# The Unapologetic Mathematician

## 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.

March 30, 2010 Posted by | Analysis, Measure Theory | 5 Comments