# The Unapologetic Mathematician

## Using Measurable Covers and Kernels I

There are a bunch of useful facts that we can prove with the help of measurable covers and measurable kernels. These will allow us to move statements we want to show about outer and inner measures to the realm of proper measures, where we can use nice things like additivity.

For example, given a pairwise disjoint sequence $\{E_i\}\subseteq\mathcal{H}(\mathcal{S})$, we can show that

$\displaystyle\mu_*\left(\bigcup\limits_{i=1}^\infty E_i\right)\geq\sum\limits_{i=1}^\infty\mu_*(E_i)$

Just pick a measurable kernel $F_i$ for each $E_i$. Then we can use countable additivity to show

$\displaystyle\sum\limits_{i=1}^\infty\mu_*(E_i)=\sum\limits_{i=1}^\infty\mu(F_i)=\mu\left(\bigcup\limits_{i=1}^\infty F_i\right)\leq\mu_*\left(\bigcup\limits_{i=1}^\infty E_i\right)$

For another, given a set $A\in\mathcal{H}(\mathcal{S})$ and a disjoint sequence $\{E_i\}\subseteq\overline{\mathcal{S}}$ whose union is $E$, then

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

This time, let $F$ be a measurable kernel of $A\cap E$, so that

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

On the other hand, $A\cap E$ is the union of the $A\cap E_i$, and so we can use the previous result to get the opposite inequality.

Next: if $E\in\overline{\mathcal{S}}$, then clearly $\mu_*(E)=\mu^*(E)=\bar{\mu}(E)$. But conversely, if $\mu_*(E)=\mu^*(E)<\infty$, then $E\in\overline{\mathcal{S}}$. To see this, let $A$ be a measurable kernel and $B$ be a measurable cover of $E$. Then we calculate

$\displaystyle\mu(B\setminus A)=\mu(B)-\mu(A)=\mu^*(E)-\mu_*(E)=0$

But $E\setminus A\subseteq B\setminus A$, so $E\setminus A\in\overline{\mathcal{S}}$ (by the completeness of $\bar{\mu}$), and thus $E=(E\setminus A)\cup A\in\overline{\mathcal{S}}$. Thus sets of finite measure in $\overline{\mathcal{S}}$ are exactly those in $\mathcal{H}(\mathcal{S})$ for which the outer and inner measures coincide.

Interestingly, notice how the last step of this proof echoes our earlier result that a set is Jordan measurable if and only if the Jordan content of its boundary is zero.

April 12, 2010 Posted by | Analysis, Measure Theory | 1 Comment