Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice?

Question

What would it imply if there was such an example?

EDIT: For instance, maybe this will help understand the kind of example I had in mind, it is known that an important feature to determine that the AC is needed in the Banach-Tarski case is the non-transitivity of rotations on Euclidean space, one might find an example of a transitive group for some given space keeping the rest equal and make the AC unnecessary?

I guess then it might be said that this transitivity will be in this particular case equivalent to the AC or some amount of it, but I guess that it would be important to show this if it hadn't been realized before.

Although the timing is a bit odd, could this possibly be related to the new PBS Infinite Series video about non-measurable sets and AC? — Asaf Karagila, Sep 09 '17 at 02:09
I often find this question irritating. Not because it's bad. It's a very natural question to ask, of course, but because it sort of mixes syntax and semantics in a terrible way. Since ZF is consistent with "all sets are measurable" or even with the far more extreme "there is no nontrivial countably additive measure which vanishes on singletons", it is clear that ZF itself cannot prove that any definable set is measurable. But your question is talking about "sets which exist, and we just can't prove they are [non-]measurable". Exist where? — Asaf Karagila, Sep 09 '17 at 02:11
(The point is that we can write a certain definition for a set to be either the least set in L which is non-measurable in the universe if such set exists, or empty otherwise. We certainly don't need to use choice to prove this set exists; but we need even more than just choice in order to prove that it is non-measurable. And yes, this is the usual "ugh, I hate talking to mathematicians/logicians/whatever" sort of answer, but it points a significant flaw in the naive formulation of the question.) — Asaf Karagila, Sep 09 '17 at 02:14
Also, just a quick search in my posts brought up a couple of related questions: https://math.stackexchange.com/q/133999/; https://math.stackexchange.com/q/1620692/; and a handful of vaguely-related threads about independence results of this flavor, or otherwise measure-related questions. — Asaf Karagila, Sep 09 '17 at 02:24
@Asaf Karagila Hi, thanks for commenting! 1) I had no idea about the PBS series, I just googled it and it was publised a day after I asked, (this might be what you refer to by odd timing which it would be more than odd if their was a direct relation due to the nature of time, but maybe you were wondering if I'm from PBS and had taken part in the making of it). — bonif, Sep 09 '17 at 09:47
Yes, that's what I had meant by 'odd timing', that maybe you're part of the production team and were wondering about this for some future episode. :) — Asaf Karagila, Sep 09 '17 at 09:48
(cont'd) But I'm concerned that in applied mathematics(which is always trying to avoid the AC) this could lead to problems with those sets, as we know that without the chance to tag them as non-measurable they lead to nonsensical consequences. Now they could be evident enough that they are esily caught since they clash with observations, but when they are not so eviden or don't lend themselves to direct verification they could lead applied math models astray. — bonif, Sep 09 '17 at 10:05
I don't understand your comment. You cannot prove that the paradoxical decomposition of the BT exists. And you cannot prove that a Vitali set exists. So what does it mean that ZF "proves that these sets exist, but not that they are non-measurable"? And to your argument that the issues are inevitable, this is exactly why mathematics is done slowly, one step at a time, and carefully. There is no royal road to foundations of mathematics. — Asaf Karagila, Sep 09 '17 at 10:10
The sense in which I mean they exist is they exist as possibly nonmeasurable sets in ZFC( see https://math.stackexchange.com/questions/142499/are-sets-constructed-using-only-zf-measurable-using-zfc ). — bonif, Sep 09 '17 at 11:13
Ok, I'll clarify here: an example of what I have in mind is that we could consider two different groups acting on their respective sets/spaces. One of them is rotations in Euclidean space, this group is known no to require AC for its action in Euclidean space,but let's say the other requires AC for its action on a certain space but this can only be acknowledged in ZFC, and yet one insists on not leaving ZF. I guess this is why standard rigorous measure theory is done in ZFC. — bonif, Sep 09 '17 at 14:15
But as I said I was concerned with science-applied math, where there is an insistence to not use and even contradict the AC and claim everything can be done without going at all to full ZFC, as if not having the possibility to "detect" and tag nonmeasurable sets would amount to their non existence so problem solved(ostrich strategy of sorts). — bonif, Sep 09 '17 at 14:15
