# The Unapologetic Mathematician

## Extending a Measure to an Outer Measure

Let $\mu$ be a measure in a ring (not necessarily an algebra) $\mathcal{R}\subseteq P(X)$, and let $\mathcal{H}=\mathcal{H}(\mathcal{R})$ be the hereditary $\sigma$-ring generated by $\mathcal{R}$. For every $E\in\mathcal{H}$, define $\displaystyle\mu^*(E)=\inf\left\{\sum\limits_{i=1}^\infty\mu(E_i)\bigg\vert\{E_i\}\subseteq\mathcal{R}, E\subseteq\bigcup\limits_{i=1}^\infty E_i\right\}$

That is, $E\in\mathcal{H}$ can be covered by a countable collection of sets in $\mathcal{R}$. For every such cover, sum up the $\mu$-measures of all the sets in the cover, and define $\mu^*(E)$ to be the greatest lower bound of such sums. Then $\mu^*$ is an outer measure, which extends $\mu$ to all of $\mathcal{H}$. Further, if $\mu$ is $\sigma$-finite, then $\mu^*$ will be too. We call $\mu^*$ the outer measure “induced by” $\mu$.

First off, if $E\in\mathcal{R}$ itself, then we can cover it with itself and an infinite sequence of empty sets. That is, $E\subseteq E\cup\emptyset\cup\emptyset\cup\dots$. Thus we must have $\mu^*(E)\leq\mu(E)+\mu(0)+\dots=\mu(E)$. On the other hand, if $E\in\mathcal{R}$ is contained in the union of a sequence $\{E_i\}\subseteq\mathcal{E}$, then monotonicity tells us that $\mu(E)\leq\sum_i\mu(E_i)$, and thus $\mu(E)\leq\mu^*(E)$. That is, $\mu(E)$ must be equal to $\mu^*(E)$ for sets $E\in\mathcal{R}$; as a set function, $\mu^*$ indeed extends $\mu$. In particular, we find that $\mu^*(\emptyset)=\mu(\emptyset)=0$.

If $E$ and $F$ are sets in $\mathcal{H}$ with $E\subseteq F$ and $\{E_i\}$ is a sequence covering $F$, then it must cover $E$ as well. Thus $\mu^*(E)$ can be at most $\mu^*(F)$, and may be even smaller. This establishes that $\mu^*$ is monotonic.

We must show that $\mu^*$ is countably subadditive. Let $E\in\mathcal{H}$ and $\{E_i\}_{i=1}^\infty\subseteq\mathcal{H}$ be sets so that $E$ is contained in the union of the $E_i$. Let $\epsilon$ be an arbitrarily small positive number, and for each $i$ choose some sequence $\{E_{ij}\}_{j=1}^\infty\subseteq\mathcal{R}$ that covers $E_i$ such that $\displaystyle\sum\limits_{j=1}^\infty\mu(E_{ij})\leq\mu^*(E_i)+\frac{\epsilon}{2^i}$

This is possible because the definition of $\mu^*$ tells us that we can find a covering sequence whose measure-sum exceeds $\mu^*(E_i)$ by an arbitrarily small amount. Then the collection of all the $E_{ij}$ constitute a countable collection of sets in $\mathcal{R}$ which together cover $E$. Thus we conclude that $\displaystyle\mu^*(E)\leq\sum\limits_{i=1}^\infty\sum\limits_{j=1}^\infty\mu(E_{ij})\leq\sum\limits_{i=1}^\infty\left(\mu^*(E_i)+\frac{\epsilon}{2^i}\right)=\sum\limits_{i=1}^\infty\mu^*(E_i)+\epsilon$

Since $\epsilon$ was arbitrary, we conclude that $\displaystyle\mu^*(E)\leq\sum\limits_{i=1}^\infty\mu^*(E_i)$

and so $\mu^*$ is countably subadditive.

Finally, if $E\in\mathcal{H}$, we can pick a cover $\{E_i\}_{i=1}^\infty\subseteq\mathcal{R}$. If $\mu$ is $\sigma$-finite, we can cover each of these sets by a sequence $\{E_{ij}\}_{j=1}^\infty\subseteq\mathcal{R}$ so that $\mu(E_{ij})<\infty$. The collection of all the $E_{ij}$ is then a countable cover of $E$ by sets of finite measure; the extension $\mu^*$ is thus $\sigma$-finite as well.

March 26, 2010 - Posted by | Analysis, Measure Theory

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