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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 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 $Ord \cup \Bbb{R}$.

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, 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.

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 $\phi$".

However, if $ZFC$ is strengthened to $ZFC+V=L$ then such a formula does exist, as pointed out in Goldstern's answer.

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 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 $Ord \cup \Bbb{R}$.

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, 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.

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 $\phi$".

However, if $ZFC$ is strengthened to $ZFC+V=L$ then such a formula does exist, as pointed out in Goldstern's answer.

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 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 $Ord \cup \Bbb{R}$.

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, 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.

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 $\phi$".

 

However, if $ZFC$ is strengthened to $ZFC+V=L$ then such a formula does exist, as pointed out in Goldstern's answer.

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 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 $Ord \cup \Bbb{R}$.

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, 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.

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 $\phi$".

However, if $ZFC$ is strengthened to $ZFC+V=L$ then such a formula does exist, as pointed out in Goldstern's answer.

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 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], and Krivine [without an inaccessible, see Demer's answer] 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 $Ord \cup \Bbb{R}$.

Indeed, as shown by Harvey Friedman, here, there is a model of $ZFC$ 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.

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 $\phi$".

However, if $ZFC$ is strengthened to $ZFC+V=L$ then such a formula does exist, as pointed out in Goldstern's answer.

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 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 $Ord \cup \Bbb{R}$.

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, 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.

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 $\phi$".

However, if $ZFC$ is strengthened to $ZFC+V=L$ then such a formula does exist, as pointed out in Goldstern's answer.