The Vitali Covering Theorem for Measurable Sets

Fine Ball Families and Almost-Everywhere Coverage

A disjoint collection of balls cannot generally cover a set point by point: boundaries and gaps may remain. The Vitali covering theorem gives a precise substitute. If balls centered at each point of a bounded measurable set are available at arbitrarily small radii, then one can select countably many pairwise disjoint balls whose union misses only a Lebesgue-null subset. The key is to choose balls by size and control the remaining points using enlarged versions of the selected balls.

Throughout, the ambient measure space is $(\mathbb{R}^n,\mathcal{L},m)$, with Lebesgue measure. It is complete and sigma-finite. Write $B(x,r)=\{y\in\mathbb{R}^n:|y-x|<r\}$ for an open ball of center $x$ and radius $r>0$, and write $3B(x,r)=B(x,3r)$. Lebesgue measure is translation invariant and scales by $m(B(x,r))=\omega_n r^n$, where $\omega_n=m(B(0,1))$. In particular, a bounded set has finite measure.

Definition: Let $E\subseteq\mathbb{R}^n$. A family $\mathcal{V}$ of open balls is a fine covering family of $E$ if every ball in $\mathcal{V}$ is centered at a point of $E$, and for each $x\in E$ and each $\delta>0$, there is a ball $B(x,r)\in\mathcal{V}$ with $0<r<\delta$. The condition concerns arbitrarily small balls centered at each point; it is stronger than merely requiring $E\subseteq\bigcup_{B\in\mathcal{V}}B$.

We will use the elementary fact that the boundary of a ball has Lebesgue measure zero. For $R>0$ and $0<\varepsilon<R$, its boundary is contained in the annulus $B(x,R+\varepsilon)\setminus B(x,R-\varepsilon)$, whose measure is $\omega_n((R+\varepsilon)^n-(R-\varepsilon)^n)$. This quantity tends to zero as $\varepsilon$ tends to zero, so the boundary is null. The same argument applies to every ball boundary below.

The Vitali Covering Theorem

Theorem (Vitali covering theorem for bounded measurable sets): Suppose $E\in\mathcal{L}$ is bounded in $\mathbb{R}^n$, and $\mathcal{V}$ is a fine covering family of $E$. Then there is a finite or countable subfamily of pairwise disjoint balls $\{B_j\}\subseteq\mathcal{V}$ such that $$ m\left(E\setminus\bigcup_j B_j\right)=0. $$ If $E=\varnothing$, the empty subfamily has this property.

Proof. Assume $E\ne\varnothing$. Restrict $\mathcal{V}$ to balls of radius less than $1$. The restricted family is still fine: for any $\delta>0$, use the defining property with $\min(\delta,1)$. Every such ball is contained in a fixed bounded neighborhood of $E$, which has finite Lebesgue measure.

Choose balls recursively. After choosing $B_1,\ldots,B_{k-1}$, call a ball in the restricted family eligible if it is disjoint from every previously chosen open ball. If there are no eligible balls, stop. Otherwise let $s_k$ be the supremum of their radii and choose an eligible ball $B_k$ with radius $r_k>s_k/2$. Such a choice is possible by the definition of supremum. This constructs a finite or countable pairwise disjoint subfamily. If the construction is infinite, countable additivity and containment in a bounded set give $$ \sum_{j=1}^{\infty}\omega_n r_j^n =\sum_{j=1}^{\infty}m(B_j) \le m(K)<\infty $$ for some bounded measurable set $K$ containing all the selected balls.

It remains to show that the part of $E$ outside the selected balls is null. First consider points outside their closed balls. If the construction stopped after finitely many choices, take such a point $x$. Its distance from the finite union of the closed selected balls is positive. By fineness, there is a ball in the restricted family centered at $x$ with radius smaller than that distance. It is disjoint from every selected ball, contradicting the stopping rule. Thus, in the finite case, $E$ is contained in the union of the selected closed balls.

Now suppose the construction is infinite, and fix $x\in E$ outside every selected closed ball. Fix a positive integer $N$. The distance from $x$ to the union of the first $N$ closed balls is positive. Fineness therefore supplies a ball $B(x,r)$ from the restricted family disjoint from those first $N$ balls. We claim that it intersects some later selected ball. If it never intersected any selected ball, it would be eligible at every stage $k>N$. The selection rule would then give $r_k>r/2$ for every $k>N$, contradicting the finiteness of $\sum_j\omega_n r_j^n$.

Let $k>N$ be the first index for which $B(x,r)$ intersects $B_k=B(c_k,R_k)$. At the stage when $B_k$ was chosen, $B(x,r)$ was eligible. Hence $R_k>r/2$. Intersection of the two open balls implies $|x-c_k|<r+R_k<3R_k$, so $x\in 3B_k$. Since $N$ was arbitrary, every such $x$ belongs to $\bigcup_{j>N}3B_j$ for every $N$. Consequently, for each $N$, $$ m\left(E\setminus\bigcup_j\overline{B_j}\right) \le \sum_{j>N}m(3B_j) =3^n\sum_{j>N}m(B_j). $$ The series on the right converges, so its tails tend to zero. The set on the left therefore has measure zero. Finally, $E\setminus\bigcup_j B_j$ is contained in that null set together with the countable union of the boundaries $\partial B_j$. Each boundary is null, so countable subadditivity gives $m(E\setminus\bigcup_jB_j)=0$. The finite case has the same conclusion because only finitely many ball boundaries are added. $\square$

Worked Applications

Worked Example: A Disjoint Selection on the Unit Interval

Take $E=[0,1]\subset\mathbb{R}$, with Lebesgue measure $m$, and let $\mathcal{V}$ consist of all open intervals $B(x,r)$ centered at $x\in E$ with $0<r<1$. This family is fine: given $x\in E$ and $\delta>0$, any radius $r$ satisfying $0<r<\min(\delta,1)$ gives a ball in $\mathcal{V}$.

