I'm following a lecture note to prove Vitali's Covering Theorem. Below is an exposition of my understanding. I have no question here, but I'm very happy to receive your suggestion.
Let $(E, d)$ be a metric space. For $x \in E$ and $r>0$, we denote the open and closed balls by $$ B(x, r) := \{y \in E : d(x, y) < r\} \quad \text{and} \quad \overline{B} (x, r) := \{y \in E : d(x, y) \le r\}. $$
Prop 1 Let $\mathcal F$ be a finite collection of open balls in $E$. There is a subcollection $\mathcal G \subset \mathcal F$ of pairwise disjoint balls such that any ball $B(x, r) \in \mathcal F$ intersects a ball $B(x', r') \in \mathcal G$ with $r' \ge r$.
Proof WLOG, we assume $\mathcal F = \{B_i=B(x_i, r_i) :i = 1, \ldots, n\}$ with $r_1 \ge r_2 \ge \cdots \ge r_n$. First, let $G_1 := B_1$. We construct $\mathcal G$ inductively as follows. For $i = 1, \ldots, n$, $B_i \in \mathcal G$ IFF $B_i$ is disjoint from every ball in $\{B_1, \ldots, B_{i-1}\} \cap \mathcal G$. If $B_i \notin \mathcal G$, then $B_i \cap B_j \neq \emptyset$ for some $B_j \in \mathcal G$ with $j \le i$. This completes the proof. $\tag*{$\blacksquare$}$
Vitali's covering lemma (finite case) Let $\mathcal{F}$ be a finite collection of open balls in $E$. Then there exists a subcollection $\mathcal{G} \subseteq \mathcal{F}$ of pairwise disjoint balls, such that $$ \bigcup \mathcal{F} \subseteq \bigcup\{3 B: B \in \mathcal{G}\} $$ where $3 B$ denotes the ball with the same center as $B$ but with 3 times the radius.
Proof Let $\mathcal G$ be given by Prop 1. Let $B(x, r) \in \mathcal F$. There is $B(x', r') \in \mathcal G$ with $r' \ge r$ that intersects $B(x, r)$. It follows that $B(x, r) \subset B(x', r'+2r) \subset B(x', 3r')$. $\tag*{$\blacksquare$}$
Prop 2 Let $\mathcal{F}$ be a collection of open balls in $E$, each with radius at most $R<\infty$. Then there exists a subcollection $\mathcal{G} \subseteq \mathcal{F}$ of pairwise disjoint balls, such that any ball $B(x, r) \in \mathcal{F}$ intersects a ball $B\left(x^{\prime}, r^{\prime}\right) \in \mathcal{G}$ with $2 r^{\prime}>r$
Proof For $n \in \mathbb N$, let $\mathcal F_{n} := \{B(x, r) \in \mathcal F : R2^{-(n+1)} <r \le R2^{-n}\}$. We define subcollection $\mathcal G_n \subset \mathcal F_n$ inductively as follows.
- By Zorn's lemma, $\mathcal F_1$ has a maximal subcollection $\mathcal G_1$ of pairwise disjoint balls.
- Assume $\mathcal G_k$ has been constructed for $1 \le k \le n$. Let $$ \mathcal H_{n} := \left \{B \in \mathcal F_{n} :B \cap C =\emptyset \text{ for all } C \in \bigcup_{k=1}^{n-1} \mathcal G_k \right \}. $$ By Zorn's lemma, $\mathcal H_{n}$ has a maximal subcollection $\mathcal G_{n}$ of pairwise disjoint balls.
Finally, let $\mathcal G := \bigcup_n \mathcal G_n$. Let $B= B(x, r) \in \mathcal F_{n}$. Then one of the following statements hold, i.e.,
- $B \in \mathcal G_{n}$.
- $B \in \mathcal H_{n} \setminus \mathcal G_{n}$. So there is $C \in \mathcal G_{n}$ such that $B \cap C \neq \emptyset$.
- $B \notin \mathcal H_n$. So there is $C \in \bigcup_{k=1}^{n-1} \mathcal G_k$ such that $B \cap C \neq \emptyset$.
In any case, $B$ intersects a ball in $\mathcal{G}$ of radius $r^{\prime}$ with $r' > R2^{-(n+1)}$. On the other hand, $r \le R2^{-n}$. The claim then follows. $\tag*{$\blacksquare$}$
Vitali's covering lemma (infinite case) Let $\mathcal{F}$ be a collection of open balls in $E$, each with radius at most $R<\infty$. Then there exists a subcollection $\mathcal{G} \subseteq \mathcal{F}$ of pairwise disjoint balls, such that $$ \bigcup \mathcal{F} \subseteq \bigcup\{5 B: B \in \mathcal{G}\}. $$
Proof Let $\mathcal G$ be given by Prop 2. Let $B(x, r) \in \mathcal F$. There is $B(x', r') \in \mathcal G$ with $2r' \ge r$ that intersects $B(x, r)$. It follows that $$ B(x, r) \subset B(x', r'+2r) \subset B(x', 5r'). \tag*{$\blacksquare$} $$
Remark Above results still hold true if we replace open balls with closed balls.
