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