# The Unapologetic Mathematician

## Properties of Metric Spaces of Measure Rings

Today we collect a few facts about the metric space $\mathfrak{S}$ associated to a measure ring $(\mathcal{S},\mu)$.

First of all, the metric $\rho$ on $\mathfrak{S}$ is translation-invariant. That is, if $E$, $F$, and $G$ are sets in $\mathcal{S}$ with finite measure, then $\rho(E\Delta G,F\Delta G)=\rho(E,F)$. Indeed, we calculate

\displaystyle\begin{aligned}\rho(E\Delta G,F\Delta G)&=\mu\left((E\Delta G)\Delta(F\Delta G)\right)\\&=\mu\left((E\Delta F)\Delta(G\Delta G)\right)\\&=\mu\left((E\Delta F)\Delta\emptyset\right)\\&=\mu(E\Delta F)\\&=\rho(E,F)\end{aligned}

Next we need to define an “atom” of a measure ring to be a minimal nonzero element. That is, it’s an element $\emptyset\neq E\in\mathcal{S}$ so that if $F\subseteq E$ then either $F=\emptyset$ or $F=E$. We also define a measure ring to be “non-atomic” if it has no atoms. As a quick result: a totally $\sigma$-finite measure ring can have at most countably many atoms, since $X$ must contain all of them, and no two of them can have a nontrivial intersection — if there were uncountably many of them, any decomposition of $X$ could only cover countably many of them.

On the other hand, we define a metric space to be “convex” if for any two distinct elements $E$ and $F$ there is an element $G$ between them. That is, $G$ is neither $E$ nor $F$, and it satisfies the equation $\rho(E,F)=\rho(E,G)+\rho(G,F)$. We assert that the metric space of a $\sigma$-finite measure ring is convex if and only if the measure ring is non-atomic.

Let $E$ and $F$ be elements of $\mathfrak{S}$. Without loss of generality we can assume that $F=\emptyset$ by using the translation-invariance of $\rho$. Indeed, we can replace $E$ with $E\Delta F$ and $F$ with $F\Delta F=\emptyset$. There will be an element $G$ between $E$ and $F$ if and only if $G\Delta F$ is between $E\Delta F$ and $\emptyset$.

So for any $E$ is there an element $G$ between $E$ and $\emptyset$? Such a $G$ will satisfy

\displaystyle\begin{aligned}\mu(E)&=\rho(E,\emptyset)\\&=\rho(E,G)+\rho(G,\emptyset)\\&=\mu(E\Delta G)+\mu(G)\\&=\mu(E\setminus G)+\mu(G\setminus E)+\mu(G\setminus E)+\mu(G\cap E)\\&=\mu(E)+2\mu(G\setminus E)\end{aligned}

This is only possible if $\mu(G\setminus E)=0$, which means that $G\setminus E=\emptyset$, and so $G\subseteq E$. But for $G$ to be between $E$ and $\emptyset$ it cannot be equal to either of them, which means that $E$ cannot be an atom for any $E$ with $\mu(E)<\infty$. Since we can decompose any $E\in\mathcal{S}$ into a countable union of elements of finite measure, no element of infinite measure can be an atom either.

August 12, 2010 Posted by | Analysis, Measure Theory | 5 Comments