# 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 | 5 Comments

## Regular Outer Measures

As usual, let $\mathcal{R}$ be a ring of sets, $\mathcal{S}(\mathcal{R})$ be the smallest $\sigma$-algebra containing $\mathcal{R}$, and $\mathcal{H}(\mathcal{R})$ be the smallest hereditary $\sigma$-algebra containing $\mathcal{R}$. We’ve asked about the relation between a measure $\mu$ on $\mathcal{R}$, the outer measure $\mu^*$ it induces on $\mathcal{H}(\mathcal{R})$, and the measure $\bar{\mu}$ we get by restricting $\mu^*$ to $\mathcal{S}(\mathcal{R})$. But for now, let’s consider what happens when we start with an outer measure on $\mathcal{H}(\mathcal{R})$.

Okay, so we’ve got an outer measure $\mu^*$ on a hereditary $\sigma$-ring $\mathcal{H}$ — like $\mathcal{H}(\mathcal{R})$. We can define the $\sigma$-ring $\overline{\mathcal{S}}$ of $\mu^*$-measurable sets and restrict $\mu^*$ to a measure $\bar{\mu}$ on $\overline{\mathcal{S}}$. And then we can turn around and induce an outer measure $\bar{\mu}^*$ on the hereditary $\sigma$-ring $\mathcal{H}(\overline{\mathcal{S}})$.

Now, in general there’s no reason that these two should be related. But we have seen that if $\mu^*$ came from a measure $\mu$ (as described at the top of this post), then $\mathcal{H}(\overline{\mathcal{S}})=\mathcal{H}(\mathcal{R})$, and the measure $\bar{\mu}^*$ induced by $\bar{\mu}$ is just $\mu^*$ back again!

When this happens, we say that $\mu^*$ is a “regular” outer measure. And so we’ve seen that any outer measure induced from a measure on a ring is regular. The converse is true as well: if we have a regular outer measure $\mu^*=\bar{\mu}^*$, then it is induced from the measure $\bar{\mu}$ on $\overline{\mathcal{S}}$. Induced and regular outer measures are the same.

Doesn’t this start to look a bit like a Galois connection?

April 2, 2010

## Measurable Covers

Again, let $\mathcal{R}$ be a ring, let $\mathcal{S}(\mathcal{R})$ be the smallest $\sigma$-ring containing $\mathcal{R}$, and let $\mathcal{H}(\mathcal{R})$ be the smallest hereditary $\sigma$-ring containing $\mathcal{R}$. Given a measure $\mu$ on $\mathcal{R}$, it induces an outer measure $\mu^*$ on $\mathcal{H}(\mathcal{R})$, which restricts to a measure $\bar{\mu}$ on $\mathcal{S}(\mathcal{R})$.

The sets in $E\in\mathcal{H}(\mathcal{R})$ are easily described — they’re everything that can be countably covered by sets in $\mathcal{R}$ — but we don’t have a measure on this collection. Instead, we need good approximations of these sets by sets in $\mathcal{S}(\mathcal{R})$, where we do have a measure. And so we say that a set $F\in\mathcal{S}(\mathcal{R})$ is a “measurable cover” of $E$ if $E\subseteq F$, and if for every $G\in\mathcal{S}(\mathcal{R})$ with $G\subseteq F\setminus E$ we have $\bar{\mu}(G)=0$. That is, $F$ may be larger than $E$, but only negligibly.

If $E\in\mathcal{H}(\mathcal{R})$ has $\sigma$-finite outer measure, then it has a measurable cover $F$ with $\bar{\mu}(F)=\mu^*(E)$. Indeed, our hypothesis is that $E$ can be written as the countable disjoint union of sets of finite outer measure, so if we can show this for sets of finite measure the $\sigma$-finite case will follow.

By the comparisons we showed last time, for every $n$ there is some set $F_n\in\mathcal{S}(\mathcal{R})$ so that $E\subseteq F_n$ and $\bar{\mu}(F_n)\leq\mu^*(E)+\frac{1}{n}$. Taking $F$ to be the intersection of this sequence, we must have

$\displaystyle \mu^*(E)\leq\bar{\mu}(F)\leq\bar{\mu}(F_n)\leq\mu^*(E)+\frac{1}{n}$

Since $n$ is arbitrary, we must have $\bar{\mu}(F)=\mu^*(E)$. If $G\in\mathcal{S}(\mathcal{R})$ fits into $F\setminus E$, then $E\subseteq F\setminus G$. And so:

$\bar{\mu}(F)=\mu^*(E)\leq\bar{\mu}(F\setminus G)=\bar{\mu}(F)-\bar{\mu}(G)\leq\bar{\mu}(F)$

Since $\bar{\mu}(F)<\infty$ we can subtract it off and find $\bar{\mu}(G)=0$.

In fact, any measurable cover $F$ must have $\bar{\mu}(F)=\mu^*(E)$. Further, any two of them have a negligible difference.

Indeed, if we have two measurable covers, $F_1$ and $F_2$, then $E\subseteq F_1\cap F_2\supseteq F_1$, which tells us that $F_1\setminus(F_1\cap F_2)\subseteq F_1\setminus E$. But since $F_1$ is a measurable cover, we conclude that $\bar{\mu}(F_1\setminus(F_1\cap F_2))=0$, and similarly that $\bar{\mu}(F_2\setminus(F_1\cap F_2))=0$. But $F_1\Delta F_2$ is the union of these two differences, and so $\bar{\mu}(F_1\Delta F_2)=0$.

If $\mu^*(E)=\infty$, then any measurable cover must have $\bar{\mu}(F)=\infty$ as well. On the other hand, if $\mu^*(E)$ is finite, then there exists at least one measurable cover with $\bar{\mu}(F)=\mu^*(E)$. Then any other measurable cover differs negligibly, and so has the same measure.

And so if the measure $\mu$ on $\mathcal{R}$ is $\sigma$-finite, then so are the measures $\bar{\mu}$ on $\mathcal{S}(\mathcal{R})$ and $\bar{\mu}$ on $\overline{\mathcal{S}}$. We’ve already seen that $\mu^*$ must be $\sigma$-finite in this situation, and so any $\mu^*$-measurable set can be covered by a countable sequence of $\mu^*$-finite sets. Applying the above results to each of the sets in this sequence gives our result.

April 1, 2010 Posted by | Analysis, Measure Theory | 4 Comments