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.

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April 1, 2010 - Posted by John Armstrong | Analysis, Measure Theory

4 Comments »

[...] -finite on , we just need to show that if has finite outer measure, then . But in this case has a measurable cover with . Since these are finite, we find that . But also has a measurable cover , with . And so we [...]

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[...] A measurable kernel is the flip side of a measurable cover. Specifically, given , a measurable kernel of is a set such that , and if for every with we [...]

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[...] Measurable Covers [...]

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[...] yesterday’s post, here are some more useful facts that we can prove with the help of measurable covers and measurable [...]

Pingback by Using Measurable Covers and Kernels II « The Unapologetic Mathematician | April 13, 2010 | Reply

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