# The Unapologetic Mathematician

## Partitions in Measure Algebras

Let $(\mathcal{S},\mu)$ be a totally finite measure algebra, and write $X$ for the maximal element. Without loss of generality, we can assume that $\mu$ is normalized so that $\mu(X)=1$.

We define a “partition” $\mathcal{P}$ of an element $E\subseteq\mathcal{S}$ to be a finite set of “disjoint” elements of $\mathcal{S}$ whose “union” is $E$. Remember, of course, that the elements of $\mathcal{S}$ are not (necessarily) sets, so the set language is suggestive, but not necessarily literal. That is, if $\mathcal{P}=\{E_1,\dots,E_k\}$ then $E_i\cap E_j=\emptyset$ and

$\displaystyle E=\bigcup\limits_{i=1}^kE_i$

The “norm” $\lvert\mathcal{P}\rvert$ of a partition $\mathcal{P}$ is the maximum of the numbers $\{\mu(E_i)\}$. If $\mathcal{P}=\{E_1,\dots,E_k\}$ is a partition of $E$ and if $F\subseteq E$ is any element of $\mathcal{S}$ below $E$, then $\mathcal{P}\cap F=\{E_1\cap F,\dots,E_k\cap F\}$ is a partition of $F$.

If $\mathcal{P}_1$ and $\mathcal{P}_2$ are partitions, then we write $\mathcal{P}_1\leq\mathcal{P}_2$ if each element in $\mathcal{P}_1$ is contained in an element of $\mathcal{P}_2$. We say that a sequence of partitions is “decreasing” if $\mathcal{P}_{n+1}\leq\mathcal{P}_n$ for each $n$. A sequence of partitions is “dense” if for every $E\in\mathcal{S}$ and every positive number $\epsilon$ there is some $n$ and an element $E_0\in\mathcal{S}$ so that $\rho(E,E_0)<\epsilon$, and $E_0$ is exactly the union of some elements in $\mathcal{P}_n$. That is, we can use the elements in a fine enough partition in the sequence to approximate any element of $\mathcal{S}$ as closely as we want.

Now, if $(\mathcal{S},\mu)$ is a totally finite, non-atomic measure algebra, and if $\{\mathcal{P}_n\}$ is a dense, decreasing sequence of partitions of $X$, then $\lim\limits_{n\to\infty}\lvert\mathcal{P}_n\rvert=0$. Indeed, the sequence of norms $\{\lvert\mathcal{P}_n\rvert\}$ is monotonic and bounded in the interval $[0,1]$, and so it must have a limit. We will assume that this limit is some positive number $\delta>0$, and find a contradiction.

So if $\mathcal{P}_1=\{E_1,\dots,E_k\}$ then at least one of the $E_i$ must be big enough that $\lvert\mathcal{P}_n\cap E_i\rvert\geq\delta$ for all $n$. Otherwise the sequence of norms would descend below $\delta$ and that couldn’t be the limit. Let $F_1$ be just such an element, and consider the sequence $\{\mathcal{P}\cap F_1\}$ of partitions of $F_1$. The same argument is just as true, and we find another element $F_2\subseteq F_1$ from the partition $\mathcal{P}_2$, and so on.

Now, let $F$ be the intersection of the sequence $\{F_n\}$. By assumption, each of the $F_n$ has $\mu(F_n)\geq\delta$, and so $\mu(F)\geq\delta$ as well. Since $(\mathcal{S},\mu)$ is non-atomic, $F$ can’t be an atom, and so there must be an $F_0\subseteq F$ with $0<\mu(F_0)<\mu(F)$. This element must be either contained in or disjoint from each element of each partition $\mathcal{P}_n$.

We can take $\epsilon$ smaller than either $\mu(F_0)$ or $\mu(F)-\mu(F_0)$. Now no set made up of the union of any elements of any partition $\mathcal{P}_n$ can have a distance less than $\epsilon$ from $F_0$. This shows that the sequence of partitions cannot be dense, which is the contradiction we were looking for. Thus the limit of the sequence of norms is zero.