# Vitali Covering

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 $\epsilon > 0$, $m$*$[E$~$\bigcup_{k=1}^{\infty}I_k] < \epsilon$. 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)?

• Note that in the Vitali covering theorem $E$ need not be measurable. Also, Royden's "Vitali Covering Lemma" yields a finite collection of intervals. – T. Eskin Dec 9 '12 at 20:43
• From my book: The Vitali Covering Lemma - 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 epsilon>0, there is a finite disjoint subcollection, Ik's, of F for which etc. How are they different? – Jake Casey Dec 10 '12 at 20:21
• In the Vitali Covering Lemma, the collection of intervals obtained is finite, while in this exercise, it is countably infinite. Note also that we only know $\mu^{*}(E)<\infty$, not necessarily that $E$ is measurable. – T. Eskin Dec 10 '12 at 22:16
• Ok, so what would be a good way to approach this problem? – Jake Casey Dec 10 '12 at 22:23