Vitali coverings on the real line

For $x\in ℝ$ and $r>0$ write

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

A Vitali covering of a set $E\subset ℝ$ is a collection $𝒱$ of closed intervals such that for $ϵ>0$ and for $x\in E$ there is some $I\in 𝒱$ with $x\in I$ and .

The following is the Vitali covering theorem.

Theorem 1 (Vitali covering theorem).

Let $U$ be an open set in $\mathrm{R}$ with , let $E\mathrm{\subset }U$, and let $\mathrm{V}$ be a Vitali covering of $E$ each interval of which is contained in $U$. Then for any $ϵ\mathrm{>}\mathrm{0}$, there are disjoint $I_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{I}_n\mathrm{\in }\mathrm{V}$ such that

Proof.

Suppose that $I_1,\mathrm{\dots },I_n\in 𝒱$ are pairwise disjoint. If $E\subset {\bigcup }_{j=1}^n{I}_j$ then $I_1,\mathrm{\dots },I_n$ 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 $𝒱$ is a Vitali covering of $E$ there is some $I\in 𝒱$ with $x\in I\subset B(x,\eta )\subset U_n$. Thus ${\delta }_n>0$ for

$${\delta }_n=sup\{\lambda (I):I\in 𝒱,I\subset U_n\},$$

and there is some $I_{n+1}\in 𝒱$ with $I_{n+1}\subset U_n$ and $\lambda (I_{n+1})>\frac{{\delta }_n}{2}$.

For $j\ge 1$ 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 $I_1,I_2,\mathrm{\dots }$ are pairwise disjoint Borel sets each contained in $U$,

and it follows from that ${\sum }_{j=M}^{\mathrm{\infty }}\lambda (J_j)\to 0$ as $M\to \mathrm{\infty }$, which with

$$\lambda \left(\bigcup _{j=M}^{\mathrm{\infty }}{J}_j\right)\le \sum _{j=M}^{\mathrm{\infty }}\lambda (J_j)$$

yields $\lambda \left({\bigcup }_{j=M}^{\mathrm{\infty }}{J}_j\right)\to 0$ as $M\to \mathrm{\infty }$.

Let $M\ge 1$. If $x\in E\setminus {\bigcup }_{j=1}^{\mathrm{\infty }}{I}_j$ 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 $𝒱$ is a Vitali covering of $E$, so there is some $I\in 𝒱$ with $x\in I$ and $I\subset B(x,\eta )\subset U_M$. Now, $\lambda (I_{j+1})>\frac{{\delta }_j}{2}$ and together imply ${\delta }_n\to 0$ as $n\to \mathrm{\infty }$, so there is some $n$ for which . 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 $I\not\subset U_N$. : if $M\ge N$ then $I\subset U_M\subset U_N$, contradicting $I\not\subset U_N$. (We shall merely use that $M\le N$.) The fact that $I\not\subset U_N$ and $I\subset U_{N-1}$ means that $I\cap I_N\ne \mathrm{\varnothing }$ and also, by the definition of ${\delta }_{N-1}$, . Write $I=[y-r,y+r]$. $I\cap I_N\ne \mathrm{\varnothing }$ tells us $y-r\le x_N+r_N$ and $y+r\ge x_N-r_N$, and tells us , hence

$$y+r\le x_N+r_N+2r\le x_N+5r_N,y-r\ge x_N-r_N-2r\ge x_N-5r_N,$$

showing that

$$x\in I=[y-r,y+r]\subset J_N\subset \bigcup _{j=M}^{\mathrm{\infty }}{J}_j.$$

This is true for each $x\in E\setminus {\bigcup }_{j=1}^{\mathrm{\infty }}{I}_j$, which means that

$$E\setminus \bigcup _{j=1}^{\mathrm{\infty }}{I}_j\subset \bigcup _{j=M}^{\mathrm{\infty }}{J}_j.$$

Because $\lambda ({\bigcup }_{j=M}^{\mathrm{\infty }}{J}_j)\to 0$ as $M\to \mathrm{\infty }$, this yields

$${\lambda }^{*}\left(E\setminus \bigcup _{j=1}^{\mathrm{\infty }}{I}_j\right)=0.$$

But $E\setminus {\bigcup }_{j=1}^n{I}_j$ is an increasing sequence of sets tending to $E\setminus {\bigcup }_{j=1}^{\mathrm{\infty }}{I}_j$, therefore

$${\lambda }^{*}\left(E\setminus \bigcup _{j=1}^n{I}_j\right)\to {\lambda }^{*}\left(E\setminus \bigcup _{j=1}^{\mathrm{\infty }}{I}_j\right)=0,n\to \mathrm{\infty },$$

so there is some $n$ such that and then $I_1,\mathrm{\dots },I_n$ satisfy the claim. ∎