Talk:Vitali set

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I'm going to merge my A subset of R which is not Lebesgue measurable into this article. I plan to keep the basic structure and everything intact, but I want to change a little notation and do a different proof.

First, the Lebesgue measure is denoted λ on the article Lebesgue measure, so I thought it was best to maintain that standard (esp. given that this is not notation for some arbitrary measure satisfying conditions, but a specific one).

As for the other proof, here are the main reasons I think it works better (and these observations came from working through both proofs on my own):

  • In Vitali's original proof (I'm assuming that's what it is), you have to choose your reps to be in the interval [0, 1]. This takes a fleeting moment to verify this is possible, but it seems more natural to be able to pick any set of reps without restriction.
  • The subset relations [0, 1] contained in Union(V_k) contained in [−1, 2] seems unnatural to me. The first containment is not difficult, but neither is it obvious (esp. with no explanation given).
I have added an explanation for this containment as it has puzzled me a lot the first time I read the article. For now, I gave the whole explanation; maybe a small hint would be sufficient to make it more obvious. Simon Lacoste-Julien 11:11, 25 Feb 2005 (UTC)
  • Instead of arriving at the contradiction by boxing in an infinite sum of identical terms between 1 and 3, it seems more natural to me to simply show that R has zero measure -- it's obvious to everyone this can't be true.
  • The other proof easily generalises to any set of positive measure, or any finite-dimensional Euclidean space (take Q^n and R^n, unit cube), e.g.

There is of course, one major drawback to the other proof -- it relies on the fact that the measure of a set is the supremum of measures of compact sets contained in it. This is a price to pay, but it's one of the main properties of Lebesgue measure, so it seems legitimate to assume to me.

Any comments?

Revolver 00:47, 5 Mar 2004 (UTC)

Actually, I'm thinking of putting the original Vitali proof back in (wikipedia is not paper?) since it doesn't depend on inner regularity (explicitly). There are also some other interesting facts from Oxtoby's book that would seem interesting here. Revolver 07:43, 8 Mar 2004 (UTC)

I don't particularly care, but maybe the one that doesn't say "compact" is more accessible to the general public. Loisel 07:51, 8 Mar 2004 (UTC)

I meant, I'll include both proofs (there is no harm in doing so); maybe Vitali's first, and then the one that is there now. I guess I don't feel quite as much unease about using something like compactness...it seems to me most people with the maturity to understand the concept of a nonmeasurable set and seek out its construction would have already been exposed to the definition of a compact set. Revolver 09:01, 8 Mar 2004 (UTC)

For some reason, there is a rush on this topic. See the recent Vitali theorem, as well non-measurable set. Some rationalisation is required.

Charles Matthews 20:26, 4 May 2004 (UTC)

I have redirected Vitali theorem to Vitali set. The content of [[Vitali theorem was this:

There exists a subset of [0,1] that is not Lebesgue measurable.

Contents

[edit] Proof

Suppose for the purpose of contradiction that every subset of [0,1] is measurable.

Consider the equivalence relation on [0,1]: x \sim y \leftrightarrow (x - y) \in \mathbb{Q}

Construct a set A by choosing a single element from each equivalence class. Notice that if q is rational then define q \oplus A := \{q+a |a \in A \}

Since the Rational Numbers in the interval [0,1) are countable we can enumerate them as below:

q_i | i \in N

Notice that \{ q_i \oplus A | i \in N \} is a partition of [0,1)

It is not difficult to see that the \oplus-invariance of the Lebesgue measure follows from the translation-invariance of the Lebesgue measure.

So by the σ-additivity of the Lebesgue measure:

1= m(A)= m(\cup_{i=1}^{\infty})\in \{0,\infty \}

This is clearly a contradition and so we are done.

Charles Matthews 08:58, 5 May 2004 (UTC)

In mathematics, the Vitali set is an elementary example of a set of real numbers that is not Lebesgue measurable. The Vitali theorem is the existence theorem that there are such sets. It is a non-constructive result. The naming is for Giuseppe Vitali.

