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If AC be accepted, then there exists a Lebesgue unmeasurable set called Vitali Set. However, I'm curious about measure valued in hyperreal numbers. Argument in disproof of unmeasurability of Vitali sets had used the fact that no positive real number countable infinite sums finite. But can the Vitali sets have infinitesimal measure? Then the countable infinite copy of its measure may sums sutible positive finite.

However there still a problem: $\mathbb Z$ is also a countable infinite set of points but has $0$ measure whereas Vitali set has non-zero. So I'm not sure if its feasible to use hyperreal measure to make Vitali set measurable.

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2 Answers 2

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Before you can ponder about hyperreal valued measures there are some subtleties to consider. If $^*\mathbb R$ is any enlargement of $\mathbb R$ then, with its natural order, it is no longer complete (that is not every bounded below set has an infimum). That means that when using hyperreals as the codomain of a measure function it is no longer clear what it would mean for the measure to be countably additive.

With that in mind let us recall the context for the Vitali sets. Looking to extend the notion of length of intervals we seek a function $\mu:\mathcal P(\mathbb R)\to V$, where $V=[0,\infty]$, which assigns to every interval its length, is $\sigma$-additive, and translation invariant. One then shows the impossibility of such a measure by using the Axiom of Choice to construct a Vitali set.

However, these conditions are tightly packed. If the Axiom of Choice does not hold then a measure as above [does] can exist (Later edit: 'does' changed to 'can' to reflect comment). If either one of translation invariance or $\sigma$-additivity is dropped then again such a measure exists. And of course if $\mathcal P(\mathbb R)$ is replaced by the smaller $\sigma$-algebra of Borel or Lebesgue sets then a measure with the other properties exists.

Now, for hyperreal valued measures (choosing $V=^{*}[0,\infty ]$) one can't state $\sigma$-additivity in any straightforward way since it requires completeness (which is not present). So, if one only requires a hyperreal valued measure to be $\sigma$-additive then of course Vitali sets become measurable as in the non-hyper case. But this is probably not quite what interests you.

A more detailed answer to your question is given in the article "Completely additive measure and integration" by Shorb, Alan McK.

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    $\begingroup$ Thanks a lot to the article, I'll read it. $\endgroup$
    – Popopo
    Dec 13, 2012 at 7:30
  • $\begingroup$ $\neg$AC does not imply that every set is measurable. $\endgroup$
    – Michael Greinecker
    Dec 13, 2012 at 17:12
  • $\begingroup$ @Michael - I thought that (slightly weak) AC is required for the existence of non-measurable sets. Please correct me if I'm wrong so I can correct my answer. $\endgroup$
    – Ittay Weiss
    Dec 13, 2012 at 20:55
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    $\begingroup$ If every set of reals is Lebesgue measurable, then AC does not hold. But in general, that AC fails to hold does not imply that every set is measurable. Even more, one cannot even prove that it is relatively consistent to ZFC that every set of reals is measurable. It is consistent reltive to ZFC+$\exists$inaccessible cardinal. See the wikipdia page for the Solovay model. $\endgroup$
    – Michael Greinecker
    Dec 13, 2012 at 23:53
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    $\begingroup$ @Michael, there are some points to add. It is impossible to prove all sets are measurable in ZF+DC without having an inaccessible in an inner model. However if one is willing to forfeit DC then it is possible to have all sets Borel, and thus measurable. One can even have that countable unions of countable sets are countable (at least for sets of real numbers), and so to some degree develop measure theoretic tools. One has to note more critically, however, that without non-measurable sets there are no free ultrafilters on $\omega$, and model theory might break a little bit. $\endgroup$
    – Asaf Karagila ♦
    Dec 14, 2012 at 0:14
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The Vitali set cannot be made measurable by extending the measure to *R (the Hyper-reals). One reason is that countable additivity is not (in general) defined on *R due its being Dedekind Incomplete (as noted in the first answer above). But that is not the only reason. Even if you could find a way around that (such as finding another extension of R which would allow countable additivity), the Vitali set could not be made measurable. The reason has to do with translation invariance.

As you know, the standard construction of Vitali sets yield a countable collection of sets which are congruent (modulo translation), but whose union is [0,1). So the goal here would be to assign each Vitali set a measure of some value e, such that e + e + ... = 1. However, these same Vitali sets may be rearranged (via translation only) so that its union is now [0,2). Or [0,3). Or [0,n). So now the same equation e + e + e + ... = 2, or = n, or whatever value you'd like. Translation invariance is what really kills this, not just countable additivity.

If you drop translation invariance in the measure, then you could indeed consistently assign a measure to each of the Vitali sets, so that their sum is correct. You wouldn't even need to go to *R, a translation variant measure with the standard reals could suffice to make all subsets of R measurable. Unfortunately, such measures are unsatisfying, because you are not guaranteed that the interval [x,y] always has measure y-x.

Likewise, if you drop countable additivity requirement (requiring only finite additivity), you can consistently define the Vitali set as having measure 0, and in fact assign measures to any subset of R.

The closest I have seen to reaching this goal is the Loeb Measure, a Nonstandard Measure with *R values, which has hyper-finite additivity. Still, its weaknesses and restrictions make it very unappealing (at least to me).

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  • $\begingroup$ Why the measure of the union of same Vitali sets can be arbitrary any positive real number? $\endgroup$
    – Popopo
    Feb 28, 2013 at 20:10
  • $\begingroup$ Okay, I will demonstrate it by shifting the Vitali sets around so that it goes from covering [0,1) to covering [0,2). As you will see below, I will do so by moving every other one to each interval. $\endgroup$
    – Jonathan Hoyle
    Feb 28, 2013 at 20:31
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    $\begingroup$ As an aside, this is reminiscent of the Banach-Tarski spherical decomposition, in which a sphere in R^3 is decomposed into non-measurable subsets, and then through translations and rotations only, recombined to produce two spheres, each with the volume of the original. The difference being that in R^3, you can do with with only a finite number of non-measurable pieces. In R^1 and R^2, you require a countably infinite number of non-measurable pieces. $\endgroup$
    – Jonathan Hoyle
    Feb 28, 2013 at 20:44
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    $\begingroup$ I understand. It is a nice proof. Hence the conclusion seems means $V$ is non-Lebesgue measurable even in other extensions of real field, such as surreals. $\endgroup$
    – Popopo
    Feb 28, 2013 at 21:15
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    $\begingroup$ Correct. The Banach-Tarski decomposition shows the problem of non-measurability exists solely with translations/rotations, having nothing to do with infinitesimals or extensions of the reals. Suppose we start with a sphere of volume 1. The B-T procedure splits this sphere into 5 pieces: A0, A1, B0, B1 and a single point, with A0 and A1 being congruent and B0 and B1 being congruent. The single point has measure 0, and the other four pieces are non measurable. However, A0 and B0 can be recombined by translation and rotation to yield the original sphere, as does A1 and B1. $\endgroup$
    – Jonathan Hoyle
    Mar 2, 2013 at 14:59

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