Maharam's problem By Michel Talagrand Dedicated to J. W. Roberts Abstract We construct an exhaustive submeasure that is not equivalent to a mea-sure. This solves problems of J. von Neumann (1937) and D. Maharam (1947). Contents 1. Introduction 2. Roberts 3. Farah 4. The construction 5. The main estimate 6. Exhaustivity 7. Proof of Theorems 1.2 to 1.4 References 1. Introduction Consider a Boolean algebra B of sets. A map : B ! R + is called a submeasure if it satis es the following properties: (?) = 0; (1.1) A B; A;B 2 B = ) (A) (B); (1.2) A;B 2 B ) (A [ B) (A) + (B): (1.3) If we have (A [ B) = (A) + (B) whenever A and B are disjoint, we say that is a ( nitely additive) measure. We say that a sequence (En ) of B is disjoint if En Em = ? whenever n 6 = m. A submeasure is exhaustive if lim n!1 (En ) = 0 whenever (En ) is a disjoint sequence in B. A measure is obviously exhaustive. Given two 982 MICHEL TALAGRAND submeasures 1 and 2 , we say that 1 is absolutely continuous with respect to 2 if (1.4) 8" > 0; 9 > 0; 2 (A) = ) 1 (A) ": If a submeasure is absolutely continuous with respect to a measure, it is exhaustive. One of the many equivalent forms of Maharam's problem is whether the converse is true. Maharam's problem: If a submeasure is exhaustive, is it absolutely continuous with respect to a measure? In words, we are asking whether the only way a submeasure can be ex-haustive is because it really resembles a measure. This question has been one of the longest standing classical questions of measure theory. It occurs in a variety of forms (some of which will be discussed below). Several important contributions were made to Maharam's problem. N. Kalton and J. W. Roberts proved [11] that a submeasure is absolutely con-tinuous with respect to a measure if (and, of course, only if) it is uniformly exhaustive, i.e. (1.5) 8" > 0; 9n; E1; : : : ;E n disjoint = ) inf in (Ei ) ": Thus Maharam's problem can be reformulated as to whether an exhaustive submeasure is necessarily uniformly exhaustive. Two other fundamental con-tributions by J.W. Roberts [15] and I. Farah [6] are used in an essential way in this paper and will be discussed in great detail later. We prove that Maharam's problem has a negative answer. Theorem 1.1. There exists a nonzero exhaustive submeasure on the algebra B of clopen subsets of the Cantor set that is not uniformly exhaustive (and thus is not absolutely continuous with respect to a measure ). Moreover, no nonzero measure on B is absolutely continuous with respect to . We now spell out some consequences of Theorem 1.1. It has been known for a while how to deduce these results from Theorem 1.1. For the convenience of the reader these (easy) arguments will be given in a self-contained way in the last section of the paper. Since Maharam's original question and the von Neumann problem are formulated in terms of general Boolean algebras (i.e., that are not a priori represented as algebras of sets) we must brie y mention these. We will denote by 0 and 1 respectively the smallest and the largest element of a Boolean algebra B, but we will denote the Boolean operations by ; [, etc. as in the case of algebras of sets. A Boolean algebra B is called -complete if any countable subset C of B has a least upper bound [C (and thus a greatest lower bound C). A submeasure on B is called continuous if whenever (An )
