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By Solovay's theorem, assuming the existence of an inaccessible cardinal, the axiom of choice is necessary to prove the existence of nonmeasurable sets. In the past, I've thought that one consequence of this theorem is that if I construct a set without using choice (or even merely using dependent choice), then I don't have to worry about it being nonmeasurable.

But now I realize that I was making an unjustified assumption. I'm not sure what the appropriate way to precisely phrase this question is, but I'm wondering: is there a non Lebesgue measurable set $E \subseteq \mathbb{R}$ which can be explicitly defined? I'm imagining that perhaps $E$ can be defined without invoking choice (unlike Vitali sets or their cousins), but then proving that $E$ is nonmeasurable requires choice. Is this possibility also ruled out by Solovay's theorem somehow?

Martin Sleziak
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William Hoza
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  • What would "explicitly defined" mean in this context? – Asaf Karagila Jul 27 '15 at 13:08
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    See also this discussion on MathOverflow. – Asaf Karagila Jul 27 '15 at 13:10
  • I intentionally left the phrase "explicitly defined" imprecise; part of this question was "what exactly should I be trying to ask here?" – William Hoza Jul 27 '15 at 13:16
  • My beef with the "explicitly defined" is that you can add a predicate to the language of $\Bbb R$ and interpret it as a non-measurable set, then you have a non-measurable set which can be explicitly defined. So I'm trying to figure out what "explicitly defined" means. What is the language we're using? What is the mathematical universe accessible for the definition? Using just any good ol' notion of "explicit" and "definable" is broad and unclear. And of course, it's a common misconception that I hear from a lot of people. And it makes me sad, to see this regression in understanding logic... – Asaf Karagila Jul 27 '15 at 13:19
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    I understand that the phrase "explicitly defined" is broad and unclear, and I understand that the correct answer to my question might vary wildly based on the interpretation of "explicitly defined." The point is, I don't know how to formalize the phrase in a reasonable way, because I am not a logician! Indeed, I do not have a good understanding of logic -- my hope was that I might learn some logic on this website! I was hoping that answers to my question would help clarify my question, as well as answering it. And it looks like the discussion on MathOverflow does exactly that. – William Hoza Jul 27 '15 at 13:26

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If you can prove that your definition defines something, then will in particular define something in Solovay's model, and that something will in particular not be a nonmeasurable set.

So the best you can hope for is an explicit definition of a subset of $\mathbb R$, such that the thing defined will in some models of ZF be a nonmeasurable set.

And this is in fact the case: Since pointwise definable models of ZFC exist, such a model will contain a nonmeasurable set and that set will have an explicit definition in the language of set theory. (In fact, as Asaf points out in comments, we don't need to appeal to pointwise-definable models here; there is a particular formula that picks out the first nonmeasurable subset of $\mathbb R$ in any model of $\mathbf V=\mathbf L$, where "first" is according to the global well-order, and this one formula will work for the $\mathbf L$ of every model of ZF).

We can even arrange that this particular definition always provably defines some subset of $\mathbb R$ -- if our element of the pointwise definable model is the only set that satisfies the formula $\phi(x)$, then $$ \{ y\in \mathbb R \mid \exists x: \phi(x)\land y\in x \} $$ will define some subset of $\mathbb R$ in every model, and it will at least sometimes be non-measurable.

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hmakholm left over Monica
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    You don't need to go to pointwise definable models. Just work in $L$ and pick the least Vitali set in the canonical $L$ ordering. – Asaf Karagila Jul 27 '15 at 13:11
  • @Asaf: Yes, that seems to give an explicit construction of a suitable $\phi$. – hmakholm left over Monica Jul 27 '15 at 13:21
  • @Henning : $;;;$ Solovay's model is not a model of ZFC, so a priori, he could hope "that the thing defined will in" all "models of" ZFC "be a nonmeasurable set." $:$ (The Krivine result rules that out.) $;;;;;;;;$ –  Jul 27 '15 at 13:24
  • @Ricky: In Solovay's original model, he actually proves that in the $\sf ZFC$ model after the collapse every ordinal definable set of reals is measurable (and in fact every set of reals definable from a countable sequence of ordinals, not just ordinals and reals; at least assuming $\sf GCH$ in the ground model). – Asaf Karagila Jul 27 '15 at 13:28
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From wikipedia: $\:$ "In particular Krivine (1969) showed there was a model
of ZFC in which every ordinal-definable set of reals is measurable, ..."
Therefore ZFC does not prove that there is a definable nonmeasurable set.

Using an explicit definable well-order on $\mathbb{R}\hspace{-0.04 in}\cap$OD to run a Vitali construction
gives an explicit set which won't be measurable if $\: \mathbb{R} \subset \hspace{.04 in}$OD$\;$.