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 $E\in\overline{\mathcal{S}}$ .

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

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