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(Edit.) With a closer reading of your question, I see that you asked for a very specific notion of definability.

If you allow the family to have size larger than continuum, there is a trivial Yes answer. Namely, let phi(x) be the assertion "x is a non-measurable set of reals". In any model of ZFC, this formula defines a family of non-measurable sets of reals, and it is not difficult to show in ZFC that there are at least continuum many such sets (for example, as in the comment of Qiachu Yuan). Thus, ZFC proves that {x | phi(x)} is a family of non-measurable sets of size at least continuum.

But if you insist that the family have size exactly the continuum, as your question clearly states, then this trivial answer doesn't work. Indeed, one can't even take the class of all Vitali sets in this case, since there are 2^continuum many sets of reals that contain exactly one point from each equivalence class for rational translations.

Qiachu Yuan's suggestion about translations of a single Vitali set does have size continuum, but there is little reason to expect the Vitali set to be definable in the way that you have requested, and so it does not provide the desired definable family.

In my earlier posted answer, I considered the possibility that you might have meant some other notion of definability, or whether parameters are allowed in the definition, and so on. And I find some of these other versions of the question to be quite interesting and subtle.

I pointed out that it is surely consistent with ZFC that there is the desired definable family of non-measurable sets, since in fact any set at all can be made definable in a forcing extension that adds no reals and no sets of reals. So you can take any family of non-measurable sets that you like and go to a forcing extension where this family is definable.

Perhaps a stronger notion of definibility would be to use the notion of projective definitions, where one wants to define the sets within the structure of the reals, using quantification only over reals and natural numbers (rather than over the entire set-theoretic universe). Thus, we want a projective formula phi(x,z), such that A_z={x | phi(x,z)} is always non-measurable for any z and all A_z are different. Such a formula would be a strong example of the phenomenon you seek.

The first answer to this way of asking the question is that it is consistent with ZFC that there is such a projective family. The reason is that I have mentioned in a number of questions and answers on this site, under the Axiom of Constructibility V=L, there is a projectively definable well-ordering of the reals. Thus, under V=L, one can projectively define a Vitali set, and then take the family of its translations. There is no need for a parameter in this definition, since a particular Vitali set can be projectively defined without parameters from the projectively definable well-ordering of the reals.

The second answer to this version of the question, however, is that under certain set-theoretic assumptions such as Projective Determinacy, every projective set of reals is Lebesgue measurable. In this case, there can be no such projectively defined family of non-measurable sets. The assumption of PD is consistent with ZFC from large cardinals, but perhaps one needs a much weaker hypothesis meerely to get every projective set measurable.

In summary, if one wants a projectively definable family of non-measurable sets, then it is independent of ZFC, if large cardinals are consistent. (Perhaps the need for large cardinals can be reduced.)

(Edit.) With a closer reading of your question, I see that you asked for a very specific notion of definability.

If you allow the family to have size larger than continuum, there is a trivial Yes answer. Namely, let $\phi(x)$ be the assertion "$x$ is a non-measurable set of reals". In any model of ZFC, this formula defines a family of non-measurable sets of reals, and it is not difficult to show in ZFC that there are at least continuum many such sets (for example, as in the comment of Qiachu Yuan). Thus, ZFC proves that $\{x\mid\phi(x)\}$ is a family of non-measurable sets of size at least continuum.

But if you insist that the family have size exactly the continuum, as your question clearly states, then this trivial answer doesn't work. Indeed, one can't even take the class of all Vitali sets in this case, since there are $2^{\mathfrak c}$ many sets of reals that contain exactly one point from each equivalence class for rational translations.

Qiachu Yuan's suggestion about translations of a single Vitali set does have size continuum, but there is little reason to expect the Vitali set to be definable in the way that you have requested, and so it does not provide the desired definable family.

In my earlier posted answer, I considered the possibility that you might have meant some other notion of definability, or whether parameters are allowed in the definition, and so on. And I find some of these other versions of the question to be quite interesting and subtle.

I pointed out that it is surely consistent with ZFC that there is the desired definable family of non-measurable sets, since in fact any set at all can be made definable in a forcing extension that adds no reals and no sets of reals. So you can take any family of non-measurable sets that you like and go to a forcing extension where this family is definable.

Perhaps a stronger notion of definibility would be to use the notion of projective definitions, where one wants to define the sets within the structure of the reals, using quantification only over reals and natural numbers (rather than over the entire set-theoretic universe). Thus, we want a projective formula $\phi(x,z)$, such that $A_z=\{x\mid\phi(x,z)\}$ is always non-measurable for any $z$ and all $A_z$ are different. Such a formula would be a strong example of the phenomenon you seek.

The first answer to this way of asking the question is that it is consistent with ZFC that there is such a projective family. The reason is that I have mentioned in a number of questions and answers on this site, under the Axiom of Constructibility V=L, there is a projectively definable well-ordering of the reals. Thus, under V=L, one can projectively define a Vitali set, and then take the family of its translations. There is no need for a parameter in this definition, since a particular Vitali set can be projectively defined without parameters from the projectively definable well-ordering of the reals.

