# The Unapologetic Mathematician

## 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 $\mu(F_1\Delta F_2)=0$.

About these ads

April 10, 2010 - Posted by | Analysis, Measure Theory

## 2 Comments »

1. […] 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 […]

Pingback by Using Measurable Covers and Kernels I « The Unapologetic Mathematician | April 12, 2010 | Reply

2. […] 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. […]

Pingback by Using Measurable Covers and Kernels II « The Unapologetic Mathematician | April 13, 2010 | Reply