Select the four intervals with centers $1/8,3/8,5/8,7/8$, each of radius $1/8$. Written out, they are $(0,1/4)$, $(1/4,1/2)$, $(1/2,3/4)$, and $(3/4,1)$. They are pairwise disjoint as open intervals. Each has measure $1/4$, so the sum of their measures is $4\cdot(1/4)=1$. Their union omits from $E$ exactly the five points $0,1/4,1/2,3/4,1$, a null set. Check: the total selected measure is $1=m([0,1])$, and the uncovered set is finite, hence has measure zero.

Worked Example: The Closed Unit Disk

In $\mathbb{R}^2$, let $E=\{x:|x|\le1\}$, and let $\mathcal{V}$ be all open balls centered at points of $E$ with positive radius. This is fine, since each center admits every sufficiently small positive radius. The single ball $B(0,1)$ belongs to $\mathcal{V}$, and its union is disjoint from no other selected ball because it is the only one selected.

The uncovered part of $E$ is the unit circle. Its measure is zero: for $0<\varepsilon<1$, it lies in the annulus between radii $1-\varepsilon$ and $1+\varepsilon$, whose area is $$ \pi(1+\varepsilon)^2-\pi(1-\varepsilon)^2 =\pi(4\varepsilon)=4\pi\varepsilon. $$ As $\varepsilon$ tends to zero, this upper bound tends to zero. The disk has area $\pi$, and the selected ball has area $\pi$, which also checks that the uncovered boundary contributes no area.

Worked Example: Computing a Tail Bound for Enlarged Intervals

Consider the pairwise disjoint intervals $I_j=(2^{-j},2^{-j}+2^{-j-2})$ in $\mathbb{R}$, for integers $j\ge1$. Their lengths are $m(I_j)=2^{-j-2}$. They are disjoint because the right endpoint of $I_{j+1}$ is $2^{-(j+1)}+2^{-(j+3)}=(5/8)2^{-j}$, which is less than the left endpoint $2^{-j}$ of $I_j$.

In one dimension, tripling an interval multiplies its length by $3$, so $m(3I_j)=3\cdot2^{-j-2}$. For $N=2$, the tail length is $$ \sum_{j=3}^{\infty}m(I_j) =\sum_{j=3}^{\infty}2^{-j-2} =\frac{1}{32}+\frac{1}{64}+\cdots =\frac{1}{16}. $$ The sum of the lengths of the tripled tail intervals is therefore $3/16$. In general, the geometric-series formula gives $\sum_{j>N}m(I_j)=2^{-N-2}$, so the tripled tail bound is $3\cdot2^{-N-2}$, which tends to zero. Check: at $N=2$, the formula gives $3\cdot2^{-4}=3/16$, matching the direct calculation. This is the tail estimate used in the theorem’s proof; it does not assert that enlarged intervals are themselves disjoint.

What the Conclusion Does—and Does Not—Say

The theorem guarantees that the original selected balls are pairwise disjoint and cover $E$ up to a null set. It does not guarantee that every point of $E$ lies in one of them. Points on selected ball boundaries may be missed, and other null sets may also remain. The proof controls these points by showing they lie in tails of enlarged selected balls, whose total measure tends to zero.

Common mistake: treating a dilated cover as a disjoint cover. A fivefold covering argument can produce disjoint balls whose enlargements cover a set; that statement alone does not show that the original balls cover almost all of it. For a concrete calculation, take $E=[0,1]$ and the two disjoint intervals $(0,1/4)$ and $(1/2,3/4)$. Their union leaves a set of measure $1/2$: the omitted portions have lengths $1/4$, $1/4$, and the endpoints have measure zero. Thus disjointness alone is insufficient. The Vitali theorem needs the fine-family hypothesis and the radius-selection argument; it concludes that the original selected balls, not just their enlargements, leave a null remainder.

Check Your Understanding

Use the definitions and proof above to answer the following questions.

Which condition in the definition of a fine covering family requires arbitrarily small balls, and why is it used in the proof?
Why does the radius-selection rule imply $R_k>r/2$ when the candidate ball $B(x,r)$ first intersects $B_k$?
In $\mathbb{R}^n$, what factor relates $m(3B)$ to $m(B)$, and where does that factor enter the tail estimate?
For the four intervals selected on $[0,1]$, list the uncovered points and compute the total measure of the selected intervals.
Explain why coverage by enlarged balls does not by itself prove the Vitali theorem’s conclusion about the original balls.

The Vitali Covering Theorem for Measurable Sets

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