The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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April 13, 2010 - Posted by | Analysis, Measure Theory

2 Comments »

  1. [...] so as well. If is any Lebesgue measurable set, the monotonicity of gives us . And then an earlier result tells us [...]

    Pingback by Non-Lebesgue Measurable Sets « The Unapologetic Mathematician | April 24, 2010 | Reply

  2. [...] since if and are two measurable subsets of with , then . The thickness of implies that , and we know [...]

    Pingback by Measurable Subspaces II « The Unapologetic Mathematician | April 28, 2010 | Reply


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