# Vitali covering lemma proof (wrong?)

after seeing some proofs of the Vitali covering lemma for the Lebesgue outer measure, I asked myself why no one used compacity since this aroses naturally in this kind of problem.

If $\mathcal {V}$ is a Vitali covering (composed intervals, not necessarilly closed) for some set $E$ such that $m*(E) < +\infty$, then picking the intervals such that if itś not open, suppose $(a,b]$, then you add $(b-\epsilon, b+\epsilon)$ to it. After this you get a open covering for $E$, then picking an compact subset $K$ of $E$ such that they differs in outer measure by an tiny $\epsilon'$, then, by compacity, you can get a finite open covering for $K$ and since itś finite you can now exclude the $\epsilon$ 's added in the beggining, resulting on a finite subcovering of $\mathcal{V}$. Is there anything wrong in this proof(idea).

• If I understood you correctly, you don't get a subcovering of $E$, only one for $K$. – Thomas Nov 29 '12 at 14:22
• However you can obtain a disjoint subcovering by this one, you just have to cut the overlapping intervals, and since there are only a finite numbers of intervals you will just lose some $\epsilon$'s. Furthermore these ones will belong to $\mathcal{V}$ since it's a Vitali covering. – user40276 Nov 30 '12 at 16:27
• I don't see how truncating the intervals gives you elements of $\mathcal{V}$. A Vitali covering only guarantees you the membership in the collection of some interval around the point smaller than a prescribed length. – Thomas Nov 30 '12 at 21:57