# The Unapologetic Mathematician

## 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