The second answer to this version of the question, however, is that under certain set-theoretic assumptions such as Projective Determinacy, every projective set of reals is Lebesgue measurable. In this case, there can be no such projectively defined family of non-measurable sets. The assumption of PD is consistent with ZFC from large cardinals, but perhaps one needs a much weaker hypothesis meerely to get every projective set measurable.

In summary, if one wants a projectively definable family of non-measurable sets, then it is independent of ZFC, if large cardinals are consistent. (Perhaps the need for large cardinals can be reduced.)

(Edit.) With my closer reading of your question, I see that you asked for a very specific notion of definability. And for this notion of definability, the answer to your question is Yes, for a trivial reason. Let phi(x) be the assertion "x is a non-measurable set of reals". In any model of ZFC, this formula defines a family of non-measurable sets of reals, and it is not difficult to show in ZFC that there are continuum many such sets (for example, as in the comment of Qiachu Yuan). Thus, ZFC proves that {x | phi(x)} is a continuum sized family of non-measurable sets. Indeed, all of them.

In my earlier posted answer, however, I considered the possibility that you might have meant some other notion of definability, or whether parameters are allowed in the definition, and so on. And I find some of these other versions of the question to be quite interesting and subtle.

I pointed out that it is surely consistent with ZFC that there is a definable family of non-measurable sets, since in fact any set at all can be made definable in a forcing extension that adds no reals. So you can take any family of non-measurable sets and go to a forcing extension where this family is definable.

A Vitali set (and there are many) may not itself be definable, so taking translates of just one such set may not produce a definable family, except in that set as a parameter. One might consider the family consisting of all translates of ANY Vitali set, however, which would be similar to the family I proposed above. It is certainly size continuum, and consists entirely of non-measurable sets.

Perhaps a stronger notion of definibility would be to use the notion of projective definitions, where one wants to define the sets within the structure of the reals, using quantification only over reals and natural numbers (rather than over the entire set-theoretic universe). Thus, we want a projective formula phi(x,z), such that A_z={x | phi(x,z)} is always non-measurable for any z and all A_z are different. Such a formula would be a strong example of the phenomenon you seek.

The first answer is that it is consistent with ZFC that there is such a projective family. The reason is that I have mentioned in a number of questions and answers on this site, under the Axiom of Constructibility V=L, there is a projectively definable well-ordering of the reals. Thus, under V=L, one can projectively define a Vitali set, and then take the family of its translations. There is no need for a parameter in this definition, since a particular Vitali set can be projectively defined without parameters from the projectively definable well-ordering of the reals.

The second answer, however, is that under certain set-theoretic assumptions such as Projective Determinacy, every projective set of reals is Lebesgue measurable. In this case, there can be no such projectively defined family of non-measurable sets. The assumption of PD is consistent with ZFC from large cardinals, but perhaps one needs a much weaker hypothesis meerely to get every projective set measurable.

In summary, if one wants a projectively definable family of non-measurable sets, then it is independent of ZFC, if large cardinals are consistent. (Perhaps the need for large cardinals can be reduced or omitted.)

(Edit.) With a closer reading of your question, I see that you asked for a very specific notion of definability.

If you allow the family to have size larger than continuum, there is a trivial Yes answer. Namely, let phi(x) be the assertion "x is a non-measurable set of reals". In any model of ZFC, this formula defines a family of non-measurable sets of reals, and it is not difficult to show in ZFC that there are at least continuum many such sets (for example, as in the comment of Qiachu Yuan). Thus, ZFC proves that {x | phi(x)} is a family of non-measurable sets of size at least continuum.

But if you insist that the family have size exactly the continuum, as your question clearly states, then this trivial answer doesn't work. Indeed, one can't even take the class of all Vitali sets in this case, since there are 2^continuum many sets of reals that contain exactly one point from each equivalence class for rational translations.

Qiachu Yuan's suggestion about translations of a single Vitali set does have size continuum, but there is little reason to expect the Vitali set to be definable in the way that you have requested, and so it does not provide the desired definable family.

In my earlier posted answer, I considered the possibility that you might have meant some other notion of definability, or whether parameters are allowed in the definition, and so on. And I find some of these other versions of the question to be quite interesting and subtle.

I pointed out that it is surely consistent with ZFC that there is the desired definable family of non-measurable sets, since in fact any set at all can be made definable in a forcing extension that adds no reals and no sets of reals. So you can take any family of non-measurable sets that you like and go to a forcing extension where this family is definable.

Perhaps a stronger notion of definibility would be to use the notion of projective definitions, where one wants to define the sets within the structure of the reals, using quantification only over reals and natural numbers (rather than over the entire set-theoretic universe). Thus, we want a projective formula phi(x,z), such that A_z={x | phi(x,z)} is always non-measurable for any z and all A_z are different. Such a formula would be a strong example of the phenomenon you seek.

