A question about "small" uncountable cardinal numbers If $X$ denotes a set, let $C(X)$ denote its cardinal number and let $P(X)$ denote its power set. There is a school of thought which considers any set having the cardinal number $C(P(\mathbb{R}))$—where $\mathbb{R}$ denotes the set of real numbers—to be too large for the intuition to grasp. This school mantains that in almost all branches of mathematics except set theory itself, there is no need to require the existence of any set whose cardinal number is greater than $C(\mathbb{R})$.
Although this viewpoint sounds plausible, I wonder whether it might lead to problems in Measure Theory. Suppose one wants to prove in $\mathsf{ZFC}$ that there  exist sets of real numbers which are not Lebesgue measurable. Does there exist any set $M$, definable in $\mathsf{ZFC}$, which satisfies the following conditions:


*

*$M$ is a subset of $P(\mathbb{R})$,

*It is provable in $\mathsf{ZFC}$ that at least one element of $M$ is not Lebesgue measurable, and

*The cardinal number of $M$ is $C(\mathbb{R})$?


I know a number of examples of sets—definable in ZFC—that satisfy (1) and (2), but none of them also satisfies (3).
 A: Here is a positive answer, unfortunately it is predicated on a hypothesis which is widely believed to be false: that the existence of inaccessible cardinals is inconsistent with ZFC. Stated in a more digestible manner, the argument below shows that some large cardinal hypotheses are necessary to obtain a negative answer to the question.
Let $V$ be a model of $ZFC$, let $\mathfrak{c}^V$ denote the cardinality of the continuum in $V$ and let $\aleph_1^V$ denote the first uncountable ordinal in $V$. In $L$, $\aleph_1^V$ must be an uncountable regular cardinal and, since there aren't any inaccessible cardinals, it must be a successor cardinal. Let $\kappa$ denote the cardinal in $L$ such that $(\kappa^+)^L = \aleph_1^V$; note that $\kappa$ is definable in $V$. In $V$, $\kappa$ is countable, so there are reals $r$ in $V$ such that $\kappa$ is countable in $L[r]$ and for such reals we necessarily have $\aleph_1^{L[r]} = \aleph_1^V$. Let $K$ be the set of all reals such that $\aleph_1^{L[r]} = \aleph_1^V$; note that $K$ is definable in $V$ and $K$ has size $\mathfrak{c}^V$. For each $r \in K$, $X_r = \mathbb{R}^{L[r]}$ is a set of reals with a canonical wellordering of order type $\aleph_1^{L[r]} = \aleph_1^{V}$. 
Following Raisonnier [A mathematical proof of S. Shelah's theorem on the measure problem and related results, Israel J. Math. 48 (1984), no. 1, 48–56; MR0768265; DOI:10.1007/BF02760523], we can associate to each $r \in K$ a filter $F_r$ and a collection of sets $\mathcal{H}_r$ of size at most $\mathfrak{c}^V$ such that either $F_r$ is rapid and hence not measurable, or at least one element of $\mathcal{H}_r$ is not measurable. (The set $\mathcal{H}_r$ consists of all sets that Raisonnier denotes $\tilde{H}(X_r)$. These are associated to $G_\delta$ sets $H \subseteq \mathbb{R}\times\mathbb{R}$ with null sections; since there are continuum many such $G_\delta$ sets, it follows that $|\mathcal{H}_r| \leq \mathfrak{c}^V$.)
Putting all this together, under the unlikely assumption that inaccessible cardinals provably do not exist in ZFC, in any model $V$ of ZFC, the definable set $$\{F_r : r \in K\} \cup \bigcup_{r \in K} \mathcal{H}_r$$ has size $\mathfrak{c}^V$ and it must contain a non measurable set.
A: This is independent. See: Harvey Friedman, On definability of nonmeasurable sets, Canad. J. Math. 32 (1980), no. 3, 653--656.