You misunderstand the proof of the BT paradox. The group action is just fine and requires no choice to be well-defined, since the groups are relatively simple enough. What you need to use choice is to pick representatives from the different orbits. These sets exists in a model of ZFC, but not necessarily in a model of ZF. — Asaf Karagila, Sep 09 '17 at 14:42
You missed my point. Picking those representatives would not be necessary if the group action was transitive, but it isn't for rotations in BTand therefore the need for the AC. — bonif, Sep 09 '17 at 16:07
I didn't say the AC was needed for the action to be well defined, I wrote that a group action might imply the use of AC but that could be not manifestly provable in ZF. — bonif, Sep 09 '17 at 16:11
Yes, I may have missed your point, but you also missed the point. Choice is not making that action transitive. Choice lets you circumvent the whole goddamn thing. — Asaf Karagila, Sep 09 '17 at 16:11
That is correct. But again I'm not saying that choice makes the action transitive, but that an paradoxical action that already implies choice(in ZFC) doesn't need it to circunvent the whole thing. In this case a group that unlike rotations on Euclidean space is transitive to begin with doesn't need the explicit use of AC. There are examples of such paradoxical actions. — bonif, Sep 09 '17 at 16:21
Okay, I'm officially lost now. You are claiming that there are group actions which are provably transitive, and some which are not. Okay, and...? Remember that the BT paradox was part of the original motivation for amenable groups. Then I don't see what the axiom of choice is even doing in this discussion, other than being a huge red herring (and unfortunately, in most "non-AC" discussions, choice ends up being a red herring)... — Asaf Karagila, Sep 09 '17 at 16:23
I'm talking about paradoxical actions and non-amenable groups. — bonif, Sep 09 '17 at 16:26
Yes. I understand that. And I am saying that choice is used to prove the existence of representatives for the orbit equivalence relation, and and without choice it's not that we cannot prove that the set of representative is non-measurable, we can outright prove it does not exist. So if you want to talk about its existence and whether or not it is measurable, then choice is a red herring here, and you should focus on other questions in ZFC, like whether or not the group is amenable. — Asaf Karagila, Sep 09 '17 at 16:28
I'm trying to get across is the following case that is not the BT case, because I'm asking to consider in the context wanting to remain in ZF a paradoxical action of a G whose action on X is transitive so it doesn't need to invoke explicitly the AC to pick one point from each orbit. The set X it acts on is nonmeasurable, right? But we have not had to use AC in this step, this must mean that somewhere in the paradoxical action of this G there is the AC(or some amount of it) implicit because we know that it or some amount of it is needed to construct a non-measurable set but then are we in ZF? — bonif, Sep 09 '17 at 16:44
"Assume there is a natural number whose successor is 0; but 0 is not a successor of any natural number by the axioms of PA; therefore there is a problem!". If you assume there is a paradoxical action which admits a choice of representatives, then you invariably assumed there is a non-measurable set. You can't quite avoid that, because it is enveloped in the definition of a paradoxical action. — Asaf Karagila, Sep 09 '17 at 16:48
Yes, that's what I'm doing, let's say I have an example of such paradoxical action, then I must be in ZFC, right? — bonif, Sep 09 '17 at 16:56
No. You can just work in ZF+"Banach--Tarski paradox", which is itself is weaker than ZF+"Hahn--Banach theorem", which is itself is weaker than ZF+"Ultrafilter lemma", which is itself is weaker than ZFC. And other than ZFC, all those mentioned cannot even prove that every infinite set of real numbers has a countably infinite subset. Not to mention, you might work in ZF+"Many various failures of choice"+"The real numbers can be well-ordered", which is also just as plausible as ZFC or ZF. — Asaf Karagila, Sep 09 '17 at 16:58
Fair enough, I intended it to mean that I cannot be in ZF only. But if I don't have evident consequences of the nonmeasurability of my set, and I don't know that this action is paradoxical because to prove it I would need more than just ZF, and I don't wnat to use that, I can fool myself into thinking that I'm still in ZF instead of ZF+something like what you mentioned. This was my concern with respect to applied math. — bonif, Sep 09 '17 at 17:05