The first answer to this way of asking the question is that it is consistent with ZFC that there is such a projective family. The reason is that I have mentioned in a number of questions and answers on this site, under the Axiom of Constructibility V=L, there is a projectively definable well-ordering of the reals. Thus, under V=L, one can projectively define a Vitali set, and then take the family of its translations. There is no need for a parameter in this definition, since a particular Vitali set can be projectively defined without parameters from the projectively definable well-ordering of the reals.

The second answer to this version of the question, however, is that under certain set-theoretic assumptions such as Projective Determinacy, every projective set of reals is Lebesgue measurable. In this case, there can be no such projectively defined family of non-measurable sets. The assumption of PD is consistent with ZFC from large cardinals, but perhaps one needs a much weaker hypothesis meerely to get every projective set measurable.

In summary, if one wants a projectively definable family of non-measurable sets, then it is independent of ZFC, if large cardinals are consistent. (Perhaps the need for large cardinals can be reduced.)

You have to specify what you mean by "definable". In which structure are you defining the family? Also, are paramters allowed in the definition?

If you mean that it should be definable from paramters in the set-theoretic universe, where ZFC holds, then the example provided by the comment of Qiaochu Yuan shows that the answer is Yes. One simply takes as a parameter a given non-measurable set.

But of course, the Vitali set (and many other non-measurable sets) is produced by an application of the Axiom of Choice, for which there is little reason to expect that it produces a definable set. It is true on general set-theoretic grounds that any set at all can be made definable without parameters in a forcing extension, so it is at least consistent that such a family could exist.

But perhaps you mean that you want the family to somehow be definable within the structure of the reals itself. That is, you want a family {A_x | x in R } such that y in A_x iff the reals satisfy phi(x,y) (or perhaps phi(x,y,z) with a real parameter z). In other words, you want the family to be uniformly projective.

In this case, my answer is that it is consistent with the axioms of ZFC that there is such a family. As I have explained in a few other question/answers on this site, if one assumes the Axiom of Constructibility V=L, then there is a projective well-ordering of the reals, and from this order one can projectively define a Vitali set, and its translates will be as desired.

However, under the set-theoretic axiom known as Projective Determinacy, which is relatively consistent with ZFC if certain large cardinals are consistent, then every projective set is Lebesgue measurable, and so there will be no such family.

So under this projective notion of definability, the answer is that it is independent of ZFC (assuming large cardinals are consistent). I am not quite sure at the moment whether the consistent of "every projective set is Lebesgue measurable" requires large cardinals.

(Edit.) With my closer reading of your question, I see that you asked for a very specific notion of definability. And for this notion of definability, the answer to your question is Yes, for a trivial reason. Let phi(x) be the assertion "x is a non-measurable set of reals". In any model of ZFC, this formula defines a family of non-measurable sets of reals, and it is not difficult to show in ZFC that there are continuum many such sets (for example, as in the comment of Qiachu Yuan). Thus, ZFC proves that {x | phi(x)} is a continuum sized family of non-measurable sets. Indeed, all of them.

In my earlier posted answer, however, I considered the possibility that you might have meant some other notion of definability, or whether parameters are allowed in the definition, and so on. And I find some of these other versions of the question to be quite interesting and subtle.

I pointed out that it is surely consistent with ZFC that there is a definable family of non-measurable sets, since in fact any set at all can be made definable in a forcing extension that adds no reals. So you can take any family of non-measurable sets and go to a forcing extension where this family is definable.

A Vitali set (and there are many) may not itself be definable, so taking translates of just one such set may not produce a definable family, except in that set as a parameter. One might consider the family consisting of all translates of ANY Vitali set, however, which would be similar to the family I proposed above. It is certainly size continuum, and consists entirely of non-measurable sets.

Perhaps a stronger notion of definibility would be to use the notion of projective definitions, where one wants to define the sets within the structure of the reals, using quantification only over reals and natural numbers (rather than over the entire set-theoretic universe). Thus, we want a projective formula phi(x,z), such that A_z={x | phi(x,z)} is always non-measurable for any z and all A_z are different. Such a formula would be a strong example of the phenomenon you seek.

The first answer is that it is consistent with ZFC that there is such a projective family. The reason is that I have mentioned in a number of questions and answers on this site, under the Axiom of Constructibility V=L, there is a projectively definable well-ordering of the reals. Thus, under V=L, one can projectively define a Vitali set, and then take the family of its translations. There is no need for a parameter in this definition, since a particular Vitali set can be projectively defined without parameters from the projectively definable well-ordering of the reals.

The second answer, however, is that under certain set-theoretic assumptions such as Projective Determinacy, every projective set of reals is Lebesgue measurable. In this case, there can be no such projectively defined family of non-measurable sets. The assumption of PD is consistent with ZFC from large cardinals, but perhaps one needs a much weaker hypothesis meerely to get every projective set measurable.

In summary, if one wants a projectively definable family of non-measurable sets, then it is independent of ZFC, if large cardinals are consistent. (Perhaps the need for large cardinals can be reduced or omitted.)