# The Unapologetic Mathematician

## 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 $\mathcal{E}$.