthened to $ZFC+V=L$ then such a formula **does** exist, as pointed out in Goldstern's answer.
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