Lemma 24.7.1 (Vitali Covering Lemma).label Let $\bracsn{B_{\real^d}(x_i, \eps_i)}_{i \in I}$ be open balls in $\real^{d}$, $U = \bigcup_{i \in I}B_{\real^d}(x_{i}, \eps)$, and $m$ be the Lebesgue measure on $\real^{d}$, then for any $c \in [0, m(U))$, there exists $J \subset I$ finite such that:
- (1)
$\bracsn{B_{\real^d}(x_j, \eps_j)}_{j \in J}$ is pairwise disjoint.
- (2)
$m\paren{\bigsqcup_{j \in J}B_{\real^d}(x_j, \eps_j)}> 3^{-d}c$.
Proof, [Lemma 3.15, Fol99]. By inner regularity of the Lebesgue measure, there exists $K \subset U$ compact such that $m(K) > c$. In which case, there exists $J_{1} \subset I$ finite such that $K \subset \bigcup_{j \in J_1}B_{\real^d}(x_{j}, \eps_{j})$.
For each $n \in \natp$, if $J_{n} \ne \emptyset$, let $j_{n} \in J_{n}$ such that $\eps_{j_n}= \max_{j \in J_n}\eps_{j}$, and
As $J$ is finite, there exists $n \in \natp$ such that $J_{n+1}= \emptyset$. The above construction then yields a pairwise disjoint family $\bracsn{B_{\real^d}(x_{i_k}, \eps_{i_k})}_{1}^{n}$.
For each $j \in J$, let $1 \le k \le n$ be the smallest $k$ such that $B_{\real^d}(x_{j}, \eps_{j}) \cap B_{\real^d}(x_{i_k}, \eps_{i_k}) \ne \emptyset$, then $j \in J_{k}$, and $\eps_{j} \le \eps_{i_k}$. Thus $B_{\real^d}(x_{j}, \eps_{j}) \subset B_{\real^d}(x_{i_k}, 3\eps_{i_k})$. Therefore
$\square$
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