The Unapologetic Mathematician

Mathematics for the interested outsider

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