The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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