# The Unapologetic Mathematician

## Extensions of Measures

Oops, forgot to post this this earlier…

We can put together what we’ve been doing recently to state the following theorem:

If $\mu$ is a $\sigma$-finite measure on the ring $\mathcal{R}$, then there is a unique measure $\bar{\mu}$ on the $\sigma$-ring $\mathcal{S}(\mathcal{R})$ extending $\mu$. That is, if $E\in\mathcal{R}\subseteq\mathcal{S}(\mathcal{R})$, then $\bar{\mu}(E)=\mu(E)$. Further, the extended measure $\bar{\mu}$ is also $\sigma$-finite.

The existence is straightforward. We can induce an outer measure, and then restrict it to get $\bar{\mu}$. It’s straightforward to verify from the definitions that $\bar{\mu}(E)=\mu(E)$. And we know that since $\mu$ is $\sigma$-finite, so is $\mu^*$, and thus $\bar{\mu}$.

What we need to show is that $\bar{\mu}$ is unique. To this end, let $\mu_1$ and $\mu_2$ be two measures on $\mathcal{S}(\mathcal{R})$ that both extend $\mu$. Let $\mathcal{M}\subseteq\mathcal{S}(\mathcal{R})$ be the class of sets on which $\mu_1$ and $\mu_2$ agree; this obviously contains $\mathcal{R}$.

Now, if one of these two measures — say $\mu_1$ — is finite, and if $\{E_i\}_{i=1}^\infty\subseteq\mathcal{M}$ is a monotone sequence of sets on which $\mu_1$ and $\mu_2$ agree, then the limit of this sequence is again in $\mathcal{M}$. Indeed, since measures are continuous, we must have

$\displaystyle\mu_1\left(\lim\limits_{i\to\infty}E_i\right)=\lim\limits_{i\to\infty}\mu_1(E_i)$

and similarly for $\mu_2$. Since $\mu_1$ is finite, and $\mu_2$ agrees with $\mu_1$ on $\mathcal{M}$, we have a sequence of finite measures $\mu_1(E_i)=\mu_2(E_i)$. The limits of these sequences must then agree, and so $\lim_iE_i\in\mathcal{M}$ as well. Thus $\mathcal{M}$ is a monotone class. Since it contains $\mathcal{R}$, it must contain $\mathcal{S}(\mathcal{R})$, and thus $\mu_1=\mu_2$.

On the other hand, neither measure may be finite. In this case, let $A\in\mathcal{R}$ be some fixed set of finite measure. Now $\mathcal{R}\cap A$ — the collection of intersections of sets in $\mathcal{R}$ with $A$ — is again a ring, and $\mathcal{S}(\mathcal{R})\cap A$ is the smallest $\sigma$-ring containing it. Restricting $\mu_1$ and $\mu_2$ to $\mathcal{S}(\mathcal{R})\cap A$ gives finite measures, and we can use the argument above.

Now every set $E\in\mathcal{S}(\mathcal{R})$ can be covered by a countable, pairwise disjoint collection of sets $A_i\in\mathcal{R}$. For each one, we have $E_i=E\cap A_i\in\mathcal{S}(\mathcal{R})\cap A_i$, and so we must find $\mu_1(E_i)=\mu_2(E_i)$. From here, countable additivity finishes the theorem.

In light of the uniqueness of this extension, we will just call the extended measure $\mu$ again, rather than $\bar{\mu}$.

April 6, 2010 - Posted by | Analysis, Measure Theory

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