From: magidin@bosco.berkeley.edu (Arturo Magidin) Subject: Re: Contructing a non-measurable set in R1 Date: 11 Nov 1999 16:21:10 GMT Newsgroups: sci.math In article <80enn3$tin$1@nnrp1.deja.com>, *Paragon* wrote: >Can someone provide a construction of a non-measurable set in R1 (the >real number line)? Here, measurable is defined as Lebesque measure. Depends on what you mean by "construction." The standard definition of the Vitali Set requires the Axiom of Choice and is non-constructive. I think that Sollovay proved that if you accept a particular negation of the Axiom of Choice (or a particular axiom which implies not(Choice), then every subset would be measurable; I know he did it at some dimension, but cannot guarantee it was in dimension one. Anyway, here's the standard "construction": Consider the unit interval; we partition it into equivalence classes by the rule x~y if and only if x-y is a rational number. It is easy to see that this is indeed an equivalence relation, so it partitions the unit interval into disjoint, nonempty sets. Let V be a set containing exactly one representative from each equivalence class (possible by the Axiom of Choice). We claim that V is non-measurable. First, note that if q and q' are two distinct rationals, and we let V+q be the set consisting of the numbers you get by adding q to each element of V, then V+q and V+q' are disjoint. Indeed, if an element x lies in both V+q and V+q', then it must equal y+q and z+q', with y,q, in V. But then y+q=z+q', so y-z=q'-q, a rational. In particular, since V containes exactly one representative from each equivalence class, and y~z, we have y=z. But then q=q', a contradiction. So all the V+q are pairwise disjoint. Consider the union of all the V+q, with q ranging over all rationals between -1 and 1. If V is measurable, then so is each V+q, and they all measure the same, because Lebesgue measure is invariant under translation. First, note that the union of all V+q is certainly contained between -1 and 2, because elements of V lie between 0 and 1, and q lies between -1 and 1. So their union is contained in the interval [If V is measurable, then so is V+q for all q, and they are all pairwise disjoint; by the sigma-additivity of Lebesgue measure, the measure of the union of the V+q's would be aleph_0 times the measure of V. But since the union is contained in [-1,2], it must be finite. Hence, if V is measurable, then its measure must be zero. Now I claim that the union of all the V+q, q ranging over all rationals between -1 and 1, contains the interval [0,1]. Indeed, let x be any real number in the interval. Then x is equivalent to some y with y in V; that means that x-y is a rational. Since both x and y are between 0 and 1, x-y is a rational, q, between -1 and 1. Therefore, x=y+q, so x lies in V+q. Therefore, since [0,1] has measure 1 and is contained in the union of all the V+q, if V is measurable then the measure of the union of all the V+q must be at least 1 because the Lebesgue measure is monotone. But we already know the measure of V (and hence of all V+q) is 0 IF V is measurable, so the measure of the union of all the V+q would be zero. This contradiction comes from assuming V is measurable. So V is not a measurable set. ====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes") ====================================================================== Arturo Magidin magidin@math.berkeley.edu ============================================================================== From: Lieven Marchand Subject: Re: Contructing a non-measurable set in R1 Date: 11 Nov 1999 18:42:55 +0100 Newsgroups: sci.math magidin@bosco.berkeley.edu (Arturo Magidin) writes: > Depends on what you mean by "construction." The standard definition of > the Vitali Set requires the Axiom of Choice and is non-constructive. I > think that Sollovay proved that if you accept a particular negation of > the Axiom of Choice (or a particular axiom which implies not(Choice), > then every subset would be measurable; I know he did it at some > dimension, but cannot guarantee it was in dimension one. It's a bit more complex than that. In the Solovay reals, the axiom of dependent choice still holds (DC) so you have many of the standard results of real analysis. Without any form of choice, the real numbers can become very weird. Jech has a chapter on that in his book about the Axiom of Choice. What Solovay proved (1964) was that the consistency of ZF + IC implied the consistency of ZF + DC + LM, with ZF = Zermelo-Fr"ankel, IC = the existence of an inaccessible cardinal and LM = all sets of reals are Lebesgue measurable. As it is fairly typical in set theory that the first proof of a result uses large cardinals since it gives you more room to work with so to speak, originally people assumed that the hypothesis of IC could be avoided. Shelah proved the reverse implication in 1980 which put an end to these assumptions. BTW: a proof in dimension n, automatically proves all lower dimensions, since you can always consider mu'(A):=mu(A x [0,1]^k), k being the difference in dimension. -- Lieven Marchand If there are aliens, they play Go. -- Lasker ============================================================================== From: apollo@persian.berkeley.edu (Apollo Hogan) Subject: Re: Contructing a non-measurable set in R1 Date: 11 Nov 1999 17:45:30 GMT Newsgroups: sci.math In article <80eqdm$2f7$1@agate.berkeley.edu>, Arturo Magidin wrote: >In article <80enn3$tin$1@nnrp1.deja.com>, >*Paragon* wrote: >>Can someone provide a construction of a non-measurable set in R1 (the >>real number line)? Here, measurable is defined as Lebesque measure. > >Depends on what you mean by "construction." The standard definition of >the Vitali Set requires the Axiom of Choice and is non-constructive. I >think that Sollovay proved that if you accept a particular negation of >the Axiom of Choice (or a particular axiom which implies not(Choice), >then every subset would be measurable; I know he did it at some >dimension, but cannot guarantee it was in dimension one. Solovay proved that if it is consistent that there is an inaccessible cardinal, then it is consistent that _every_ set of reals (in R^1) is measurable. This implies that there are no non-measurable sets in R^n for any n. In fact, you can get the Axiom of Determinacy true, which implies that every set of reals is measurable, has the Baire property, has the perfect set property, etc. (The idea is to take a model of ZFC+(exists inaccessible K), add K-many Cohen reals, then look at HOD({R}\union R), where R=reals in V[G]. This model has the desired properties, it even satisfies dependent choice.) --Apollo Hogan student U.C.Berkeley