The Unapologetic Mathematician

Mathematics for the interested outsider

Measurable Kernels

A measurable kernel is the flip side of a measurable cover. Specifically, given E\in\mathcal{H}(\mathcal{S}), a measurable kernel of E is a set F\in\mathcal{S} such that F\subseteq E, and if for every G\in\mathcal{S} with G\subseteq E\setminus F we have \mu(G)=0. And, as it happens, every set E\in\mathcal{H}(\mathcal{S}) has a measurable kernel.

To find it, let \hat{E} be a measurable cover of E. Then let N be a measurable cover of \hat{E}\setminus E, and set F=\hat{E}\setminus N\in\mathcal{S}. Since N contains \hat{E}\setminus E, we find

\displaystyle F=\hat{E}\setminus N\subseteq\hat{E}\setminus(\hat{E}\setminus E)=E.

If G\subseteq E\setminus F, then

\displaystyle G\subseteq E\setminus(\hat{E}\setminus N)=E\cap N\subseteq N\setminus(\hat{E}\setminus E)

Since N was picked to be a measurable cover of \hat{E}\setminus E, we conclude that \mu(G)=0, as we hoped.

Now if F is a measurable kernel of E, then \mu(F)=\mu_*(E). Indeed, since F\subseteq E, we have \mu(F)\leq\mu_*(E). If this inequality is strict then \mu(F)<\infty, and there must be some F_0\in\mathcal{S} with F_0\subseteq E and \mu(F_0)>\mu(F). But F_0\setminus F\subseteq E\setminus F, while \mu(F_0\setminus F)\geq\mu(F_0)-\mu(F)>0, contradicting the fact that F was chosen to be a measurable kernel of E.

The symmetric difference of any two measurable kernels is negligible. Given two measurable kernels F_1 and F_2, we know that F_1\subseteq F_1\cup F_2\subseteq E. This implies that (F_1\cup F_2)\setminus F_1\subseteq E\setminus F_1, and thus \mu((F_1\cup F_2)\setminus F_1)=0. Similarly, \mu((F_1\cup F_2)\setminus F_2)=0, and thus \mu(F_1\Delta F_2)=0.

April 10, 2010 Posted by | Analysis, Measure Theory | 2 Comments