Despite the terminology, there are many Vitali sets. They are constructed using the axiom of choice, and for reasons too complex to discuss here, Vitali sets are impossible to describe explicitly.

I can see this being easily confusing to someone reading about this for the first time, the first paragraph says it is a nonconstructive result (true, it uses choice in essential way); then the next paragraph talks about the 'construction' of these sets (also true, we are 'constructing' a specific example, albeit using choice to make the "construction") The distinction or use of the term here seems beyond the scope of analysis and into set theory, logic, and philosophy. I understand what is being said, but this self-contradiction may confuse people. Revolver 128.111.88.227 3 days to defense!

[edit] Deleted "another construction"

I deleted this entire section (lets not reproduce it here unless the discussion requires it) for the following reasons:

  • This is not another construction, at best this is another proof of the same construction
  • The other proof is simpler and doesn't require to know much about the technicalities of Lebesgue measure. In other words, it shows that no translation invariant countably additive functional on all the sets of R exists.
  • The allusion to Banach-Tarski paradox is irrelevant. Banach-Tarski is a deep result because it shows that a non-Abelian free group can be embedded in SL(3). The variations necessary in Vitali's technique (if any) are minor.

Gadykozma 12:40, 22 Sep 2004 (UTC)

[edit] Question:Evasions by means of functionals?

Have attempts been made to evade the Vitali result be redefining the concept of measure? Any of these attempts successful? Here's the train of thought leading to my question: Once upon a time, the Dirac delta function was on poor mathematical footing, and requires generalized functions (link distribution (mathematics)) and what-not to make it "legit". So, the "obvious" but naive statement about Vitali is to say that its wrong to take the measure to be "just a number", it should be an "infinitesimal" of some sort. For example, one says "the rationals are a set of measure zero", shouldn't one say "the rationals are a set of measure infinitesimal"? Naively, I guess this contradicts the continuum hypothesis, so naively, this won't work. However, Conway has a stable-full of wild and crazy infinitesimals... any attempts along those lines?

I'm not sure why you say this would contradict the continuum hypothesis (the rationals would still be a countable set, after all). Furthermore, although a natural enticing idea, I don't know that infinitesimals are the answer here, at least if one aims to preserve countable addivity. I imagine you want to say that the Vitali sets V_k used in the proof each have infinitesimal measure e. But then e < 1/2 and e < 1/4 and e < 1/8, etc., so that e + e + e + ... < 1/2 + 1/4 + 1/8 + ... = 1. And similarly we can show that e + e + e + ... < 1/n for every natural n. Thus, we see that e + e + e + e + ... is itself infinitesimal. We still will have that x + x + x + ... is either infinitesimal or infinite in all cases, which, unless I misunderstand what you intend to do, means we will still have the Vitali result. -Chinju 19:24, 19 July 2006 (UTC)

Another possiblity: instead of taking \mu:\sigma\to\mathbb{R}^+ where σ is some (topology-like) collection of sets (for example, sigma-algebra), what if, instead, we took \mu:\sigma\to \mathcal{F} where \mathcal{F} is the set of generalized functions? I'd be curious to read about attempts to evade this result (aside from just chucking out the axiom of choice). linas 03:37, 30 August 2005 (UTC)

[edit] "the rationals are a set of measure zero" contradiction.

I don't agree with this unproven assertion. In fact, if you are saying that you are introducing a contradiction, by saying that the subset of rationals in a interval is Lebesgue-measurable. Just consider this property of Lebesgue-measurable sets:

  • if a subset X of a Lebesgue-measurable set E is Lebesgue-measurable, so is its complement E\X over E

Now use:

  • E = [a,b] (the closed interval of reals between a and b): it is Lebesgue-measurable and its measure is |b-a|
  • X = rationals in E

If X is measurable, and given the measure 0, then E\X, the set of irrationals over [a,b], is measurable and has measure |b-a|.

