We construct an exhaustive submeasure that is not equivalent to a measure.
This solves problems of J. von Neumann (1937) and D. Maharam (1947)

Dedicated To J. W. Roberts

Michel Talagrand

English

arXiv

Talagrand, Michel

Numérisation de Documents Anciens Mathématiques

C. R. Acad. Sci. Paris, Ser. I 342 (2006) 501–503http://france.elsevier.com/direct/CRASS1/Functional AnalysisMaharam’s problemMichel TalagrandÉquipe d’analyse de l’institut mathématique, université Paris VI, boîte 186, 4, place Jussieu, 75230 Paris cedex 05, FranceReceived 27 January 2006; accepted 30 January 2006Available online 3 March 2006Presented by Gilles PisierAbstractWe solve Maharam’s problem [D. Maharam, An algebraic characterization of measure algebras, Ann. Math. 48 (1947) 154–167.[3]], also known as the Control Measure Problem. We construct a non-zero exhaustive submeasure on the algebra of clopen setsof the Cantor set that is not absolutely continuous with respect to a measure. To cite this article: M. Talagrand, C. R. Acad. Sci.Paris, Ser. I 342 (2006). 2006 Académie des sciences. Published by Elsevier SAS. All rights reserved.RésuméLe problème de Maharam. Nous construisons une sous-mesure exhaustive non nulle sur l’algèbre des ouverts-fermés de l’en-semble de Cantor qui n’est absolument continue par rapport à aucune mesure. Ceci résout un problème posé par D. Maharam[D. Maharam, An algebraic characterization of measure algebras, Ann. Math. 48 (1947) 154–167. [3]], en 1947, aussi connu sousle nom de problème des mesures de contrôle. Pour citer cet article : M. Talagrand, C. R. Acad. Sci. Paris, Ser. I 342 (2006). 2006 Académie des sciences. Published by Elsevier SAS. All rights reserved.1. IntroductionGiven a Boolean algebra B of sets, a map ν :B → R+ is called a submeasure if it satisfies the following properties:(1.1) ν(∅) = 0,(1.2) A ⊂ B , A,B ∈ B ⇒ ν(A)  ν(B),(1.3) A,B ∈ B ⇒ ν(A ∪ B)  ν(A) + ν(B).If moreover ν(A∪B) = ν(A)+ν(B) whenever A and B are disjoint ν is called a measure. We say that ν is exhaustiveif limn→∞ ν(En) = 0 for each disjoint sequence (En) of B (that is, En∩Em = ∅ if n = m). It is obvious that a measureis exhaustive. Given two submeasures ν1 and ν2, we say that ν1 is absolutely continuous with respect to ν2 if∀ε > 0, ∃α > 0, ν2(B) < α ⇒ ν1(B)  ε.In other words, for a certain function f : R+ → R+ continuous at zero, we have ν1(B)  f (ν2(B)). If a submeasureis absolutely continuous with respect to a measure, it is exhaustive. Whether the converse is true has been a longE-mail address: spinglass@talagrand.net (M. Talagrand).1631-073X/$ – see front matter  2006 Académie des sciences. Published by Elsevier SAS. All rights reserved.doi:10.1016/j.crma.2006.01.026502 M. Talagrand / C. R. Acad. Sci. Paris, Ser. I 342 (2006) 501–503standing classical problem of measure theory. It amount to ask whether the only reason for which a submeasure canbe exhaustive is because it really looks like a measure. We show that the answer to this question is negative.Theorem 1.1. There exists a non-zero exhaustive submeasure on the algebra B of clopen sets of the Cantor set that isnot absolutely continuous with respect to a measure. Moreover, there is no non-zero measure on B that is absolutelycontinuous with respect to ν.The importance of the Control Measure Problem largely stems from the many equivalent forms that occur naturallyin a variety of questions, see [2].2. The constructionThis construction elaborates on fundamental contributions by J.W. Roberts [4] and I. Farah [1].Consider the set T = ∏n1{1, . . . ,2n}. The algebra B of the clopen sets of T is isomorphic to the algebra of theclopen sets of the Cantor set {0,1}n. We denote by Bn the algebra generated by the coordinates of rank  