The Unapologetic Mathematician

Mathematics for the interested outsider

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 »

  1. […] the is all of . Thus the countable union of these closed subsets has an interior point. But since is a complete metric space, it is a Baire space as well. And thus one of the must have an interior point as […]

    Pingback by Associated Metric Spaces and Absolutely Continuous Measures I « The Unapologetic Mathematician | August 16, 2010 | Reply

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