Let $E$ be a set of finite outer measure and $F$ a collection of closed, bounded intervals that cover $E$ in the sense of Vitali. Show that there is a countable disjoint collection {${I_k}$}$_{k=1}^\infty$ of intervals in $F$ for which
$m$*$[E$~$\bigcup_{k=1}^{\infty}I_k] = 0$.
This is the same as the Vitali Covering Lemma except for each $\epsilon > 0$, $m$*$[E$~$\bigcup_{k=1}^n I_k] < \epsilon$. In that proof, my book (Royden, 4th) uses
Theorem 11: Let $E$ be any set of real numbers. Then, each of the following four assertions is equivalent to the measurability of E.
i) For each $\epsilon > 0$, there is an open set $O$ containing $E$ for which $m$*$(O$~$E)<\epsilon$.
ii) There is a $G_\delta$ set $G$ containing $E$ for which $m$*$(G$~$E)=0$.
iii), iv) etc.
Would this proof boil down to i) -> ii)?
Best Answer
Let E be a set of finite outer measure and :F a collection of closed, bounded intervals that covers E in the sense of Vitali. Then for each E > 0, there is a finite disjoint subcollection (Ik}k=1 of :F for which m* [E"- U It] < E. k=1 (2) P