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 $\epsilon&gt;0$, there exists an interval $I\in\mathfrak{F}$ with $x\in I$ and length $\ell(I)&lt;\epsilon$. 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 $m^{*}\left(E\sim\bigcup_{k=1}^{n}I_{k}\right)&lt;\epsilon$ for each $\epsilon&gt;0$. 

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

However, there was nothing helpful in the comments.
The argument sketched is from Mattila's Geometry of Sets and Measures in Euclidean Spaces. I have omitted some details.


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 $r&gt;0$. (You can show this recursively.)
Assume that $E\subseteq [0,1]$.
Take $\varepsilon&gt;0$ 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 $\delta&gt;0$ 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 $E_1=E\setminus \bigcup_{n=1}^k I(x_n,r_n)$.
Repeat the process infinitely. At each iteration you obtain a set $E_n$ such that $m^*E_n&lt; c^n m^* E \to 0$ as $n\to\infty$. From here you can get the desired conclusion.
The argument sketched is from Mattila's Geometry of Sets and Measures in Euclidean Spaces. I have omitted some details.


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 $r&gt;0$. (You can show this recursively.)
Assume that $E\subseteq [0,1]$.
Take $\varepsilon&gt;0$ 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 $\delta&gt;0$ 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 $E_1=E\setminus \bigcup_{n=1}^k I(x_n,r_n)$.
Repeat the process infinitely. At each iteration you obtain a set $E_n$ such that $m^*E_n&lt; c^n m^* E \to 0$ as $n\to\infty$. From here you can get the desired conclusion.