By: Walt

Walt — Fri, 13 Apr 2012 06:49:35 +0000

Thanks for the correction.

By: Michael

Michael — Wed, 11 Apr 2012 09:01:24 +0000

That interpretation of Maharam’s theorem is not quite correct. The theorem says that all atomless measure algebras are of this form, where a measure algebra is obtained by taking the quotient with respect to null sets and the notion of isomorphism is a boolean sigma-isomorphism. To see the difference, we can construct a probability space with a measure algebra equilvalent to Lebesgue measure on the unit interval but larger cardinality. Let K be a cardinal larger than [0,1]. Let X=K x [0,1]. We can interpret X as a space partitioned into slices of the form {(k,r):r fixed}. The set of slices corresponds to [0,1]. If B is a Borel set in [0,1], we let B_X={K x B}. The sets of the form $B_X$ form a sigma-algebra identical to the usual Borel sigma-algebra but with points replaced by copies of K. We define a measure mu_X on them by letting mu_X(B_X)=mu(B). Then the constructed measure space is equivalent to the unit interval with Lebesgue measure as a measure algebra but has a larger cardinality. Taking the completion doesn’t change the measure algebra, but the space will contain many more null sets than there are Lebesgue measurable sets in [0,1].

Comments on: Maharam’s Theorem

By: Walt

By: Michael