# The Unapologetic Mathematician

## Completions of Measures

We’ve shown that we can uniquely extend a $\sigma$-finite measure $\mu$ on a ring $\mathcal{R}$ to a unique $\sigma$-finite measure on the $\sigma$-ring $\mathcal{S}(\mathcal{R})$. But, of course, we actually found that we could restrict the outer measure $\mu^*$ to the $\sigma$-ring $\overline{\mathcal{S}}$ of $\mu^*$-measurable sets, which may be larger than $\mathcal{S}(\mathcal{R})$. Luckily, we can get this extra ground without having to go through the outer measure.

What’s the essential difference? What do we know about $\overline{\mathcal{S}}$ that we don’t know about $\mathcal{S}(\mathcal{R})$? The measure on $\overline{\mathcal{S}}$ is complete. The smaller $\sigma$-ring $\mathcal{S}(\mathcal{R})$ may not contain all negligible sets.

So, let’s throw them in; if $\mu$ is a measure on a $\sigma$-ring $\mathcal{S}$, define $\overline{\mathcal{S}}$ to be the class of all sets $E\Delta N$, where $E\in\mathcal{S}$, and $N$ is a negligible with respect to the measure $\mu$, or “$\mu$-negligible”. This collection $\overline{\mathcal{S}}$ is a $\sigma$-algebra, and the set function $\bar{\mu}$ defined on $\overline{\mathcal{S}}$ by $\bar{\mu}(E\Delta N)=\mu(E)$ is a complete measure called the “completion” of $\mu$.

First, given sets $E\in\mathcal{S}$ and $N\subseteq A\in\mathcal{S}$ with $\mu(A)=0$, we have the two equations

\displaystyle\begin{aligned}E\cup N&=(E\setminus A)\Delta\left(A\cap(E\cup N)\right)\\E\Delta N&=(E\setminus A)\cup\left(A\cap(E\Delta N)\right)\end{aligned}

These tell us that any set that can be written as the symmetric difference of a set in $\mathcal{S}$ and a measurable set can also be written as the union of two other such sets, and vice versa. That is, we can also characterize $\overline{\mathcal{S}}$ as the class of sets of the form $E\cup N$ instead of $E\Delta N$.

This characterization makes it clear that $\overline{\mathcal{S}}$ is closed under countable unions. Indeed, just write each set in a sequence as a union of a set $E_i$ from $\mathcal{S}$ and a $\mu$-negligible set $N_i$. The countable union of the $E_i$ is still in $\mathcal{S}$, and the countable union of the negligible sets is still negligible. Thus $\overline{\mathcal{S}}$ is a $\sigma$-ring.

Now, if we have two ways of writing a set in $\overline{\mathcal{S}}$, say $E_1\Delta N_1=E_2\Delta N_2$, then we also have the equation $E_1\Delta E_2=N_1\Delta N_2$. And, therefore, $\mu(E_1\Delta E_2)=\mu(N_1\Delta N_2)=0$. Therefore, $\mu(E_1)=\mu(E_2)$, and the above definition of $\bar{\mu}$ is unambiguous.

Using the characterization of $\overline{\mathcal{S}}$ in terms of unions, it’s easy to verify that $\bar{\mu}$ is a measure. We only need to check countable additivity, which is perfectly straightforward; the union of a pairwise disjoint sequence of sets $\{E_i\cup N_i\}$ is formed by taking the union of the $E_i$ and the $N_i$. The measure $\mu$ is countably additive on the $E_i$, and the union of the $N_i$ is still negligible.

Finally, we must show that $\bar{\mu}$ is complete. But if we have a set $E\cup N$ with $\bar{\mu}(E\Delta N)=0$, then $\mu(E)=0$ and $N\subseteq A$ with $\mu(A)=0$. Thus any subset $M\subseteq E\cup N$ is also a subset of $E\cup A$, with $\mu(E\cup A)=0$. Writing it as $\emptyset\Delta M$, we find $M\in\overline{\mathcal{S}}$, and so $\bar{\mu}$ is complete.

There’s just one fly in the ointment: we don’t really know that this complete measure is the same as the one we get by restricting from $\mu^*$ to $\mu^*$-measurable sets. It turns out that if $\mu$ is $\sigma$-finite on $\mathcal{R}$, then the completion of the extension of $\mu$ to $\mathcal{S}(\mathcal{R})$ is the same as this restriction.

Since we’ve been using $\overline{\mathcal{S}}$ for the completion, let’s write $\mathcal{S}^*$ for the class of $\mu^*$-measurable sets. Since $\mu^*$ restricted to $\mathcal{S}^*$ is complete, it follows that $\overline{\mathcal{S}}\subseteq\mathcal{S}^*$, and that $\bar{\mu}$ and $\mu^*$ coincide on $\overline{\mathcal{S}}$. We just need to show that $\mathcal{S}^*\subseteq\overline{\mathcal{S}}$.

In light of the fact that $\mu^*$ is also $\sigma$-finite on $\mathcal{S}^*$, we just need to show that if $E\in\mathcal{S}^*$ has finite outer measure, then $E\in\overline{\mathcal{S}}$. But in this case $E$ has a measurable cover $F$ with $\mu^*(F)=\mu(F)=\mu^*(E)$. Since these are finite, we find that $\mu^*(F\setminus E)=0$. But $F\setminus E$ also has a measurable cover $G$, with $\mu(G)=\mu^*(F\setminus E)=0$. And so we can write $E=(F\setminus G)\cup(E\cap G)$, showing that $E\in\overline{\mathcal{S}}$.

April 6, 2010 Posted by | Analysis, Measure Theory | 3 Comments

## 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 | 5 Comments