n. For n  1,q  2n, we consider the set Sn,q = {y ∈ T ; yn = q} and its complement Scn,q = {y ∈ T ; yn = q}. We setα(k) = 1(k + 5)3 ; N(k) = (k + 6)1+(k+5)3 .There is nothing magic about these choices, what is really required is that the sequence α(k) decreases fast enoughand the sequence N(k) increases fast enough. For k  1, we consider the class Dk ⊂ B× R+ consisting of all couples(X,w) such thatX =⋂n∈ISn,q(n), w  2−k(N(k)card I)α(k)where I ⊂ N, card I  N(k) and q(n) is an arbitrary element of {1, . . . ,2n}. We set D = ⋃k1 Dk .For a finite subset F = {(A1,w1), . . . , (An,wn)} of B × R+, let us set∪F =⋃nA; w(F) =∑nw.Given a subset C of B × R+, we define a submeasure ϕC byϕC(A) = inf{w(F); A ⊂ ∪F ; F ⊂ C}.Let ψ denote the submeasure ϕD . Following Farah [1], given a submeasure θ on T , and X ∈ B, we say that X is(m,n, θ)-thin if the following condition holds. For each atom A of Bm, there exists a set C ∈ Bn, C ⊂ A, C ∩ X = ∅,with θ(π−1A (C))  1, where πA is the map T → A that send y = (yi)i1 to z ∈ A given by zi = qi for i  m, zi = yifor i > m, where q1, . . . , qm are such that A = {z ∈ T ; ∀i  m, zi = qi}. Given X ∈ B and a finite set I we say thatX is (I, θ)-thin if it is (m,n, θ)-thin whenever m < n, m,n ∈ I .Given an integer p, we proceed to the construction, by decreasing induction over k of classes Ck,p and submea-sures ϕk,p . For k = p, we set Cp,p = D, ϕp,p = ϕD = ψ . Having constructed Ck+1,p and ϕk+1,p , we setEk,p ={(X,w) ∈ B × R+; ∃I, X is (I, ϕk+1,p)-thin, card I  N(k), w  2−k(N(k)card I)α(k)},and we set Ck,p = Ek,p ∪ Ck+1,p , ϕk,p = ϕCk,p .Consider an ultrafilter U on N. We define the classes Ek by(X,w) ∈ Ek ⇔{p; (X,w) ∈ Ek,p} ∈ U .We then define the classes Ck by Ck = D∪⋃k E, and the submeasures νk = ϕCk . One can then prove the following:Theorem 2.1. For each k the submeasure νk satisfies the requirements of Theorem 1.1.M. Talagrand / C. R. Acad. Sci. Paris, Ser. I 342 (2006) 501–503 503Roughly speaking, one can say that, even though all the submeasures νk are exhaustive this becomes harder andharder to check as k increases. Given a disjoint sequence (Ei) of B, it requires the same effort (as measured by howfar one has to go into the sequence (Ei)) to prove that lim supi→∞ νk(Ei)  a whenever a > 2−k than to prove thatlim supi→∞ νk(Ei)  2−k .The main difficulty in the proof of Theorem 2.1 is showing that the submeasures νk do not vanish. The mainingredient for this proof is an estimate of the type ϕk,p(T )  Cp > 0, which is obtained by decreasing inductionover k. The proof that νk is exhaustive follows the lines of [1]. As for the fact that νk is not absolutely continuous withrespect to any measure, this property is shared by any non-zero submeasure ν  ψ . Indeed, such a submeasure hasthe property that for some a > 0 and n large enough, ν(Scn,q)  a for all q  2n. To see that we simply observe thatfor any k, any set I with card I = N(k), if for n ∈ I we choose q(n)  2n, then0 < ν(T ) <∑n∈Iν(Scn,q(n)) + ν( ⋂n∈ISn,q(n))and the last term is  2−k , so if 2−k < ν(T )/2, we have∑n∈I ν(Scn,q(n))  ν(T )/2 which suffices. Also, wheneverν  ψ , there is no non-zero measure that is absolutely continuous with respect to ν, because one can show that sucha property is already true for ψ .References[1] I. Farah, Example of ε-exhaustive pathological measures, Fund. Math. 181 (2004) 257–272.[2] D.H. Fremlin, Measure Theory, Vol. 3: Measure Algebras, second ed., Torres Fremlin, 2004.[3] D. Maharam, An algebraic characterization of measure algebras, Ann. Math. 48 (1947) 154–167.[4] J.W. Roberts, Maharam’s problem, in: Kranz, Labuda (Eds.), Procedings of the Orlicz Memorial Conference, 1991, unpublished.

Maharam's problem

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