# The Unapologetic Mathematician

## Using Measurable Covers and Kernels II

Following yesterday’s post, here are some more useful facts that we can prove with the help of measurable covers and measurable kernels.

If $E$ and $F$ are disjoint sets in $\mathcal{H}(\mathcal{S})$, then

$\displaystyle\mu_*(E\cup F)\leq\mu_*(E)+\mu^*(F)\leq\mu^*(E\cup F)$

Here, we take $A$ to be a measurable cover of $F$ and $B$ to be a measurable kernel of $E\cup F$. The difference $B\setminus A$ must be contained in $E$, and so

$\displaystyle\mu_*(E\cup F)=\mu(B)=\mu(B\setminus A)+\mu(A)\leq\mu_*(E)+\mu^*(F)$

On the other hand, we can take $A$ to be a measurable kernel of $E$ and $B$ to be a measurable cover of $E\cup F$. Now the difference $B\setminus A$ is contained in $F$, and we find

$\displaystyle\mu^*(E\cup F)=\mu(B)=\mu(A)+\mu(B\setminus A)\geq\mu_*(E)+\mu^*(F)$

Now, if $E\in\overline{\mathcal{S}}$, then for every $A\subseteq X$ whatsoever, we have

$\displaystyle\mu_*(A\cap E)+\mu^*(A^c\cap E)=\bar{\mu}(E)$

We can take $A\cap E$ and $A^c\cap E$ and stick them into the previous result to find

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

But since $E\in\overline{\mathcal{S}}$, we know that $\mu_*(E)=\mu^*(E)=\bar{\mu}(E)$, and this establishes our result.

Interestingly, we can use this method of inner measures as an alternative approach to our extension theorems. If $\mu$ is a $\sigma$-finite measure on a ring $\mathcal{R}$, and if $\mu^*$ is the induced outer measure on $\mathcal{H}(\mathcal{R})$, then for every set $E\in\mathcal{R}$ of finite measure and every $A\in\mathcal{H}(\mathcal{R})$ we have

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

Then if $E$ and $F$ are two sets in $\mathcal{R}$ such that $A\cap E=A\cap F$, then we find

$\mu(E)-\mu^*(A^c\cap E)=\mu(F)-\mu^*(A^c\cap F)$

and so we can use this formula as the definition of the inner measure $\mu_*$. Then we can define a set $E\in\mathcal{H}(\mathcal{R})$ with $\mu^*(E)<\infty$ to be $\mu^*$-measurable if the inner and outer measures match: $\mu_*(E)=\mu^*(E)$. And from here, the rest of the theory is as before.

April 13, 2010 Posted by | Analysis, Measure Theory | 2 Comments