Let $m^*$ denote the Lebesgue outer measure and when the set is measurable, define Lebesgue measure $m:=m^*$.
Prove the following:
Take any set $E\subset\mathbb R^n$ (not necessarily measurable), for any $\varepsilon>0$, there exist open (thus measurable) set $U$, such that
$$m(U)\leqslant m^*(E)+\varepsilon $$
I think if we can
construct a finite open cover of $\partial E$, i.e. $\partial E\subseteq \bigcup_{j\in J} W_j$, such that $m(W_j)<\frac{\varepsilon}{|J|}$ (?)
Then define $U:=\text{int}(E)\cup\left( \bigcup_{j\in J}W_j\right)$, it is open because it is a finite union of open sets. And $U=E\cup\left( \bigcup_{j\in J}W_j\right)$, so by finite subaddativity:
$$m(U)\leqslant m^*(E)+\sum_{j\in J}m(W_j)<m^*(E)+\varepsilon.$$
Is the construction (?) really possible? If so, how to make a rigorous argument? The reason why I believe it works is that $m^*(E)=m^*(\overline{E})$
Thanks!
Best Answer
We are not concerned with the lebesgue measureability of E because we won’t be taking an actual measure on it, just $m^*$. Don’t worry about the boundary of E whatsoever, and chase the definition of $m^{*}$, which is the infemum of the sum of measures of collections of balls (or open sets) which cover E. By definition, if $\epsilon$ > 0, there exists a countable collection of open sets ${O_i}$ with $E \subset \cup_i O_i$ and $m^*(E) \leq \Sigma_i m(O_i) < m^*(E)+ \epsilon$
Then $O = \cup_i O_i$ is an open measureable set, with $E \subset O$ so that, by subadditivity $m^*(E) \leq m^*(O) = m(O) \leq \Sigma_i m(O_i) < m^*(E)+\epsilon$.