Noah Schweber · Answer 1 · 2017-09-06T19:37:24.737

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No, there is not - it is consistent with ZF that every set of reals is Lebesgue measurable, so some amount of the axiom of choice is needed.

This is even true if we assume dependent choice, a certain reasonably strong choice property (enough to do most useful theorems from analysis): this was proved by Solovay. He needed an additional assumption, though: that the theory ZFC+"There is an inaccessible cardinal" is consistent. It was later shown by Shelah to be necessary: we can prove that ZF+"There is an inaccessible cardinal" is consistent if and only if ZF+DC+"All sets of reals are measurable" is consistent.

In lieu of choice, I believe the consistency of ZF+"All sets of reals are measurable" from merely the consistency of ZF (the weakest possible assumption here) was proved by Truss. However, it's worth pointing out that in ZF alone, "measurable" doesn't necessarily mean what you think. For example, it's consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, and in that context basic analytic properties like measurability go absolutely bonkers. So really the right base theory to look at, in my opinion, is ZF+DC; and then it's a very interesting fact that an inaccessible is needed.

edited Sep 06 '17 at 19:37

answered Sep 06 '17 at 19:30

Noah Schweber

245,398

+1. I'd like to point out that on the other hand you can have non-measurable sets in $ZF\neg C$. It really depends on which $ZF\neg C$ system you're talking about, there are a lot of them and some of them are quite different from others. – Ian Sep 06 '17 at 19:31
@Noah Schweber:well, "some amount of the axiom of choice" but not the explicit axiom itself might be shown to be implicit or hidden and unbeknownst so far in some assumption in certain mathematical structure. On the other hand I think that ZF+Hahn-Banach theorem is enough for the Banach-Tarski theorem proof, so that might be what you are referring to by "some amount of the AC is needed". In a way this is an answer in the affirmative of my question given that it was meant to refer to the explicit AC. I might edit the question to make it clearer what I was asking. – bonif Sep 06 '17 at 22:03
@bonif Ah, are you just asking for a proof which doesn't appear to invoke Choice? (It will have to invoke something which can't be proved in ZF alone, so ...) – Noah Schweber Sep 06 '17 at 22:10
@Noah Schweber Yes, that doesn't invoke it explicitly, so I guess I'm considering the possibility that certain math results that don't need proofs outside ZF might have implicitly certain amounts of AC. The fact that ZF is consistent with AC and its negation is attributed to its independence but it might also mean that is not strong enough to deal with some element of inconsistency in some of its implications. – bonif Sep 06 '17 at 22:41
I took the question (as stated in the title) to mean, is there a set $S$ whose existence can be proved without choice, so that the axiom of choice is only needed to prove nonmeasurability of $S.$ That's probably not what the OP had in mind, and I don't know how you would make it precise. – bof Sep 07 '17 at 02:06
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@bof Well, the natural way to say that would be to ask for a formula $\varphi$ such that (i) ZF proves that there is exactly one set of reals satisfying $\varphi$ and (ii) ZFC proves that that set is nonmeasurable. I think that's perfectly precise, and I don't know the answer off the top of my head. (If we replace C with V=L, then it has an easy positive answer since V=L gives canonical nonmeasurable sets (e.g. the $<_L$-least such), but C alone doesn't provide that kind of canonical structure.) – Noah Schweber Sep 07 '17 at 13:51

score 6 · Accepted Answer · answered Jan 07 '18 at 17:41

Yes.

Or more precisely,

Theorem: We can explicitly construct in ZF a set $S$ that ZF cannot prove to be measurable.

The set $S$ could then be proven non-measurable by assuming a suitable choice principle. So the role of choice is not in the existence of the set $S$, but in the proof it is nonmeasurable.

The proof of this theorem involves the constructible universe. In particular, one can explicitly define in ZF a well-ordering on the class of constructible sets.

One can construct $S$ by carrying out one's favorite construction of a non-measurable set inside the constructible universe, using the aforementioned explicitly constructible well-ordering instead of invoking the well-ordering theorem.

One cannot prove in ZF that $S$ is actually a non-measurable set. However, you can do so if you further assume the axiom of constructibility.

bonif · Answer 3 · 2017-09-08T23:17:26.563

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I think bof's interpretation of my original question is reasonable. I'm compelled to the following argument as an anwer to it in those terms but I'm certainly no set theorist so any correction is welcome: It seems to me that an affirmative answer would imply that ZF is not powerful enough to produce proofs about set measurability without the aid of the AC, probably because the AC (or the relavant part of it that bears on measurability) may be a direct consequence of the primitive notion of set that is needed to construct the ZF axioms and therefore ZF without an explicit extra axiom like AC cannot prove the existence of such an unmeasurable set. Unfortunately this would mean this affirmative answer could not be proved within ZF so it ultimately has to appeal to the AC wich would actually answer the original answer in the negative, leading to a situation of undecidability of the question within ZF to avoid contradiction.

edited Sep 08 '17 at 23:17

answered Sep 08 '17 at 09:45

bonif

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This sounds like a really-over-complicated way of saying "The existence of a non-measurable set is independent of ZF," which is exactly what I said in my answer - or is there something more here? – Noah Schweber Sep 10 '17 at 15:50
You are right that it amounts to basically this, if anything the something more that I might mention is that, using reciprocity derived from LEM, the existence of nonmeasurable sets being independent of ZF seems to suggest that the existence of measurable sets may also be independent of ZF. See also: https://math.stackexchange.com/questions/142499/are-sets-constructed-using-only-zf-measurable-using-zfc – bonif Sep 10 '17 at 20:21
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No, ZF easily proves that e.g. the emptyset is measurable; the problem is entirely with non-measurability. This kind of reciprocity is not at all a consequence of LEM. – Noah Schweber Sep 10 '17 at 20:25
You are right. For nonempy finite sets it is hard for me not to link existence of this property and its negation. – bonif Sep 10 '17 at 21:02

Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice?

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