Talk:Non-measurable set

Opening sentence

Why does the opening sentence restrict us to finite measures? There are certainly nonmeasurable subsets of the real line which are contained in no set of finite measure. —Preceding unsigned comment added by 68.185.170.23 (talk) 04:29, 28 January 2009 (UTC)[reply]

I agree. I see no reason why this restriction is made. I think it should read "In mathematics, a non-measurable set is a subset of a set with non-zero measure...", since any subset of a set with zero measure is defined as measurable (at least in Lebesgue measure). --67.193.128.233 (talk) 18:12, 8 March 2009 (UTC)[reply]

Nonsense in the article

quote : ...Vitali's theorem which basically states that you can take an interval of length 1, dissect it into pieces, move the pieces around and get an interval of length 2 (sometimes this result is called the Hausdorff paradox).

This is nonsense, you cannot get an interval of length 2. For dimensions 1 and 2 Banach proved existence of finitely additive translation invariant measures. He actually constructed them. Whoever wrote this should actually learn what Hausdorff-Tarsky Paradox is about. It is in R^3 ! Vitali's set is NEVER called the Hausdorff paradox.

—The preceding unsigned comment was added by 206.188.85.61 (talk) 19:04, 2 May 2007 (UTC).[reply]

Previous version

Here is the text of this page before my changes:

In measure theory, a non-measurable set is one which does not belong to the algebra of measurable sets of some measurable space. The usual context for this is stated in terms of translation invariant measures on the set of reals R.

Theorem. There is no countably additive measure on all subsets of R which is translation invariant and is finite and non vanishing on [0,1].

This follows from Theorem D, Section 16, of the Halmos reference below. That theorem requires use of the axiom of choice. It follows immediately from this theorem, that non-measurable sets exist for any countably additive translation invariant measure which is finite and non vanishing on [0,1]. This is true in particular, for the algebra of Caratheodory-measurable subsets of R relative to the outer measure defined as follows:

\phi (A)=\inf \left\{\sum _{i=1}^{\infty }\operatorname {length} (I_{i})\right\}

where the infimum above is taken over all countable covers of A by intervals {I_i}_i.

Non-measurable sets are highly non-constructive and mathematicians consider them to be extremely pathological.

A slight extension of the above theorem is as follows:

Theorem. There is no countably additive measure on all subsets of Rⁿ which is translation invariant and is finite and non vanishing on the n-dimensional cube [0,1]ⁿ.

The decomposition of the unit ball in R³ into five disjoint subsets as stated in the Banach-Tarski paradox involves non-measurable sets.

For an explicit "construction" of a non-measurable set, see Vitali set.

References

Paul R. Halmos, Measure Theory, D. van Nostrand Co., 1950.

I removed this text because:

We have plenty of pages describing this topic from a technical point of view.
This one just states a few theorems with no explanation and a link to the horrible book of Halmos. The exposition in Vitali set is much nicer.

Not theorems - all open problems!

I believe all of the above have only been proved with the axiom of choice. AFAIK, the existence of Lebesgue-unmeasurable sets in ZF (and ZF with the negation of AC) is still an open question.

Therefore, the page is quite misleading. If I'm wrong, can someone write explicitly that it has been proved, and give a reference?

--Clausen 07:47, 9 Oct 2004 (UTC)

No, you are not wrong, they depend on the axiom of choice. This is not an open problem, though. It is known that no such construction can be done without it.
The page explicitly states that. Read it all through the end.

Gadykozma 12:03, 9 Oct 2004 (UTC)

This is not quite right. There are plenty of sets that can be constructed without the axiom of choice which may or may not be measurable. You can add the axiom that these sets are measurable without deriving a contradiction, but it's not true that they are measurable within ZF (or ZFC). -- Walt Pohl 07:08, 22 September 2005 (UTC)[reply]

It is known that the existence of a nonmeasurable set of reals cannot be proved in ZF. To be precise, if there is such a proof, then the theory "ZFC+there exists an inaccessible cardinal" is inconsistent; while some would claim that this conclusion is not "known" to be false, you could say the same thing about the consistency of ZFC itself, or even much weaker theories (say PA).