One problem immediately appears with this definition: given the two measures of E and X\E, you will think that the set of irrationals over an interval are infinitily more dense than the set of rationals over the same interval. But:

  • For any two distinct rationals q1 and q2 in E, there's an infinite number of distinct irrational x between q1 and q2 which is in E\X
    • (this matches the idea that the irrationals are dense).
  • For any two distinct irrationals x1 and x2 in E\X, you can extremely easily find an infinite number of distinct rationals q between x1 and x2; just consider:
    • n=\left\lfloor\frac{|x_1-x_2|}{|b-a|}\right\rfloor which divides the interval [a,b] into a finite number n of equal-sized half-open subintervals so that x1 and x2 belong to two successive subintervals; now these subintervals are easily countable by their integer index between 0 and n-1, it's easy to determine a matching rational q using this integer-numbered scale. and you can then find an infinite number of rationals, by further subdividing this scale in as many subscales as wanted (just subdivide this scale using the set of prime integers as divisors).
    • For this reason, saying that irrationals are more dense than rationals in the same interval is a non-sense; why then giving the measure 0 to the subset of rationals, and the non-zero measure |b-a| to the set of irrationals?

In my opinion, the subset of rationals, even though they are easily countable over [a,b], can't have a zero measure; if it had a measure, it should be the same as the subset of irrationals, and given the property of additivity, the measure of these two subsets should be |b-a|/2 for each; but the two results above indicate that it's impossible to decide which of the two subsets is more dense than the other; for this reason, I tend to think that the subsets of rationals over a closed real interval [a,b] and the subsets of irrationals over the same interval are BOTH non Lebesgue-measurable, given that the Lebesgue-measure is defined only over the set of finite unions of real intervals.

You may desire to have a measure over the set of rationals, but this can't have the same definition as on the set of reals. Don't call this a Lebesgue-measure. For this reason, the sentence "the rationals are a set of measure zero" seems false to me. For me it violates the initial lebesgue condition that the subset to measure must be defined with a finite union of continuous intervals, even if these intervals are reduced to degenerate singletons for each rational. Remember that the set of rationals and the set of irrationals are NOT continuous, so they intersect with the Lebesgue-measurable sets only through an infinite number of singletons!

Another consequence is that you will need to define distinct Lebesgue-related measures for finite unions of rational intervals or for finite unions of irrational intervals, and that these measures are not additive,because they don't count the same thing. This should apply also to the property related to complements: the complement sets must be defined over the same kind of numbers (rationals, irrationals, or real). And why not, you could define related measure over integer ranges, or ranges of Z/nZ fields. Don't say they are the same measure!

So in this area don't try to measure potatoes with the same measure as counting food, when you also need to count fruits, tomatoes, carots... even though many of these sets are included in each other. You are mixing the inclusiuon of subsets and the compatibility of their measures. verdy_p 14:53, 8 August 2006 (UTC)

Check any elementary real analysis textbook for the proof that Q is a null set. This implies lebesgue measure zero. In fact, any countable subset of R has lebesgue measure 0.

The irrationals are uncountable and thus cannot have measure zero.

Also, your assertion that rationals and irrationals should each have(b-a)/2 measure on any interval [a,b] is absurd. —The preceding unsigned comment was added by 64.107.168.75 (talk • contribs) 15:43, 12 September 2006 (UTC)

OK, so you got most of it right, but "null set" as you're using it is actually defined as "set of Lebesgue measure 0", so saying it "implies" it is, while true, a bit odd. More seriously, it's not true that an uncountable set can't have zero measure. The Cantor set (the original, middle-thirds, version) has the cardinality of the continuum, but has Lebesgue measure zero. --Trovatore 19:52, 12 September 2006 (UTC)

OH MY GOD I FORGOT ABOUT THE CANTOR SET, THAT SLIPPED MY MIND, YOU THINK YOU'RE SO GREAT WITH YOUR PHD? THE PROOF THAT YOU ARE A NERD IS LEFT TO YOU AS AN EXERCISE. —The preceding unsigned comment was added by 64.107.168.75 (talk • contribs) 14:34, 14 September 2006 (UTC)

