$\inf m_*(O)$ for Open Set $O$

measure-theoryreal-analysis

I am trying to understand $\inf m_*(O)$ for Open Set $O$ correctly since it took some time for me to straighten out the infimum in $m_*(O)$.

In the Exterior Measure section by Stein and Shakarchi (2009), they state:

"If $E\subset\mathbb{R}^d$, then $m_*(E)=\inf m_*(O)$ where the infimum is taken over all open sets $O$ containing $E$."

My Question:

  1. Is my understanding of $\inf m_*(O)$ correct?

$$\inf m_*(O)=\inf\{m_*(O):E\subset O\}$$
$$\inf m_*(O)=\inf\{\inf\{\sum_j|Q_j|:O\subset\bigcup_j Q_j,Q_j^s \text{ are closed cubes}\}:E\subset O\}$$
Thus,

$\inf m_*(O)$ is the infimum of the set of the greatest lower bounds of open sets that each contain $E$.

  1. Is my interpretation of the above observation correct?

You start with $E$. Consider a collection of open sets that contain E. To obtain the outer measure $E$, for each open set containing E, find the outer measure itself by the usual method of fitting the small closed cubes. We put these $m_*(O)^s$, which are numbers, into a set and take an infimum of this set. This should equal to $m_*(E)$. In other words, you are looking a sequence of nested open covers that contain $E$ and for each of this, you can obtain the exterior measure. But we want to find the infimum in this sequence, then it would be exactly the outer measure of $E$.

  1. From the above, if we have a negligible volume set $E$, how do we infer if $m_*(E)=0$, then for every $\epsilon>0$, there exists an open set $O$ with $E\subset O$ and $m_*(O)\leq\epsilon$? In other words, how do we infer the existence of such open set from the above statement?

Reference:
$\textit{Real Analysis: Measure Theory, Integration, and Hilbert Spaces}$. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.

Best Answer

You observation $1,2$ are correct.

For $3$ let $E$ with $m_*(E)=0$ and $\epsilon >0$

Then exists a sequence $Q_n$ of closed cubes,such that $E \subset \bigcup_n Q_n$ and $$m_*(E) \leq \sum_n |Q_n| \leq \frac{\epsilon}{2}$$

For every $Q_n$ we can find an open cube $A_n \supset Q_n$ such that $|A_n| \leq |Q_n|+\frac{\epsilon}{2^{n+1}}$

Then $E \subset \bigcup_n A_n$

and so for the open set $O:=\bigcup_n A_n$ we have $$m_*(O) \leq \sum_n|A_n| \leq \sum_n(|Q_n|+\frac{\epsilon}{2^{n+1}}) \leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

Related Question