Does there exist a ZFC-definable set $S$ of real numbers which is not a Borel set, but is Lebesgue-measurable? I understand there are indeed Lebesgue-measurable sets which are not Borel, but every such proof I have seen is nonconstructive. Perhaps there are no ZFC-definable examples.
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The existence of a non-Borel set of reals also depends on the axiom of choice, so the construction is necessarily non-constructive. – David Gao Jun 25 '24 at 22:50
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1If your notion of constructive is not too strict, then a famous example is at What is the Lebesgue measure of Luzin's non-Borel set of reals? In fact, this example "just misses" being Borel, and is nowhere close to being non-measurable, since it's an analytic set and there is a huge gap between analytic sets and non-measurable sets -- see my comments here and Is there a $\sigma$-algebra on $\mathbb{R}$ strictly between the Borel and Lebesgue algebras? – Dave L. Renfro Jun 25 '24 at 23:01
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2@DavidGao That's not correct, see my answer. – Noah Schweber Jun 25 '24 at 23:03
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In fact we can do this rather easily! There are several "concrete" examples of non-Borel analytic (hence measurable) sets. For example, given a relation $R$ on $\mathbb{N}$, let $r_R$ be the real whose $\langle a,b\rangle$th bit in binary expansion is a $1$ iff $R(a,b)$ holds (the map $R\mapsto r_R$ isn't quite injective but that's fine). Then the set $\{r_R:$ $R$ is ill-founded$\}$ is analytic but not Borel. Another cute example is due to Schmerl.
Why doesn't this contradict the non-ZF-provable-existence of a non-Borel set, as David Gao suggests in a comment above? Well, the issue is that there are two natural ways to think about "Borel-ness," and while equivalent over $\mathsf{ZFC}$ they are not equivalent over $\mathsf{ZF}$. See this old answer of mine. Basically, $\mathsf{ZF}$ can prove that each of the above examples is not "Borel-coded," but cannot prove that every Borel set has a code; the moral is that in the absence of choice, "smallest $\sigma$-algebra containing the open sets" is a terrible definition of Borel-ness.
Noah Schweber
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1Ah, that’s a great answer. I suppose my understanding of the matter was incorrect. Thanks for correcting my misperception. – David Gao Jun 25 '24 at 23:26
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