Sorry, that result is too trivial to be worthy of my attention even as an exercise. You know, with me being so great with my PhD, and all. (But what about you, dude? I mean, you're going to an NCAA Division III school.) --Trovatore 04:55, 15 September 2006 (UTC)

It seems a little odd that the page explains how a proof by contradiction works, but it does not explain much more advanced notions. For example, I conjecture that an individual not familiar with proofs by contradiction would not find it obvious that the equivalence classes described in the article partition R.


[edit] Variation

I like the following slight variation, exhibiting a Vitali set which is indeed a linear subspace over Q. It seems to me simpler, although less elementary, than the usual construction. It is true that the usual construction proves more, as observed above by Gadycozma (not only V is not a Lebesgue measurable set, but it could not be measurable for any translation invariant extension of the Lebesgue measure). Anyway:

Remember the following property of the Lebesgue measure (where A-A:={a-a' | a in A , a' in A}) :

Fact: if A is a measurable subset of R^n with positive Lebesgue measure, then A-A is a neighbourhood of 0.

In particular, if G is any (additive) subgroup of R, and G is measurable, then either G is a null set or G=R. On the other hand, if G has countable index, that is, R is countable union of the cosets of G, then G is neither a null set, nor, of course, G=R. Thus we have:

Fact: no subgroup of R with countable index is measurable.

A consequence of the Zorn Lemma is that there are such subgroups: for instance, a complementary linear subspace V of Q in R, where R is considered as a linear space over Q. PMajer 20:11, 23 October 2007 (UTC)

[edit] Vitali sets and the Continuum Hypotheses

To a layman like me unmeasurable sets on the real line are about as close to a true paradox one can get without calling it a true paradox. By a true paradox I mean something really substantial like Russels paradox which provoked the creation of axiomatic set theory. I guess one can rule out plain errors in the definition of various measures. They would surely have been found a century ago - if there were any.

The Axiom of Choice (AC) is clearly involved in this, but AC has been around for so long time and seems to be more or less universally accepted so I take it that AC is not to be held responsible for the sets it produces. (To me a world without AC would be even more wierd than Vitali sets.)

Is there a mathematician around that can give a hint as to wether the Continuum Hypothesis (CH) could possibly be involved?

Here's my question: Is it at all possible that the phenomenon coul´d be explained, in layman terms, that what actually has happened is that CH may be false, i.e for instance, \mathfrak c = \aleph_2 instead of \aleph_1 as in CH, and that measures handle sets with cardinality \aleph_2 and \aleph_0 properly but runs into trouble when they encounter sets with cardinality \aleph_1 that somewhat mysteriously appear because AC does the job for us in the process of creation of the sets? YohanN7 (talk) 20:49, 27 November 2007 (UTC)

No, the cardinality of the Vitali set is always \mathfrak c, no matter what \mathfrak c happens to be in terms of alephs.
However something that seems a little like the reverse of your question has an affirmative answer: It's consistent with ZFC (and actually follows from some principles that some people believe to be actually true) that \mathfrak c=\aleph_2, but all sets of cardinality \aleph_1 are measurable (because they have measure 0). --Trovatore (talk) 21:32, 27 November 2007 (UTC)

Ok, thanks Trovatore. That was a clear answer to a blurry question. The CH is a truly fascinating subject! I did look into the low-level stuff on measures a bit (Borel, not Lebesgue, but anyway) and can at least see now that measurability could be "detroyed" with some effort.217.208.31.144 (talk) 17:05, 28 November 2007 (UTC)

[edit] no excuses (:

"for reasons too complex to discuss here"

  • not good enough (: anything and everything can be fitted into Wikipedia. Why not create Vitali sets are impossible to explicitly describe? --82.130.14.173 (talk) 20:47, 22 January 2008 (UTC)