# Vitali coverings on the real line

Jordan Bell
March 10, 2016

For $x\in\mathbb{R}$ and $r>0$ write

 $B(x,r)=\{y\in\mathbb{R}:|y-x|

Let $\lambda$ be Lebesgue measure on the Borel $\sigma$-algebra of $\mathbb{R}$ and let $\lambda^{*}$ be Lebesgue outer measure on $\mathbb{R}$.

A Vitali covering of a set $E\subset\mathbb{R}$ is a collection $\mathcal{V}$ of closed intervals such that for $\epsilon>0$ and for $x\in E$ there is some $I\in\mathcal{V}$ with $x\in I$ and $0<\lambda(I)<\epsilon$.

The following is the Vitali covering theorem.11 1 Klaus Bichteler, Integration – A Functional Approach, p. 161, Lemma 10.5; John J. Benedetto and Wojciech Czaja, Integration and Modern Analysis, p. 179, Theorem 4.3.1; Russell A. Gordon, The Integrals of Lebesgue, Denjoy, Perron, and Henstock, p. 52, Lemma 4.6.

###### Theorem 1 (Vitali covering theorem).

Let $U$ be an open set in $\mathbb{R}$ with $\lambda(U)<\infty$, let $E\subset U$, and let $\mathcal{V}$ be a Vitali covering of $E$ each interval of which is contained in $U$. Then for any $\epsilon>0$, there are disjoint $I_{1},\ldots,I_{n}\in\mathcal{V}$ such that

 $\lambda^{*}\left(E\setminus\bigcup_{j=1}^{n}I_{j}\right)<\epsilon.$
###### Proof.

Suppose that $I_{1},\ldots,I_{n}\in\mathcal{V}$ are pairwise disjoint. If $E\subset\bigcup_{j=1}^{n}I_{j}$ then $I_{1},\ldots,I_{n}$ satisfy the claim, and otherwise, let

 $U_{n}=U\setminus\bigcup_{j=1}^{n}I_{j},$

and there exists some $x\in E\cap U_{n}$. As $x\in U_{n}$ and $U_{n}$ is open, there is some $\eta>0$ such that $B(x,\eta)\subset U_{n}$ and then as $\mathcal{V}$ is a Vitali covering of $E$ there is some $I\in\mathcal{V}$ with $x\in I\subset B(x,\eta)\subset U_{n}$. Thus $\delta_{n}>0$ for

 $\delta_{n}=\sup\left\{\lambda(I):I\in\mathcal{V},I\subset U_{n}\right\},$

and there is some $I_{n+1}\in\mathcal{V}$ with $I_{n+1}\subset U_{n}$ and $\lambda(I_{n+1})>\frac{\delta_{n}}{2}$.

For $j\geq 1$ write $I_{j}=[x_{j}-r_{j},x_{j}+r_{j}]$ and let $J_{j}=[x_{j}-5r_{j},x_{j}+5r_{j}]$, namely $J_{j}$ is concentric with $I_{j}$ and $\lambda(J_{j})=5\lambda(I_{j})$. Then, as the intervals $I_{1},I_{2},\ldots$ are pairwise disjoint Borel sets each contained in $U$,

 $\sum_{j=1}^{\infty}\lambda(J_{j})=5\sum_{j=1}^{\infty}\lambda(I_{j})=5\lambda% \left(\bigcup_{j=1}^{\infty}I_{j}\right)\leq 5\lambda(U)<\infty$

and it follows from $\sum_{j=1}^{\infty}\lambda(J_{j})<\infty$ that $\sum_{j=M}^{\infty}\lambda(J_{j})\to 0$ as $M\to\infty$, which with

 $\lambda\left(\bigcup_{j=M}^{\infty}J_{j}\right)\leq\sum_{j=M}^{\infty}\lambda(% J_{j})$

yields $\lambda\left(\bigcup_{j=M}^{\infty}J_{j}\right)\to 0$ as $M\to\infty$.

Let $M\geq 1$. If $x\in E\setminus\bigcup_{j=1}^{\infty}I_{j}$ then $x\in E\setminus\bigcup_{j=1}^{M}I_{j}$ and so $x\in U_{M}$, and as $U_{M}$ is open there is some $\eta>0$ with $B(x,\eta)\subset U_{M}$. But $x\in E$ and $\mathcal{V}$ is a Vitali covering of $E$, so there is some $I\in\mathcal{V}$ with $x\in I$ and $I\subset B(x,\eta)\subset U_{M}$. Now, $\lambda(I_{j+1})>\frac{\delta_{j}}{2}$ and $\sum_{j=1}^{\infty}\lambda(I_{j})<\infty$ together imply $\delta_{n}\to 0$ as $n\to\infty$, so there is some $n$ for which $\delta_{n}<\lambda(I)$. By the definition of $\delta_{n}$ as a supremum, this means that $I\not\subset U_{n}$ and so it makes sense to define $N$ to be a minimal positive integer such that $I\not\subset U_{N}$. $M: if $M\geq N$ then $I\subset U_{M}\subset U_{N}$, contradicting $I\not\subset U_{N}$. (We shall merely use that $M\leq N$.) The fact that $I\not\subset U_{N}$ and $I\subset U_{N-1}$ means that $I\cap I_{N}\neq\emptyset$ and also, by the definition of $\delta_{N-1}$, $\lambda(I)\leq\delta_{N-1}<2\lambda(I_{N})$. Write $I=[y-r,y+r]$. $I\cap I_{N}\neq\emptyset$ tells us $y-r\leq x_{N}+r_{N}$ and $y+r\geq x_{N}-r_{N}$, and $\lambda(I)<2\lambda(I_{N})$ tells us $2r<4r_{N}$, hence

 $y+r\leq x_{N}+r_{N}+2r\leq x_{N}+5r_{N},\qquad y-r\geq x_{N}-r_{N}-2r\geq x_{N% }-5r_{N},$

showing that

 $x\in I=[y-r,y+r]\subset J_{N}\subset\bigcup_{j=M}^{\infty}J_{j}.$

This is true for each $x\in E\setminus\bigcup_{j=1}^{\infty}I_{j}$, which means that

 $E\setminus\bigcup_{j=1}^{\infty}I_{j}\subset\bigcup_{j=M}^{\infty}J_{j}.$

Because $\lambda(\bigcup_{j=M}^{\infty}J_{j})\to 0$ as $M\to\infty$, this yields

 $\lambda^{*}\left(E\setminus\bigcup_{j=1}^{\infty}I_{j}\right)=0.$

But $E\setminus\bigcup_{j=1}^{n}I_{j}$ is an increasing sequence of sets tending to $E\setminus\bigcup_{j=1}^{\infty}I_{j}$, therefore

 $\lambda^{*}\left(E\setminus\bigcup_{j=1}^{n}I_{j}\right)\to\lambda^{*}\left(E% \setminus\bigcup_{j=1}^{\infty}I_{j}\right)=0,\qquad n\to\infty,$

so there is some $n$ such that $\lambda^{*}\left(E\setminus\bigcup_{j=1}^{n}I_{j}\right)<\epsilon$ and then $I_{1},\ldots,I_{n}$ satisfy the claim. ∎