Proof of The Vitali Covering Theorem in $\mathbb{R}^n$

Throughout put $\kappa=\lambda_{n}(\bar{B}(0,1))$. The ball $\bar{B}(0,1)$ is closed, hence lies in $\mathcal{B}(\mathbb{R}^{n})$, and it is bounded; by Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure it contains an open ball of positive measure and has finite measure, so $0<\kappa<\infty$. For $y\in\mathbb{R}^{n}$ and real $r>0$ we have $\bar{B}(y,r)=\{y+rz:z\in\bar{B}(0,1)\}$, by the absolute homogeneity of the norm recorded in Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$; hence, by the scaling and translation invariance of $\lambda_{n}$ (Scaling of Lebesgue Measure and the Lebesgue Integral on $\mathbb{R}^n$ and Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$),

Step 1: restricting the family to a bounded open set. By outer regularity of the outer measure there is an open set $O\subseteq\mathbb{R}^{n}$ with $E\subseteq O$ and $\lambda_{n}(O)\le\lambda_{n}^{\ast}(E)+1<\infty$. Put $\mathcal{F}'=\{(y,r)\in\mathcal{F}:\bar{B}(y,r)\subseteq O\}$.

We record a fact used twice below. Let $C\subseteq\mathbb{R}^{n}$ be closed and let $x\in E$ with $x\notin C$. Then there is $(y,r)\in\mathcal{F}'$ with $x\in\bar{B}(y,r)$ and $\bar{B}(y,r)\cap C=\varnothing$. Indeed, $O\setminus C$ is open and contains $x$, so there is a real $\rho>0$ with $\{z:\lVert z-x\rVert<\rho\}\subseteq O\setminus C$. Because $\mathcal{F}$ finely covers $E$, there is $(y,r)\in\mathcal{F}$ with $x\in\bar{B}(y,r)$ and $r<\rho/3$. For $z\in\bar{B}(y,r)$ the triangle inequality of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ gives $\lVert z-x\rVert\le\lVert z-y\rVert+\lVert y-x\rVert\le 2r<\rho$, since $x\in\bar{B}(y,r)$. Hence $\bar{B}(y,r)\subseteq O\setminus C$, which is what was claimed. Taking $C=\varnothing$ shows in particular that $\mathcal{F}'$ is nonempty whenever $E$ is.

Step 2: the greedy construction. We define pairs $(y_{1},r_{1}),(y_{2},r_{2}),\dots$ in $\mathcal{F}'$ with pairwise disjoint closed balls, recursively. Suppose $k\ge0$ and $(y_{1},r_{1}),\dots,(y_{k},r_{k})$ have been chosen with pairwise disjoint closed balls (for $k=0$ nothing has been chosen). Put

The set $C_{k}$ is closed, since its complement $\bigcap_{i\le k}(\mathbb{R}^{n}\setminus\bar{B}(y_{i},r_{i}))$ is a finite intersection of open sets and hence open by Metric Open Sets Form a Topology.

If $E\subseteq C_{k}$ the construction stops; the conclusion then holds with the $k$ pairs chosen so far and any $\varepsilon>0$, because $E\setminus C_{k}=\varnothing$ has outer measure $0$. Otherwise pick $x\in E\setminus C_{k}$; by Step 1 there is a member of $\mathcal{A}_{k}$, so $\mathcal{A}_{k}\ne\varnothing$. Every $(y,r)\in\mathcal{A}_{k}$ has $\bar{B}(y,r)\subseteq O$, so by monotonicity of $\lambda_{n}$ and $(\ast)$, $r^{n}\kappa\le\lambda_{n}(O)$; hence the set of radii occurring in $\mathcal{A}_{k}$ is a nonempty set of positive reals bounded above, and it has a least upper bound $s_{k}$ by the least upper bound property. Since $s_{k}/2<s_{k}$, the number $s_{k}/2$ is not an upper bound, so we may choose $(y_{k+1},r_{k+1})\in\mathcal{A}_{k}$ with

Its ball is disjoint from $\bar{B}(y_{1},r_{1}),\dots,\bar{B}(y_{k},r_{k})$, so the enlarged family again has pairwise disjoint balls.

Assume from now on that the construction does not stop, so that it yields a sequence $((y_{i},r_{i}))_{i\in\mathbb{N}}$ in $\mathcal{F}'$ with pairwise disjoint closed balls.

Step 3: the radii tend to zero. The balls $\bar{B}(y_{i},r_{i})$ are pairwise disjoint members of $\mathcal{B}(\mathbb{R}^{n})$ contained in $O$, so for every $M\in\mathbb{N}$ finite additivity and monotonicity of $\lambda_{n}$ (claims 1 and 2 of Basic Properties of a Measure) give

Thus the increasing sequence of partial sums is bounded above; by A Bounded Monotone Sequence of Real Numbers Converges it converges, to $S$ say, with $S\le\lambda_{n}(O)$.

Consequently $r_{i}^{n}\kappa\to0$. Indeed, if not, there would be a real $\delta>0$ and, for every $M$, some $i>M$ with $r_{i}^{n}\kappa\ge\delta$; picking such indices successively would make the partial sums exceed any bound, contradicting their convergence. Since $\kappa>0$ and $t\mapsto t^{n}$ is strictly increasing on the positive reals, this gives $r_{i}\to0$.

Step 4: choice of $N$ and the fivefold dilates. Let $\varepsilon>0$. Since the partial sums converge to $S$, there is $N\in\mathbb{N}$ with

We claim

Let $x\in E$ with $x\notin C_{N}$. By Step 1 there is $(y,r)\in\mathcal{F}'$ with $x\in\bar{B}(y,r)$ and $\bar{B}(y,r)\cap C_{N}=\varnothing$, that is $(y,r)\in\mathcal{A}_{N}$.

The ball $\bar{B}(y,r)$ must meet $\bar{B}(y_{i},r_{i})$ for some $i>N$. For otherwise $(y,r)\in\mathcal{A}_{k}$ for every $k$, whence $r\le s_{k}<2r_{k+1}$ for every $k$; letting $k$ grow and using $r_{i}\to0$ gives $r\le0$, contradicting $r>0$.

Let $i$ be the least index with $i>N$ and $\bar{B}(y,r)\cap\bar{B}(y_{i},r_{i})\ne\varnothing$; such a least index exists by The Natural Numbers Are Well Ordered. By minimality $\bar{B}(y,r)$ is disjoint from $\bar{B}(y_{1},r_{1}),\dots,\bar{B}(y_{i-1},r_{i-1})$, that is $(y,r)\in\mathcal{A}_{i-1}$, so $r\le s_{i-1}<2r_{i}$. Choose $w\in\bar{B}(y,r)\cap\bar{B}(y_{i},r_{i})$. Then, by the triangle inequality,

so $x\in\bar{B}(y_{i},5r_{i})$, proving the claim.

Step 5: conclusion. By monotonicity, countable subadditivity and agreement on Borel sets of $\lambda_{n}^{\ast}$, together with $(\ast)$ applied to the balls $\bar{B}(y_{i},5r_{i})$,

where the sum over $i>N$ is the least upper bound of its partial sums, each of which is bounded by $\varepsilon/5^{n}$ by the choice of $N$. The pairs $(y_{1},r_{1}),\dots,(y_{N},r_{N})$ therefore have the required properties.