Sets Measurable by an Outer Measure II

Yesterday, we showed that — given an outer measure $\mu^*$ on a hereditary $\sigma$ -ring $\mathcal{H}$ — the collection of $\mu^*$ -measurable sets $\overline{\mathcal{S}}$ forms a ring. In fact, it forms a $\sigma$ -ring. That is, given a countably infinite sequence ${E_i}_{i=1}^\infty$ of $\mu^*$ -measurable sets, their union $E$ is also $\mu^*$ -measurable. Even better if the $E_i$ are pairwise disjoint, then

$\displaystyle\mu^*(A\cap E)=\mu^*\left(A\cap\bigcup\limits_{i=1}^\infty E_i\right)=\sum\limits_{i=1}^\infty\mu^*(A\cap E_i)$

To see this, we start with a disjoint sequence and an equation we came up with yesterday:

$\displaystyle\begin{aligned}\mu^*(A\cap(E_1\cup E_2))&=\mu^*(A\cap E_1\cap E_2)+\mu^*(A\cap E_1\cap E_2^c)+\mu^*(A\cap E_1^c\cap E_2)\\&=\mu^*(A\cap E_1)+\mu^*(A\cap E_2)\end{aligned}$

We can keep going like this, adding in more and more of the $E_i$ :

$\displaystyle\mu^*\left(A\cap\bigcup\limits_{i=1}^nE\right)=\sum\limits_{i=1}^n\mu^*(A\cap E_i)$

for every natural number $n$ . This finite union $F_n=\bigcup_{i=1}^n E_i$ is $\mu^*$ -measurable. This, along the fact that $F^c\subseteq E^c$ , tells us that

$\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap F_n)+\mu^*(A\cap F^c)\\&\geq\sum\limits_{i=1}^n\mu^*(A\cap E_i)+\mu^*(A\cap E^c)\end{aligned}$

This is true for every $n$ , so we may pass to the limit and use the countable subadditivity of $\mu^*$

$\displaystyle\begin{aligned}\mu^*(A)&\geq\sum\limits_{i=1}^\infty\mu^*(A\cap E_i)+\mu^*(A\cap E^c)\\&\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)\end{aligned}$

But this is enough to show that $E$ is $\mu^*$ -measurable, and so $\overline{\mathcal{S}}$ is closed under countable disjoint unions. And this shows that $\overline{\mathcal{S}}$ is closed under countable unions in general, by our trick of replacing a sequence by a disjoint sequence with the same partial unions.

Since the previous inequalities must then actually be equalities, we see that we must have

$\displaystyle\sum\limits_{i=1}^\infty\mu^*(A\cap E_i)+\mu^*(A\cap E^c)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)$

It’s tempting to simply subtract $\mu^*(A\cap E^c)$ from both sides, but this might be an infinite quantity. Instead, we’ll simply replace $A$ with $A\cap E$ , which has the same effect of giving us

$\displaystyle\sum\limits_{i=1}^\infty\mu^*(A\cap E_i)\geq\mu^*(A\cap E)$

as we claimed.

If we replace $A$ by $E$ in this equation, we find that — when restricted to the $\sigma$ -ring $\overline{\mathcal{S}}$ — $\mu^*$ is actually countably additive. That is, if we define $\bar{\mu}(A)=\mu^*(A)$ for $A\in\overline{\mathcal{S}}$ , then $\bar{\mu}$ is actually a measure. Even better, it’s a complete measure.

Indeed, if $E\subseteq F$ and $\bar{\mu}(F)=\mu^*(F)=0$ , then $\mu^*(E)=0$ as well. We must show that $E$ is actually $\mu^*$ -measurable, and so $\bar{\mu}(E)$ exists and equals zero. But we can easily see that for any $A\in\mathcal{H}$

$\displaystyle\mu^*(A)=\mu^*(E)+\mu^*(A)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)$

and this is enough to show that $E$ is $\mu^*$ -measurable, and thus that $\bar{\mu}$ is complete.

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March 30, 2010 - Posted by John Armstrong | Analysis, Measure Theory

5 Comments »

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[...] hereditary -ring containing . Given a measure on , it induces an outer measure on , which restricts to a measure on [...]

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[...] existence is straightforward. We can induce an outer measure, and then restrict it to get . It’s straightforward to verify from the definitions that . And we know that since [...]

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[...] the essential difference? What do we know about that we don’t know about ? The measure on is complete. The smaller -ring may not contain all negligible [...]

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