# The Unapologetic Mathematician

## Completeness of the Metric Space of a Measure Space

Our first result today is that the metric space associated to the measure ring of a measure space $(X,\mathcal{S},\mu)$ is complete.

To see this, let $\{E_n\}$ be a Cauchy sequence in the metric space $\mathfrak{S}$. That is, for every $\epsilon>0$ there is some $N$ so that $\rho(E_m,E_n)<\epsilon$ for all $m,n>N$. Unpacking our definitions, each $E_n$ must be an element of the measure ring $(\mathcal{S},\mu)$ with $\mu(E_n)<0$, and thus must be (represented by) a measurable subset $E_n\subseteq X$ of finite measure. On the side of the distance function, we must have $\mu(E_m\Delta E_n)<\epsilon$ for sufficiently large $m$ and $n$.

Let’s recast this in terms of the characteristic functions $\chi_{E_n}$ of the sets in our sequence. Indeed, we find that $\chi_{E_m\Delta E_n}=\lvert\chi_{E_m}-\chi_{E_n}\rvert$, and so $\displaystyle\mu(E_m\Delta E_n)=\int\chi_{E_m\Delta E_n}\,d\mu=\int\lvert\chi_{E_m}-\chi_{E_n}\rvert\,d\mu$

that is, a sequence $\{E_n\}$ of sets is Cauchy in $\mathfrak{S}(\mu)$ if and only if its sequence of characteristic functions $\left\{\chi_{E_n}\right\}$ is mean Cauchy. Since mean convergence is complete, the sequence of characteristic functions must converge in mean to some function $f$. But mean convergence implies convergence in measure, which is equivalent to a.e. convergence on sets of finite measure, which is what we’re dealing with.

Thus the limiting function $f$ must — like the characteristic functions in the sequence — take the value $0$ or $1$ almost everywhere. Thus it is (equivalent to) the characteristic function of some set. Since $f$ must be measurable — as the limit of a sequence of measurable functions — it’s the characteristic function of a measurable set, which must have finite measure since its measure is the limit of the Cauchy sequence $\{\mu(E_n)\}$. That is, $f=\chi_E$, where $E\in\mathfrak{S}(\mu)$, and $E$ is the limit of $\{E_n\}$ under the metric of $\mathfrak{S}(\mu)$. Thus $\mathfrak{S}(\mu)$ is complete as a metric space.

August 9, 2010 Posted by | Analysis, Measure Theory | 1 Comment