# The Unapologetic Mathematician

## 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