# The Unapologetic Mathematician

## Measures and Induced Measures

If we start with a measure $\mu$ on a ring $\mathcal{R}$, we can extend it to an outer measure $\mu^*$ on the hereditary $\sigma$-ring $\mathcal{H}(\mathcal{R})$. And then we can restrict this outer measure to get an actual measure $\bar{\mu}$ on the $\sigma$-ring $\overline{\mathcal{S}}$ of $\mu^*$-measurable sets. And so we ask: how does the measure $\mu$ relate to the measure $\bar{\mu}$.

First, we will show that any set $E\in\mathcal{R}$ is $\mu^*$-measurable. Then, since $\overline{\mathcal{S}}$ is a $\sigma$-ring containing $\mathcal{R}$ it will also contain the smallest such $\sigma$-ring $\mathcal{S}(\mathcal{R})$.

So, if $E\in\mathcal{R}$ and $A\in\mathcal{H}(\mathcal{R})$, and if $\epsilon>0$, then the definition of the induced outer measure $\mu^*$ tells us that there exists a sequence $\{E_i\}_{i=1}^\infty\subseteq\mathcal{R}$ so that $A$ is contained in their union and

\displaystyle\begin{aligned}\mu^*(A)+\epsilon&\geq\sum\limits_{i=1}^\infty\mu(E_i)\\&=\sum\limits_{i=1}^\infty\left(\mu(E_i\cap E)+\mu(E_i\cap E^c)\right)\\&\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)\end{aligned}

Since this holds for all $\epsilon$, we conclude that $\mu^*(A)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)$, and thus that $E$ is $\mu^*$-measurable.

Now $\bar{\mu}$ is defined for every set in $\mathcal{S}(\mathcal{R})$. And since it’s a measure on each of the rings $\mathcal{S}(\mathcal{R})$ and $\overline{\mathcal{S}}$, we can induce outer measures from each! But since these are $\sigma$-rings, life is a little easier. For one thing, the hereditary $\sigma$-ring each one generates is just $\mathcal{H}(\mathcal{R})$ again; for another, the induced outer measure of a set $E$ is just the infimum of the measure of any set containing $E$. And, as it turns out, both of the induced outer measures are exactly $\mu^*$! That is, for any $E\in\mathcal{H}(\mathcal{R})$ we find:

\displaystyle\begin{aligned}\mu^*(E)&=\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\mathcal{S}(\mathcal{R})\right\}\\&=\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\overline{\mathcal{S}}\right\}\end{aligned}

Indeed, by definition we have

\displaystyle\begin{aligned}\mu^*(E)&=\inf\left\{\sum\limits_{i=1}^\infty\mu(E_i)\bigg\vert E\subseteq\bigcup\limits_{i=1}^\infty E_i,\{E_i\}\subseteq\mathcal{R}\right\}\\&\geq\inf\left\{\sum\limits_{i=1}^\infty\bar{\mu}(E_i)\bigg\vert E\subseteq\bigcup\limits_{i=1}^\infty E_i,\{E_i\}\subseteq\mathcal{S}(\mathcal{R})\right\}\\&=\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\mathcal{S}(\mathcal{R})\right\}\\&\geq\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\overline{\mathcal{S}}\right\}\end{aligned}

since everything in the first set is also in the second, and everything in the third set is also in the fourth. The equality in the middle holds because every sequence in $\mathcal{S}(\mathcal{R})$ can be replaced by a disjoint sequence, and the sum of the measures can then be replaced by the measure of the disjoint union, and so we only ever need to use one set in $\mathcal{S}(\mathcal{R})$.

But since $\bar{\mu}(F)=\mu^*(F)$ for every $F\in\overline{\mathcal{S}}$, we must have $\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\overline{\mathcal{S}}\right\}\geq\mu^*(E)$. Thus all the inequalities above are equalities, as we claimed.

March 31, 2010 Posted by | Analysis, Measure Theory | 2 Comments