Completeness of Boolean Rings
We have a notion of “completeness” for boolean rings, which is related to the one for uniform spaces (and metric spaces), but which isn’t quite the same thing. We say that a Boolean ring is complete if every set of elements of has a union.
A complete Boolean ring is clearly a Boolean -algebra. It’s an algebra, because we can get a top element by taking the union of all the elements of together, and it’s a -ring because countable unions are included under all unions. The converse is a little hairier.
First off, every totally finite measure algebra is complete. That is, if — and thus for all , then every collection of elements of will have a union.
We let be the collection of all finite unions of elements in , all of which exist since is a Boolean ring. We let be the (finite) supremum of the measures of all elements in . Since this is a finite supremum, there must be some increasing sequence so that increases to . I say that the union of the sequence — which exists because is a -ring — is the union of all of .
Indeed, we must have by the continuity of the measure . Take any set and consider the difference , which is the amount by which extends past . Our assertion is that this is always . Define the sets , which is a disjoint union since is disjoint from . Each of the is in , and the limit of the sequence is . This set has measure
but since each of the is bounded above by then so is their limit. Thus , and thus , since we consider all negligible sets to be the same.
The same is true for totally -finite measure algebras. Indeed, we can break such a measure algebra up into a countable collection of finite measure algebras by breaking into a countable number of elements so that . We define the finite measure algebra to consist of the intersections of elements of with . Then given any collection we consider its image under such intersections to get . What we said above shows that each of these collections has an intersection , and their union in is the union of the original collection .
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