# 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.

August 24, 2010 Posted by | Analysis, Measure Theory | 1 Comment