The Unapologetic Mathematician

Mathematics for the interested outsider

Inner Measures

A quick one for today.

In analogy with the outer measure \mu^* induced on the hereditary \sigma-ring \mathcal{H}(\mathcal{S}) by the measure \mu on the \sigma-ring \mathcal{S}, we now define the “inner measure” \mu_* that \mu induces on the same hereditary \sigma-ring \mathcal{H}(\mathcal{S}). We’ve seen that the outer measure is

\displaystyle\mu^*(E)=\inf\{\mu(F)\vert E\subseteq F\mathcal{S}\}

Accordingly, the inner measure is a set function defined by

\displaystyle\mu_*(E)=\sup\{\mu(F)\vert E\supseteq F\mathcal{S}\}

In a way, the properties of \mu_* are “dual” to those of \mu^*. The easy ones are the same: it’s non-negative, monotone, and \mu_*(0)=0.

We could also define \mu_* in terms of the completed measure. Since \mathcal{S}\subseteq\overline{\mathcal{S}}, it’s clear

\displaystyle\mu_*(E)=\sup\{\mu(F)\vert E\supseteq F\mathcal{S}\}\leq\sup\{\bar{\mu}(F)\vert E\supseteq F\overline{\mathcal{S}}\}

On the other hand, the definition of the completion says that for every F\in\overline{\mathcal{S}} there is a G\in\mathcal{S} with G\subseteq F and \mu(G)=\bar{\mu}(F), and so this is actually an equality.

April 8, 2010 Posted by | Analysis, Measure Theory | 3 Comments