# The Unapologetic Mathematician

## Sets Measurable by an Outer Measure I

An outer measure $\mu^*$ on a hereditary $\sigma$-ring $\mathcal{H}$ is nice and all, but it’s not what we really want, which is a measure. In particular, it’s subadditive rather than additive. We want to fix this by restricting to a nice collection of sets within $\mathcal{H}$.

Every set $E$ splits every other set into two pieces: the part that’s in $E$ and the part that’s not. What we want to focus on are the sets that split every other set additively. That is, sets $E\in\mathcal{H}$ so that for every set $A\in\mathcal{H}$ we have

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

We call such sets “$\mu^*$-measurable”. Actually, to show that $E$ is $\mu^*$-measurable we just need to show that $\mu^*(A)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)$ for every $A\in\mathcal{H}$, because the opposite inequality follows from the subadditivity of $\mu^*$.

This condition seems sort of contrived at first, and there’s really not much to justify it at first besides the foregoing handwaving. But we will soon see that this definition turns out to be useful. For one thing, the collection $\overline{\mathcal{S}}\subseteq\mathcal{H}$ of $\mu^*$-measurable sets is a ring!

The proof of this fact is straightforward, but it feels like pulling a rabbit out of a hat, so follow closely. Given sets $\mu^*$-measurable sets $E$ and $F$, we need to show that their union $E\cup F$ and difference $E\setminus F=E\cap F^c$ are both $\mu^*$-measurable as well. Saying that $E$ is $\mu^*$-measurable means that for every $A\in\mathcal{H}$ we have

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

Saying that $F$ is $\mu^*$-measurable means that for every $A\in\mathcal{H}$ we have

\displaystyle\begin{aligned}\mu^*(A\cap E)&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)\\\mu^*(A\cap E^c)&=\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

We can take each of these and plug them into the first equation to find the key equation

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)\\&+\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

Now this key equation works for $A\cap(E\cup F)$ as well as $A$. We know that $(E\cup F)\cap E=E$ and $(E\cup F)\cap F=F$, but $(E\cup F)\cap E^c\cap F^c=\emptyset$. So, sticking $A\cap(E\cup F)$ into the key equation we find

\displaystyle\begin{aligned}\mu^*(A\cap(E\cup F))&=\mu^*(A\cap(E\cup F)\cap E\cap F)+\mu^*(A\cap(E\cup F)\cap E\cap F^c)\\&+\mu^*(A\cap(E\cup F)\cap E^c\cap F)+\mu^*(A\cap(E\cup F)\cap E^c\cap F^c)\\&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)+\mu^*(A\cap E^c\cap F)\end{aligned}

But the three terms on the right are the first three terms in the key equation. And so we can replace them and write

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap(E\cup F))+\mu^*(A\cap E^c\cap F^c)\\&=\mu^*(A\cap(E\cup F))+\mu^*(A\cap(E\cup F)^c)\end{aligned}

which establishes that $E\cup F$ is $\mu^*$-measurable! Behold, the rabbit!

Let’s see if we can do it again. This time, we take $A\cap(E\setminus F)^c=A\cap(E^c\cup F)$ and stick it into the key equation. We find

\displaystyle\begin{aligned}\mu^*(A\cap(E\setminus F)^c)&=\mu^*(A\cap(E^c\cup F)\cap E\cap F)+\mu^*(A\cap(E^c\cup F)\cap E\cap F^c)\\&+\mu^*(A\cap(E^c\cup F)\cap E^c\cap F)+\mu^*(A\cap(E^c\cup F)\cap E^c\cap F^c)\\&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

Again we can find the three terms on the right of this equation on the right side of the key equation as well. Replacing them in the key equation, we find

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap(E\setminus F)^c)+\mu^*(A\cap(E\setminus F)^c)\end{aligned}

which establishes that $E\setminus F$ is $\mu^*$-measurable as well!

March 29, 2010 Posted by | Analysis, Measure Theory | 12 Comments