Shea Stuart

Answered

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 ${\mu}^{\ast}(E)<\mathrm{\infty}$, then for each $\epsilon $ existx $A\in \mathcal{A}$ such that ${\mu}^{\ast}(A\mathrm{\u25b3}E)<\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?

If $E\in {\mathcal{M}}_{{\mu}^{\ast}}$ and satisfacies ${\mu}^{\ast}(E)<\mathrm{\infty}$, then for each $\epsilon $ existx $A\in \mathcal{A}$ such that ${\mu}^{\ast}(A\mathrm{\u25b3}E)<\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?

Answer & Explanation

iskakanjulc

Expert

2022-07-10Added 18 answers

By definition of the outer measure, there is a countable collection $\{{Q}_{j}\}$ of elements in $\mathcal{A}$ such that

$\sum _{j=1}^{\mathrm{\infty}}\mu ({Q}_{j})\le {\mu}^{\ast}(E)+\epsilon .$

Since ${\mu}^{\ast}(E)<\mathrm{\infty}$, the series above converges. Hence there is an $N$ so that

$\sum _{j=N+1}^{\mathrm{\infty}}\mu ({Q}_{j})<\frac{\epsilon}{2}.$

Consider $A=\bigcup _{j=1}^{N}{Q}_{j}$.

$\sum _{j=1}^{\mathrm{\infty}}\mu ({Q}_{j})\le {\mu}^{\ast}(E)+\epsilon .$

Since ${\mu}^{\ast}(E)<\mathrm{\infty}$, the series above converges. Hence there is an $N$ so that

$\sum _{j=N+1}^{\mathrm{\infty}}\mu ({Q}_{j})<\frac{\epsilon}{2}.$

Consider $A=\bigcup _{j=1}^{N}{Q}_{j}$.

Addison Trujillo

Expert

2022-07-11Added 6 answers

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

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