 Shea Stuart

2022-07-09

Let $X$ be a set, $\mathcal{A}$ a ring of subsets of $X$, $\mu :\mathcal{A}\to {\overline{\mathbb{R}}}_{\ge 0}$ a premeasure and ${\mu }^{\ast }$ the outer measure generated by $\mu$. (By Caratheodory)
If $E\in {\mathcal{M}}_{{\mu }^{\ast }}$ and satisfacies ${\mu }^{\ast }\left(E\right)<\mathrm{\infty }$, then for each $\epsilon$ existx $A\in \mathcal{A}$ such that ${\mu }^{\ast }\left(A\mathrm{△}E\right)<\epsilon$
I try to test this, my first idea was to give a cover of $E$, I just don't know if I can find said cover in $\mathcal{A}$, as the measure is not finite, so the extension is not unique, someone can give me a hint how to proceed? iskakanjulc

Expert

By definition of the outer measure, there is a countable collection $\left\{{Q}_{j}\right\}$ of elements in $\mathcal{A}$ such that
$\sum _{j=1}^{\mathrm{\infty }}\mu \left({Q}_{j}\right)\le {\mu }^{\ast }\left(E\right)+\epsilon .$
Since ${\mu }^{\ast }\left(E\right)<\mathrm{\infty }$, the series above converges. Hence there is an $N$ so that
$\sum _{j=N+1}^{\mathrm{\infty }}\mu \left({Q}_{j}\right)<\frac{\epsilon }{2}.$
Consider $A=\bigcup _{j=1}^{N}{Q}_{j}$. Expert

Be careful, since ${\mu }^{\ast }\left(E\right)=inf\left\{\sum _{j=1}^{\mathrm{\infty }}\mu \left({Q}_{j}\right):\left\{{Q}_{j}{\right\}}_{j=1}^{\mathrm{\infty }}\subseteq \mathcal{A},E\subseteq \bigcup _{j=1}^{\mathrm{\infty }}{Q}_{j}\right\}$ is not $\mathrm{\infty }$, ${\mu }^{\ast }\left(E\right)+\epsilon \left(>{\mu }^{\ast }\left(E\right)\right)$ is not an lower bound for $\left\{\sum _{j=1}^{\mathrm{\infty }}\mu \left({Q}_{j}\right):\left\{{Q}_{j}{\right\}}_{j=1}^{\mathrm{\infty }}\subseteq \mathcal{A},E\subseteq \bigcup _{j=1}^{\mathrm{\infty }}{Q}_{j}\right\}$, and so there exists $\left\{{Q}_{j}{\right\}}_{j=1}^{\mathrm{\infty }}\subseteq \mathcal{A}$ with $E\subseteq \bigcup _{j=1}^{\mathrm{\infty }}{Q}_{j}$ such that $\sum _{j=1}^{\mathrm{\infty }}\mu \left({Q}_{j}\right)<{\mu }^{\ast }\left(E\right)+\epsilon$. In other words, the condition ${\mu }^{\ast }\left(E\right)<\mathrm{\infty }$ is necessary for this step.