Walt Pohl is correct if he means what I think he means, but the way he phrases it is somewhat unfortunate--"measurable within ZF" is meaningless, a category error. ZF is syntactic, not semantic. He's right, though, that you can give a definition for a set, such that ZF neither proves nor refutes the measurability of the set thus defined. --Trovatore 05:54, 29 September 2005 (UTC)[reply]

Borel or Lebesgue?

"The measurable sets on the line are formed by countable unions and intersections of intervals"

AFAIK these are Borel-sets, while the article talks about Lebesgue non-measurable sets (the Borel non-measurable sets aren't that interesting because their existence is trivial and is not "a source of great controversy"). bungalo (talk) 10:09, 17 November 2010 (UTC)[reply]

I am not sure that the lead of this article (that "gives a general overview of the concept") must be so exact. Anyway, I have added the missing reservations. Boris Tsirelson (talk) 14:27, 17 November 2010 (UTC)[reply]

Well, really, that's the problem with this sort of article. To be honest I think this article should not exist. It should be merged somewhere. Wikipedia should never involve itself in "lies to children". --Trovatore (talk) 19:43, 17 November 2010 (UTC)[reply]

That is, Wikipedia should include some scientific literature, but not some popular science. Really? Why? Boris Tsirelson (talk) 22:06, 17 November 2010 (UTC)[reply]

There is nothing wrong with trying to make things clear to readers of limited background. We just can't permit that to be an excuse for saying things that aren't true. --Trovatore (talk) 22:12, 17 November 2010 (UTC)[reply]

More nonsense

There is perpetual confusion between saying the negation of X is consistent, and jumping to saying it is true. The joker is, of course, the need for Solovay to assume the existence of an inaccessible cardinal... this article needs drastic rewriting in this regard. — Preceding unsigned comment added by 98.109.239.253 (talk) 03:24, 5 February 2012 (UTC)[reply]

I'm not sure I understand your complaint. I don't see anywhere in the article that claims it is true that all sets are measurable (I'm guessing that's your "negation of X"?), only that it's true in the Solovay model. And yes, if there's no inaccessible, then there's no Solovay model. So what? --Trovatore (talk) 20:13, 5 February 2012 (UTC)[reply]

Oh wait a minute, maybe I see. Is your complaint that claim that if the existence of an inaccessible is consistent, then the Solovay model exists? I have to agree that that is suboptimally worded. In context, the point is that if it's consistent that there's an inaccessible, then it's consistent that all sets of reals are measurable. It follows that there's a model that satisfies "all sets of reals are measurable". However, it doesn't follow that that model is the Solovay model. I kind of think "more nonsense" is putting things a little strongly, but it's true that the text is not quite correct as worded. Any suggestions for fixing it while still getting the point across, and hopefully without overly complicating the text? --Trovatore (talk) 21:21, 5 February 2012 (UTC)[reply]

countability in example

In the example of the unit circle, it says "the circle gets partitioned into a countable collection of disjoint sets." Aren't there uncountably many disjoint orbits? I see the same thing on the axiom of choice page.

173.25.54.191 (talk) 22:48, 19 November 2012 (UTC)[reply]

Yes, there are uncountably many disjoint orbits (each of which is countable). But that's not the partition the text is talking about. If you have a set that picks out exactly one point from each orbit, then the collection of all translates of that set is a partition into a countable collection of disjoint sets. Maybe this point can be stated more clearly; any suggestions? --Trovatore (talk) 23:09, 19 November 2012 (UTC)[reply]

Man you're fast! I figured it out and came back clear my comment. Instead of "In other words, the circle gets partitioned . . . ", could we say "The set of those translates partitions the circle . . . "? I'll take the liberty and let you veto it.

173.25.54.191 (talk) 01:32, 20 November 2012 (UTC)[reply]

Looks fine. Except, is everyone going to understand what translate means as a noun? Might have to think about that... --Trovatore (talk) 02:27, 20 November 2012 (UTC)[reply]

Example

The article says

Consider the unit circle S, and the action on S by a group G consisting of all rational rotations. Namely, these are rotations by angles which are rational multiples of π. Here G is countable (more specifically, G is isomorphic to

\mathbb {Q} /\mathbb {Z}

) while S is uncountable.

