# Are sets constructed using only ZF measurable using ZFC?

Suppose $S$ is a subset of $\mathbb{R}$ which can be defined without using the axiom of choice, i.e. which can be proved to exist using only the axioms of ZF. Does it follow that $S$ is measurable?

We know that ZF + “All subsets of $\mathbb{R}$ are Lebesgue measurable” is consistent (assuming ZF is), but the claim doesn’t follow from this alone, since there could be a proof that $S$ is not measurable which uses choice.

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To see the “no” answer, there is a formula $\phi(x)$ such that a real $r$ satisfies $\phi$ iff $r$ is in the constructible universe $L$. Similarly, there is a formula defining a well-ordering $<_L$ of $L$, and therefore there is a formula $\psi$ such that $\psi(r)$ holds iff $r$ is a real in $L$, and it belongs to the $<_L$-first non-Lebesgue measurable subset of ${\mathbb R}$ in $L$. Such a set exists, since (provably in ZF) $L$ satisfies ZFC, and it is a theorem of ZFC that there are sets that are not measurable.
Now, it is consistent with ZF that $V=L$, that is, that every set is constructible. But then the set of reals defined by $\psi$ would not be measurable. This shows that one cannot prove in ZF (or ZFC) that any ZF-definable set of reals is measurable.
On the other hand, one of the most significant results in set theory is the theorem of ZFC + large cardinals that any reasonably definable set of reals is measurable; for example, any set of reals in the universe $L({\mathbb R})$, and this certainly includes all sets “naturally definable”.