The Unapologetic Mathematician

Mathematics for the interested outsider

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 \mathcal{R} is complete if every set of elements of \mathcal{R} has a union.

A complete Boolean ring is clearly a Boolean \sigma-algebra. It’s an algebra, because we can get a top element X by taking the union of all the elements of \mathcal{R} together, and it’s a \sigma-ring because countable unions are included under all unions. The converse is a little hairier.

First off, every totally finite measure algebra (\mathcal{S},\mu) is complete. That is, if \mu(X)<\infty — and thus \mu(E)<\infty for all E\in\mathcal{S}, then every collection \mathcal{E} of elements of \mathcal{S} will have a union.

We let \hat{\mathcal{E}} be the collection of all finite unions of elements in \mathcal{E}, all of which exist since \mathcal{S} is a Boolean ring. We let \alpha be the (finite) supremum of the measures of all elements in \hat{\mathcal{E}}. Since this is a finite supremum, there must be some increasing sequence \{E_n\} so that \mu(E_n) increases to \alpha. I say that the union E of the sequence \{E_n\} — which exists because \mathcal{S} is a \sigma-ring — is the union of all of \mathcal{E}.

Indeed, we must have \mu(E)=\alpha by the continuity of the measure \mu. Take any set E_0\in\mathcal{E} and consider the difference D=E_0\setminus E, which is the amount by which E_0 extends past E. Our assertion is that this is always \emptyset. Define the sets D_n=E_n\uplus D, which is a disjoint union since D is disjoint from E. Each of the D_n is in \hat{\mathcal{E}}, and the limit of the sequence is E\uplus D. This set has measure

\displaystyle\mu(E\uplus D)=\mu(E)+\mu(D)=\alpha+\mu(D)

but since each of the \mu(D_n) is bounded above by \alpha then so is their limit. Thus \mu(D)=0, and thus D=\emptyset, since we consider all negligible sets to be the same.

The same is true for totally \sigma-finite measure algebras. Indeed, we can break such a measure algebra up into a countable collection of finite measure algebras by breaking X into a countable number of elements X_n so that \mu(X_n)<\infty. We define the finite measure algebra \mathcal{S}_n to consist of the intersections of elements of \mathcal{S} with X_n. Then given any collection \mathcal{E}\subseteq\mathcal{S} we consider its image under such intersections to get \mathcal{E}_n\subseteq\mathcal{S}_n. What we said above shows that each of these collections has an intersection E_n\in\mathcal{S}_n, and their union in \mathcal{S} is the union of the original collection \mathcal{E}.


August 10, 2010 - Posted by | Algebra, Ring theory

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