My question is how $\mathbb {Q} /\mathbb {Z}$ can be a group? What is its identity element and group operation then? And what is the isomorphism between these sets? I believe the author meant $\mathbb {Q} /\{0\}$ (or rather $\mathbb {Q} \setminus \{0\}$ ). I am right?

94.193.53.206 (talk) 11:26, 24 January 2013 (UTC)[reply]

The quotient of any abelian group G by its subgroup H has a canonical structure of a group, and Q/Z is no different. The identity element is still 0 and the group operation is addition. On the unit circle in C, the group operation is multiplication. The isomorphism is given by the exponential function

e^{ix}

(up to adjusting the constant in the exponent). Tkuvho (talk) 15:14, 24 January 2013 (UTC)[reply]

Contradiction

This article states that

[Lebesgue-measurable] sets are rich enough to include every conceivable definition of a set that arises in standard mathematics

but then (in the 'Example') section, gives a definition of a set that is not measurable. Since measure theory is certainly 'standard mathematics' and we have just conceived of the definition of a non-measurable set, the initial assertion must be false.

J.Gowers 14:17, 3 September 2019 (UTC)[reply]

Not quite so. Note the words "Using the axiom of choice, we could pick a single point from each orbit" in the "Example". This is not quite a definition of a set; the set is still not specified; instead, its property is specified (to contain exactly one point in every orbit); the choice axiom just postulates existence of such sets; but there are quite a lot of them. A subtlety, indeed. I'd prefer to write "proof of existence" rather than "example", but I know that many mathematicians are not scrupulous at this point. And indeed, no one expert in measure theory or set theory is able to specify a single nonmeasurable set. Boris Tsirelson (talk) 17:39, 3 September 2019 (UTC)[reply]

Actually I think the quoted text is a bit problematic. It follows for example from V=HOD that there is a formula φ such that the reals satisfying φ form a non-measurable set. (And while V=L is clearly false, I don't think there's any such consensus against V=HOD.) What you can say is that there's no definition that provably in ZFC defines a non-measurable set of reals, but that is a much weaker claim than the one in the article. --Trovatore (talk) 19:01, 3 September 2019 (UTC)[reply]

Yes. And moreover, some models of ZFC are "pointwise definable" (see here Sect. "Pointwise definability"), that is, everything is definable without parameters!

But indeed, a mathematician not involved into such deep logic usually by a "true" statement means "provable in ZFC", and so, by "example" means what you wrote finally. Boris Tsirelson (talk) 19:13, 4 September 2019 (UTC)[reply]

Yes, but Boris, those models are obviously not the intended model, because they're countable. And non-logician mathematicians rarely think about ZFC explicitly, so I'm skeptical that it has much to do with what they mean by "true".

In any case, regardless of what you personally meant by "example", the current text says [t]hese sets are rich enough to include every conceivable definition of a set that arises in standard mathematics, which is not clear even by your reading. If you want to read assertions as claiming that something is provable in ZFC, then the statement should be read as that it's provable in ZFC that there is no definition that picks out a non-measurable set, and that statement is simply false.

The only possible way to save it is if the word "standard" in "standard mathematics" is doing more work than is immediately apparent. I think it's true, for example, that you can't define such a set by quantifying only over the reals. You probably have to quantify over, say, sets of reals, or ordinals below ω₂. You can make the argument, I guess, that none of that is "standard mathematics", but this would be a longer discussion. --Trovatore (talk) 22:03, 4 September 2019 (UTC)[reply]

Yes. Then probably we should retreat to sources. It would be interesting "by quantifying only over the reals" or something like that, but is any such theorem available? Or, on the other hand, we may wonder: (a) is there a published claim that such-and-such explicitly defined set of reals is non-measurable? And if it is, then (b) is it published in analysis, measure theory, or logic? Or maybe we could write that no one of well-known proofs of existence (of non-measurable sets) singles out just one set (unless one global choice function is treated as given once and for all, as in Bourbaki)? Boris Tsirelson (talk) 04:16, 5 September 2019 (UTC)[reply]