# Vitali Cover and Null-Sets

I have the following problem that I am stuck on.

Let $E$ be a set of finite outer measure and let $\mathfrak{F}$ be a collection of closed, bounded intervals that covers $E$ in the sense of Vitali. Show that there is a countable disjoint collection $\{I_{k}\}_{k=1}^{\infty}$ of intervals in $\mathfrak{F}$ for which $m^{*}\left(E\sim \bigcup_{k=1}^{\infty}I_{k}\right)=0$.

Here, $m^{*}$ is the Lebesgue outer measure.

I know that because $\mathfrak{F}$ is a Vitali cover of $E$, for all $x\in E$ and for all $\epsilon>0$, there exists an interval $I\in\mathfrak{F}$ with $x\in I$ and length $\ell(I)<\epsilon$. I also know that by the Vitali Covering Lemma, there exists a finite disjoint subcollection $\{I_{k}\}_{k=1}^{n}$ of $\mathfrak{F}$ such that $m^{*}\left(E\sim\bigcup_{k=1}^{n}I_{k}\right)<\epsilon$ for each $\epsilon>0$.

Intuitively, it makes sense to me that as we pass from a finite subcollection to a countable subcollection, the outer measure of the complement should tend to zero. However, I can seem to work out all the details of this. Thanks in advance for any help!

EDIT: There seems to have been the same question asked at this link:

Vitali Covering

The argument sketched is from Mattila's Geometry of Sets and Measures in Euclidean Spaces. I have omitted some details.

1. In general, suppose that $A\subseteq\mathbb{R}$ is bounded and that $V=\{I_i\}_{i\in J}$ is a Vitali covering of $A$ whose elements are intervals. There is a countable, pairwise disjoint sub-collection of $V$, $\{I(x_n;r_n)\}_{n\in\mathbb{N}}$, such that $$\bigcup_{i\in J} I_i \subseteq \bigcup_{n=1}^{\infty} I_n(x_n,5r_n).$$ Here $I(x;r)$ is the interval centered at $x$ with radius $r>0$. (You can show this recursively.)
2. Assume that $E\subseteq [0,1]$.
3. Take $\varepsilon>0$ and $O\subseteq \mathbb{R}$ open such that $$E\subseteq O, \quad m^*E \leq m^* O \leq (1+\varepsilon) m^*E.$$ Take $\{I_n\}_{n}$ as in the 1. and note that $$E\subseteq \bigcup_{i\in J} I_i\subseteq \bigcup_{n} I(x_n,5r_n),$$ so $5^{-1} m^* E \leq \sum_{n} m I(x_n,r_n)$. Hence, for $\delta>0$ small and $k\in\mathbb{N}$ we have $$\delta m^* E \leq \sum_{n=1}^k m I(x_n,r_n).$$ Thus, there is some $c=c(\varepsilon,\delta)\in (0,1)$ such that $$m^*\left( E\setminus \bigcup_{n=1}^k I(x_n,r_n) \right) \leq m^*\left(O\setminus \bigcup_{n=1}^k I(x_n,r_n) \right)\leq c m^* E.$$ Call $E_1=E\setminus \bigcup_{n=1}^k I(x_n,r_n)$.

4. Repeat the process infinitely. At each iteration you obtain a set $E_n$ such that $m^*E_n< c^n m^* E \to 0$ as $n\to\infty$. From here you can get the desired conclusion.