There exists an exhaustive submeasure that is not equivalent to a measure

Analysis 1 Minute

You need to know: Boolean algebra of sets.

Background: Let {\cal B} be a Boolean algebra of sets. A map \nu:{\cal B} \to {\mathbb R} is called a submeasure if the following holds: (i) \nu(\emptyset) = 0, (ii) \nu(A) \leq \nu(B) for every A,B \in {\cal B} such that A \subset B, and (iii) \nu(A \cup B) \leq \nu(A) + \nu(B) for every A,B \in {\cal B}. A submeasure \nu is called a (finitely additive) measure if \nu(A \cup B) = \nu(A) + \nu(B) whenever sets A,B \in {\cal B} are disjoint. A submeasure \nu is called exhaustive if \lim\limits_{n\to \infty}\nu(E_n)=0 for every sequence \{E_n\}_{n=1}^\infty of elements of {\cal B} such that E_n \cap E_m = \emptyset whenever n\neq m. We say that submeasure \nu_1 is absolutely continuous with respect to a submeasure \nu_2 if for every \epsilon>0 there exists an \alpha>0 such that \nu_1(A) \leq \epsilon for every A \in {\cal B} with \nu_2(A) \leq \alpha.

The Theorem: On 27th January 2006, Michel Talagrand submitted to arxiv a paper in which he proved the existence of a Boolean algebra {\cal B} of sets, and a nonzero exhaustive submeasure \nu on it, which is not absolutely continuous with respect to any measure.

Short context: Any measure is exhaustive. Moreover, if a submeasure is absolutely continuous with respect to a measure, it is exhaustive. One of the many equivalent formulations of famous Maharam’s problem is whether the converse is true. The Theorem gives a negative answer to this question.

Links: Free arxiv version of the original paper is here, journal version is here. See also Section 8.13 of this book for an accessible description of the Theorem.

Go to the list of all theorems

Unknown's avatar

Published by Bogdan Grechuk

Published

Leave a comment