# The Unapologetic Mathematician

## The Metric Space of a Measure Ring

Let $(\mathcal{S},\mu)$ be a measure ring. We’ve seen how we can get a metric space from a measure, and the same is true here. In fact, since we’ve required that $\mu$ be positive — that $\mu(E)=0$ only if $E=\emptyset$ — we don’t need to worry about negligible elements.

And so we write $\mathfrak{S}=\mathfrak{S}(\mu)$ for the metric space consisting of the elements $E\in\mathcal{S}$ with $\mu(E)<\infty$. This has a distance function $\rho$ defined by $\rho(E,F)=\mu(E\Delta F)$. We also write $\mathfrak{S}=\mathfrak{S}(\mu)$ for the metric associated with the measure algebra associated with the measure space $(X,\mathcal{S},\mu)$. We say that a measure space or a measure algebra is “separable” if the associated metric space is separable.

Now, if we set \displaystyle\begin{aligned}f(E,F)&=E\cup F\\g(E,F)&=E\cap F\end{aligned}

then $f:\mathfrak{S}\times\mathfrak{S}\to\mathfrak{S}$, $g:\mathfrak{S}\times\mathfrak{S}\to\mathfrak{S}$, and $\mu:\mathfrak{S}\to\mathbb{R}$ itself are all uniformly continuous.

Indeed, if we take two pairs of sets $E_1$, $E_2$, $F_1$, and $F_2$, we calculate \displaystyle\begin{aligned}\mu\left((E_1\cup F_1)\setminus(E_2\cup F_2)\right)&=\mu\left((E_1\cup F_1)\cap(E_2^c\cap F_2^c)\right)\\&=\mu\left((E_1\cup F_1)\cap(E_2^c\cap F_2^c)\right)\\&=\mu\left((E_1\cap E_2^c\cap F_2^c)\cup(F_1\cap E_2^c\cap F_2^c)\right)\\&\leq\mu(E_1\cap E_2^c\cap F_2^c)+\mu(F_1\cap E_2^c\cap F_2^c)\\&\leq\mu(E_1\setminus E_2)+\mu(F_1\setminus F_2)\end{aligned}

Similarly, we find that $\mu\left((E_2\cup F_2)\setminus(E_1\cup F_1)\right)\leq\mu(E_2\setminus E_1)+\mu(F_2\setminus F_1)$. And thus \displaystyle\begin{aligned}\rho\left((E_1\cup F_1),(E_2\cup F_2)\right)&=\mu\left((E_1\cup F_1)\Delta(E_2\cup F_2)\right)\\&=\mu\left((E_1\cup F_1)\setminus(E_2\cup F_2)\right)+\mu\left((E_2\cup F_2)\setminus(E_1\cup F_1)\right)\\&\leq\mu(E_1\setminus E_2)+\mu(F_1\setminus F_2)+\mu(E_2\setminus E_1)+\mu(F_2\setminus F_1)\\&=\mu(E_1\Delta E_2)+\mu(F_1\Delta F_2)\\&=\rho(E_1,E_2)+\rho(F_1,F_2)\end{aligned}

And so if we have control over the distance between $E_1$ and $E_2$, and the distance between $F_1$ and $F_2$, then we have control over the distance between $E_1\cup F_1$ and $E_2\cup F_2$. The bounds we need on the inputs uniform, and so $f$ is uniformly continuous. The proof for $g$ proceeds similarly.

To see that $\mu$ is uniformly continuous, we calculate \displaystyle\begin{aligned}\left\lvert\mu(E)-\mu(F)\right\rvert&=\left\lvert\left(\mu(E)-\mu(E\cap F)\right)-\left(\mu(F)-\mu(F\cap E)\right)\right\rvert\\&=\left\lvert\mu\left(E\setminus(E\cap F)\right)-\mu\left(F\setminus(F\cap E)\right)\right\rvert\\&=\left\lvert\mu(E\setminus F)-\mu(F\setminus E)\right\rvert\\&\leq\mu(E\setminus F)+\mu(F\setminus E)\\&=\mu(E\Delta F)\\&=\rho(E,F)\end{aligned}

Now if $(X,\mathcal{S},\mu)$ is a $\sigma$-finite measure space so that the $\sigma$-ring $\mathcal{S}$ has a countable set of generators, then $\mathfrak{S}(\mu)$ is separable. Indeed, if $\{E_n\}$ is a countable sequence of sets that generate $\mathcal{S}$, then we may assume (by $\sigma$-finiteness) that $\mu(E_n)<\infty$ for all $n$. The ring generated by the $\{E_n\}$ is itself countable, and so we may assume that $\{E_n\}$ is itself a ring. But then we know that for every $E\in\mathfrak{S}(\mu)$ and for every positive $\epsilon$ we can find some ring element $E_n$ so that $\rho(E,E_n)=\mu(E\Delta E_n)<\epsilon$. Thus $\{E_n\}$ is a countable dense set in $\mathfrak{S}(\mu)$, which is thus separable.

August 6, 2010 Posted by | Analysis, Measure Theory | 5 Comments