TAILIEUCHUNG - Đề tài "Maharam's problem "

We construct an exhaustive submeasure that is not equivalent to a measure. 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 to References 1. Introduction Consider a Boolean algebra B of sets. | Annals of Mathematics Maharam s problem By Michel Talagrand Annals of Mathematics 168 2008 981-1009 Maharam s problem By Michel Talagrand Dedicated to J. W. Roberts Abstract We construct an exhaustive submeasure that is not equivalent to a measure. 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 to References 1. Introduction Consider a Boolean algebra B of sets. A map V B R is called a submeasure if it satisfies the following properties V 0 A c B A B 2B v A V B A B 2B V A u B v A V B . If we have v A u B V A v B whenever A and B are disjoint we say that V is a finitely additive measure. We say that a sequence En of B is disjoint if En Em whenever n m. A submeasure is exhaustive if limn 1 V En 0 whenever En is a disjoint sequence in B. A measure is obviously exhaustive. Given two 982 MICHEL TALAGRAND submeasures n1 and n2 we say that n1 is absolutely continuous with respect to n2 if 8 0 3a 0 n2 A a n1 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 exhaustive 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 continuous with respect to a measure if and of course only if it is uniformly exhaustive . 8 0 3n E1 En disjoint inf n Ei i n Thus Maharam s problem can be reformulated as to whether an exhaustive submeasure is necessarily .

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