> [!theorem] > > Let $\bracs{B(x_i, \eps_i)}_{i \in I}$ be a family of balls in $\mathbb R^d$ and $U = \bigcup_{i \in I}B(x_i, \eps)$, then for any $c < m(U)$, there exists $\seqf{i_k} \subset I$ such that: > 1. $\seqf{B(x_{i_k}, \eps_{i_k})}$ are pairwise disjoint. > 2. $m\paren{\bigsqcup_{k = 1}^n B(x_{i_k}, \eps_{i_k})} > 3^{-d}c$. > > *Proof*. Since $U \subset \real^d$ is [[Open Set|open]], by the inner regularity of the [[Lebesgue Measure|Lebesgue measure]], there exists $K \subset U$ [[Compactness|compact]] such that $m(K) > c$. By compactness of $K$, there exists $J \subset I$ finite such that $\bigcup_{i \in J}B(x_j, \eps_j) \supset K$. > > Let $J_1 = J$. For each $n \in \nat$, if $J_{n} \ne \emptyset$, let $i_n \in J_n$ such that $\eps_{i_n} \ge \eps_j$ for all $j \in J_n$, and define > $ > J_{n+1} = \bracs{j \in J: B(x_j, \eps_j) \cap \bigcup_{k = 1}^{n}B(x_{i_k}, \eps_{i_k}) = \emptyset} > $ > Since $J$ is finite, there exists $n \in \nat$ at which point $J_{n+1} = \emptyset$. By construction, $\seqf{B(x_{i_k}, \eps_{i_k})}$ are pairwise disjoint. > > For any $j \in J$, let $1 \le k \le n$ be the smallest $k$ such that $B(x_j, \eps_j) \cap B(x_{i_k}, \eps_{i_k}) \ne \emptyset$, then $j \in J_k$, and $\eps_j \le \eps_{i_k}$. Since the balls intersect, $B(x_j, \eps_j) \subset B(x_{i_k}, 3\eps_{i_k})$. In which case, > $ > \begin{align*} > c < m(K) &\le m\paren{\bigcup_{j \in J}B(x_j, \eps_j)} \le m\paren{\bigcup_{k = 1}^n B(x_{i_k}, 3\eps_{i_k})} \\ > &\le \sum_{k = 1}^nm(B(x_{i_k}, 3\eps_{i_k})) = 3^{d}\sum_{k = 1}^n m(B(x_{i_k}, \eps_{i_k})) > \end{align*} > $