The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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

5 Comments »

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