Let $A \subset E$. A collection $\mathcal{F}$ of closed balls with strictly positive radii in $E$ is a Vitali cover of $A$ if for each $(x, \delta) \in A \times \mathbb R_{>0}$, there is some ball $B=B(a, r) \in \mathcal{F}$ such that $r<\delta$ and $x \in B$. In other words, every point in $A$ is covered by an arbitrarily small ball in $\mathcal{F}$.
Vitali's Covering Theorem Let $A \subseteq \mathbb{R}^n$ be a set (not necessarily measurable), and let $\mathcal{F}$ be a Vitali cover of $A$. Then there exists a subcollection $\mathcal{G} \subseteq \mathcal{F}$ of pairwise disjoint balls, such that $$ \lambda \left (A \setminus \bigcup \mathcal{G} \right)=0. $$ Here $\lambda$ is the Lebesgue measure.
Proof Let $r(B)$ be the radius of $B \in \mathcal F$. WLOG, we assume that $r(B) \le 1$ for all $B \in \mathcal{F}$. Let $\mathcal{G}$ be given by Vitali's covering lemma (infinite case). We claim that $\mathcal{G}$ satisfies the required condition. It suffices to show that $A' :=(A \setminus \bigcup \mathcal{G}) \cap B(0, R)$ is $\lambda$-null for every $R>0$.
Fix $\varepsilon>0$. Let $\mathcal{G}^{\prime} := \{B \in \mathcal G : B \cap B(0, R) \neq \emptyset\}$. Then $\operatorname{diam} (\bigcup \mathcal G') <\infty$ and thus $\lambda (\bigcup \mathcal G') < \infty$. Hence there exists $\delta>0$ such that $$ \sum_{B \in \mathcal{G}^{\prime}, r(B)<\delta } \lambda(B) <\varepsilon 5^{-n} . $$
Those balls in $\mathcal{G}^{\prime}$ with radii at least $\delta$ have Lebesgue measure at least $\lambda(B(0, \delta))>0$, so there are at most finitely many such balls. Hence their union $K$ is a closed set.
Since $\mathcal{F}$ is a Vitali cover, for any point $x \in A'$ there exists a ball $B_x \in \mathcal{F}$ such that $x \in B_x \subseteq B(0, R)$ and $B_x \cap K=\emptyset$. By Vitali's covering lemma (infinite case), $B_x$ intersects a ball $C_x \in \mathcal{G}$ such that $B_x \subseteq 5 C_x$; It follows from $B_x \subseteq B(0, R)$ that $C_x \in \mathcal{G}^{\prime}$. Since $B_x \cap K =\emptyset$, we get $r(C_x) <\delta$. Finally, $$ \lambda(A') \leq \sum_{x \in A'} \lambda\left(B_x\right) \leq \sum_{x \in A'} \lambda\left(5 C_x\right) \leq 5^n \sum_{B \in \mathcal{G}^{\prime}, r(B)<\delta } \lambda(B)<\varepsilon . $$
This completes the proof. $\tag*{$\blacksquare$}$
Let $\lambda^*$ be the outer Lebesgue measure.
Corollary Let $A \subseteq \mathbb{R}^n$ be a set (not necessarily measurable) with $\lambda^*(A)<$ $\infty$, and let $\mathcal{F}$ be a Vitali cover of $A$. Then for all $\varepsilon>0$, there exists a finite subcollection $\mathcal{G}^{\prime} \subseteq \mathcal{F}$ of pairwise disjoint balls, such that $$ \lambda^*\left(A \setminus \bigcup \mathcal{G}^{\prime}\right)<\varepsilon. $$
Proof Let $O$ be an open subset of $\mathbb{R}^n$ such that $A \subseteq O$ and $\lambda(O)<\lambda^*(A)+$ $1<\infty$. Then the subcollection of balls $\mathcal{F}^{\prime} \subseteq \mathcal{F}$ which are contained in $O$ still form a Vitali cover of $A$. By Vitali's covering theorem, there exists a subcollection $\mathcal{G} \subseteq \mathcal{F}^{\prime}$ of pairwise disjoint balls such that $\lambda(A \setminus \bigcup \mathcal{G})=0$. For $\delta>0$, let $\mathcal{G}_\delta :=\{B \in \mathcal{G}: r(B) \geq \delta\}$. Clearly, $\mathcal{G}_\delta$ is finite. Because $\lambda (\bigcup \mathcal{G}) <\infty$, there is $\delta_0>0$ such that $$ \lambda \left (\bigcup \mathcal{G}_{\delta_0} \right ) > \lambda \left (\bigcup \mathcal{G} \right) -\varepsilon. $$
This implies $$ \lambda \left ( \bigcup \mathcal{G} \setminus \bigcup \mathcal{G}_{\delta_0} \right ) < \varepsilon. $$
Then $$ \begin{align} \lambda^* \left (A \setminus \bigcup \mathcal{G}_{\delta_0} \right) &\le \lambda^* \left ( A \cap \left ( \bigcup \mathcal{G} \setminus \bigcup \mathcal{G}_{\delta_0} \right ) \right ) + \lambda^* \left ( A \setminus \bigcup \mathcal{G} \right) \\ &\le \lambda \left ( \bigcup \mathcal{G} \setminus \bigcup \mathcal{G}_{\delta_0} \right ) + \lambda^* \left ( A \setminus \bigcup \mathcal{G} \right) \\ \\ &< \varepsilon. \end{align} $$
This completes the proof. $\tag*{$\blacksquare$}$
