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What is the least level of the constructable hierarchy that contains a non-measurable (Lebesgue) subset of $2^\omega$. If it makes a difference assume we are working inside L (V=L).

I'm pretty sure it must be bigger than $\omega_1^{CK}$ since every Borel set should be measurable. And I think enough large cardinals (projective determinacy) imply that every projective set is measurable. So I guess I'm looking for both the least ordinal this can happen for and the least ordinal it is guaranteed to occur by.

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As an amusing aside chatGPT tried to tell me every set in L was measurable until I reminded it that L was a model of ZFC. 

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    $\begingroup$ It has to be bigger than $\omega_1^{CK}$ simply because it has to be uncountable. So it has to be bigger than $\omega_1$, since all sets in $L_{\omega_1^L}$ are countable in $L$. $\endgroup$
    – Joel David Hamkins
    Commented Sep 11 at 0:11
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    $\begingroup$ It does make a difference whether you assume $V=L$ or not, since if $\omega_1^L$ is countable in $V$, then every set of reals in $L$ is countable and hence measurable, and in this case there is no such $\alpha$. $\endgroup$
    – Joel David Hamkins
    Commented Sep 11 at 0:12
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    $\begingroup$ I think the question makes more sense if it is "what is the least $\alpha$ such that $L_{\alpha}$ contains an $L_{\alpha}$-nonmeasurable set". $\endgroup$
    – 喻 良
    Commented Sep 11 at 6:23
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    $\begingroup$ What did Chatty say in response? Has his training made him believe in Woodin cardinals? $\endgroup$
    – Monroe Eskew
    Commented Sep 11 at 9:10
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    $\begingroup$ @JoelDavidHamkins I also don't know a general rule about keeping or deleting "e" when appending "able" to a verb, but here's a reasonable-sounding approximation: Keep the "e" iff deleting it wold change the (natural) pronunciation of the original verb. For example, deleting "e" from "force" would produce "forcable", which I think most people would pronounce as "forkable". $\endgroup$
    – Andreas Blass
    Commented Sep 14 at 2:25

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The question makes the most sense if we assume $V=L$, since otherwise it could be that every set of reals in $L$ is countable and hence measurable.

If $V=L$, the answer is $\omega_1+1$. Every set in $L_{\omega_1^L}$ is countable in $L$ and hence measurable, and so it cannot be $\omega_1$ or less. But at stage $\omega_1$, the set $2^\omega$ is definable in $L_{\omega_1}$, and we can also define the $L$-order on these reals. So we can define a Vitali set at this level. Namely, we can define the tail-equivalence relation on binary sequences, eventual agreement, and I can pick the $L$-least member of each equivalence class, all definably over $L_{\omega_1^L}$. This set is not measurable using the coin-flipping probability measure on $2^\omega$ in $L$, since we can consider the sets that arise by flipping finitely many digits. These are disjoint, for all the countably many ways to flip digits differently, but the resulting sets all have the same measure as the original, and they cover $2^\omega$, contradiction, just as in the usual Vitali argument.

So the nonmeasurable sets show up at the same time as the whole set $2^\omega$ itself, namely, in $L_{\omega_1+1}$.

It seems that this argument works provided that $(2^\omega)^L$ has positive measure, since it will be covered by countably infinitely many disjoint copies of a fixed set, all with the same measure. And if $(2^\omega)^L$ does not have positive measure, then either it itself is nonmeasurable and exists at this level, or it is measure zero, in which case there will be no nonmeasurable sets in $L$. So the answer is $\omega_1^L+1$, without assuming $V=L$, provided that there is indeed some nonmeasurable set in $L$.

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    $\begingroup$ Regarding the cases mentioned in the last paragraph: $(2^{\omega})^L$ (as well as any $\omega$-model of $\mathrm{RCA}_0$) is either measure 0, maximally nonmeasurable (inner measure 0 and outer measure 1), or equal to $2^{\omega},$ since it is closed under dyadic translation and from a real $r \not \in A=(2^{\omega})^L,$ we can construct infinitely many disjoint translations of $A.$ $\endgroup$
    – Elliot Glazer
    Commented Sep 11 at 3:31
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    $\begingroup$ When does the projective code (if that is a thing) for a nonmeasurable set appear? $\endgroup$
    – new account
    Commented Sep 11 at 4:01
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    $\begingroup$ Just to be clear, you answered the question “what is the smallest $α$ such that $L_α$ contains a subset of $2^ω$ which is not measurable in the universe ($V=L$)?”; another possible interpretation of the (to me ambiguous) question might have been “what is the smallest $α$ such that $L_α$ $\vDash$ «there exists a subset of $2^ω$ which is not measurable»?”. I don't know what can be said about the latter $α$, but it is obviously countable. $\endgroup$
    – Gro-Tsen
    Commented Sep 11 at 13:05
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    $\begingroup$ Yes, that is the question I answered. Your alternative question, also suggested by user 喻 良 on the main post, is subject to what else you want true in $L_\alpha$. For example, it will be sensitive to how exactly you define the measure and how you define what it means to be measurable. You should have a specific theory, such as $\text{ZFC}^-$. But in this case, once you specify the theory, the answer will be connected with the least $\gamma$ for which $L_\gamma$ satisfies that theory. For these reasons, I don't find that version of the question as robust. $\endgroup$
    – Joel David Hamkins
    Commented Sep 11 at 13:13
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    $\begingroup$ If $L_\gamma$ satisfies enough of ZFC, then the argument in this answer should work in $L_\gamma$ to show that a nonmeasurable-in-$L_\gamma$ set appears in $L_{{\omega_1}^{L_\gamma}}+1$. But what is "enough of ZFC"? You certainly need the existence of $\omega_1$ for the statement to make sense. I'd expect that to be equivalent, in models of the form $L_\gamma$, to the existence of $\mathcal P(\omega)$. Without that part of the power set axiom, you'd be looking for a nonmeasurable class of reals, which should appear at level $\gamma+1$ (just after the end of the universe). $\endgroup$
    – Andreas Blass
    Commented Sep 11 at 15:07

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