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Is there a formula $\phi$ in the language of set theory such that $$ \text{ZFC proves } \exists x \in \mathbb{R}:\text{ the set }A_x​:=\{y\in\mathbb{R}:\phi(x,y)\} \text{ is not Lebesgue measurable?} $$ In other words, is there a process that takes a countable amount of information as input and, given the right input, outputs a Lebesgue non-measurable set of real numbers?

Of course if one strengthens ZFC with, say V=L then this is possible. In ZFC + (V=L) one can construct non-measurable sets without needing to make any arbitrary choices.

I am asking this question also in connection with descriptive set theory. We know that it is (probably) consistent with ZFC that all projective sets are measurable, so one would have to go beyond the projective hierarchy to find such a $\phi$. How far beyond this hierarchy do we at least have to go in order to have any chance at this?

If we did the opposite, namely adding the negation of this as an axiom schema to ZFC, so $$ \text{ZFC} + \forall x\in \mathbb{R}:\{y\in\mathbb{R} : \phi(x,y)\}\text{ is Lebesgue measurable}, $$ with the restriction that $x$ and $y$ are the only free variables of $\phi$, is there any chance that this formal system might be consistent?

My original question is equivalent to this system being inconsistent (since any inconsistency proof can only use finitely many formulas $\phi_i$, $i=1,\ldots,n$ and the resulting definable relations on $\mathbb{R}$ can be mapped to relations from $]i,i+1[$ to $\mathbb{R}$ and then rolled into one by taking union). Intuitively this axiom scheme says that any set of reals definable from countable information is Lebesgue measurable. In particular this formal system proves that all projective sets are measurable, but it obviously proves much more.

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Any set of reals that is defined using real and ordinal parameters is not provably nonmeasurable: Solovay showed that in $V[G]$ (where $G$ is $Coll(\omega,<\kappa)$ generic and $\kappa$ is inaccessible) every such definable set is measurable.

In fact, Friedman proved an even stronger result that it is consistent (relative to an inaccessible) that every family of sets of reals definable using real and ordinal parameters of size $<2^{2^{\aleph_0}}$ contains only measurable sets.

So to "find" a nonmeasurable set, you need to either allow parameters from $2^ℝ$ or to look into families of cardinality $2^{2^{\aleph_0}}$; unfortunately, both cases are pretty trivial.

Friedman, Harvey, On definability of nonmeasurable sets, Can. J. Math. 32, 653–656 (1980). ZBL0487.03027.

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    $\begingroup$ It can only be the case that a $\mathfrak{c}$-sized parameter is necessary to define a nonmeasurable set under CH. It's a theorem of ZFC that some subset of $\omega_1$ defines a nonmeasurable set of reals, as I show in my answer. $\endgroup$ Commented Nov 10 at 1:06
  • $\begingroup$ @ElliotGlazer I didn't know that, thanks. It is a bit surprising to me because in Friedman strengthened result we have that $2^{\frak c}$ can be arbitrarily big $\endgroup$ Commented Nov 10 at 14:41
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A nonmeasurable set of reals can be defined with $\omega_1$ bits of information.

Let $s$ be an injective $\omega_1$-sequence of reals. Work in $M=L(s, \mathbb{R}) \models \mathrm{ZF}+\mathrm{DC}+\aleph(\mathbb{R}) > \omega_1,$ where there is a nonmeasurable set of reals $S$ by Raissonnier and Shelah's theorem ([1] and [2]).

$S$ is definable from $s,$ some real parameter $r,$ and an ordinal $\alpha.$ We can define from $(s, r)$ some nonmeasurable set $S'$ as the first nonmeasurable set in $\mathrm{OD}^M_{s, r}.$ Then $S'$ is also nonmeasurable and definable from $(s, r)$ in $V.$

[1] Shelah, Saharon, Can you take Solovay’s inaccessible away?, Isr. J. Math. 48, 1-47 (1984). ZBL0596.03055.

[2] Raisonnier, Jean, A mathematical proof of S. Shelah’s theorem on the measure problem and related results, Isr. J. Math. 48, 48-56 (1984). ZBL0596.03056.

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