Vitali Covering Theorem

Step 1: $P_n$ has at least one maximal element.

Let $\{\mathcal{T}_i\}_{i \in I}$ be any chain in $P_n$, and set $\mathcal{T} = \bigcup_{i \in I} \mathcal{T}_i$. Clearly $\mathcal{T}$ is an upper bound of $\{\mathcal{T}_i\}_{i \in I}$. We show that no two elements of $\mathcal{T}$ intersect, so that $\mathcal{T} \in P_n$.

Suppose for contradiction that there exist $A, B \in \mathcal{T}$ with $A \cap B \neq \varnothing$. If $A$ and $B$ both lie in a single $\mathcal{T}_i$, then $\mathcal{T}_i \notin P_n$, a contradiction. Otherwise, suppose $\mathcal{T}_a, \mathcal{T}_b \in \{\mathcal{T}_i\}_{i \in I}$ contain $A$ and $B$ respectively. Since $\{\mathcal{T}_i\}$ is a chain, either $\mathcal{T}_a \leq \mathcal{T}_b$ or $\mathcal{T}_b \leq \mathcal{T}_a$, so $A$ and $B$ both belong to one of them — contradicting that collection being in $P_n$. Therefore $\mathcal{T}$ has no two elements intersecting.

By Zorn's Lemma, $P_n$ has at least one maximal element, say $K_n \in P_n$, for each $n \in \mathbb{N}$.

Step 2: Construction of $G$.

For $n = 0$, let $G_0 = K_0$. For $n = 1$, let

For $n > 1$, let

Set $G = \bigcup_{n=0}^\infty G_n$. Then $G$ is a collection of disjoint closed balls in $\mathcal{F}$.

Step 3: $G$ is countable.

Fix $n \in \mathbb{N}$. Since $(X,d)$ is separable, there exists a countable dense set $S \subseteq X$. Let $\varepsilon = 2^{-n-1}R$, and let $I_n \subseteq X$ be the set of centers of balls in $G_n$.

For each $x \in I_n$ with $B(x, r_x) \in G_n$, we have $\varepsilon < r_x$. Since $S$ is dense in $X$, there exists $s_x \in S$ such that $s_x \in B(x, \varepsilon) \subseteq B(x, r_x)$.

We claim $s_x \neq s_y$ for all $x \neq y$ in $I_n$. Indeed, if $s_x = s_y$, then $B(x, r_x) \cap B(y, r_y) \neq \varnothing$, contradicting disjointness of $G_n$. So the map $x \mapsto s_x$ is injective from $I_n$ into $S$, hence $I_n$ is countable, and therefore $G_n$ is countable. Since $G = \bigcup_{n=0}^\infty G_n$ is a countable union of countable sets, $G$ is countable.

Step 4: Covering estimate.

Let $B(x,r) \in \mathcal{F}$. Then there is $n \in \mathbb{N}$ such that $B(x,r) \in \mathcal{F}_n$. We show $B(x,r) \subseteq \bigcup_{B(c,p) \in G} B(c, 5p)$.

Case 1: $B(x,r) \notin K_n$. Then $B(x,r)$ must intersect some ball in $K_n$, otherwise $\{B(x,r)\} \cup K_n$ would be a larger disjoint collection in $P_n$, contradicting maximality of $K_n$. Let $B(y, r_1) \in K_n$ intersect $B(x,r)$.

If $B(y, r_1) \in G_n$: since $R2^{-n-1} < r_1 \leq R2^{-n}$, we have $r < 2r_1$. For all $t \in B(x,r)$,

Hence $B(x,r) \subseteq B(y, 5r_1)$.

If $B(y, r_1) \notin G_n$: there exists $B(z, r_2) \in G_i$ with $B(y, r_1) \cap B(z, r_2) \neq \varnothing$ for some $i < n$. Taking $i = n-1$, we have $r < r_2$ and $r_1 < r_2$. Then

and for all $t \in B(y, r_1)$: $d(z,t) \leq 2r_2 + r_1 \leq 3r_2 < 5r_2$, so $B(y, r_1) \subseteq B(z, 5r_2)$. Moreover,

and for all $t \in B(x,r)$: $d(z,t) \leq 4r_2 + r \leq 5r_2$. Hence $B(x,r) \subseteq B(z, 5r_2)$.

Case 2: $B(x,r) \in K_n$. If $B(x,r) \in G_n$ we are done. Otherwise there exists $B(y, r_1) \in G_i$ intersecting $B(x,r)$ for some $i < n$. Taking $i = n-1$, we have $r < r_1$. For all $t \in B(x,r)$:

Therefore $B(x,r) \subseteq B(y, 5r_1)$.

In all cases, every ball in $\mathcal{F}$ is contained in a $5$-dilate of some ball in $G$, which shows