TITLE: Can I cover a compact set by balls {B} such that {2B} has bounded overlap?
QUESTION [5 upvotes]: Suppose I have a compact set $K \subset B_1(0) \subset \mathbb{R}^n$. Can I always find a family of open balls $\{B_{r_j}(x_j)\}$ such that

$x_j \in K$ and $B_{r_j}(x_j) \subset B_1(0)$ for each $j$;
$K \subset \bigcup_j B_{r_j}(x_j)$; and
The collection $\{B_{2r_j}(x_j)\}$ has bounded overlap, that is to say that there is some number $N = N(n)$ such that each point of $\bigcup_j B_{2r_j}(x_j)$ lies in at most $N$ balls from the collection. In other words,
$$
\sum_j \mathbf{1}_{B_{2r_j}(x_j)} \leq N
$$


The motivation behind a question like this is a fairly common situation where at each point $y \in K$ I can prove an estimate like
$$
\int_{B_{r/2}(y)} f \leq C\int_{B_r(y)} g,
$$
but what I really want is an estimate of the form
$$
\int_{\{x\ :\ \mathrm{dist}  (x,K) < r/10\}} f \leq C \int_{\{x\ :\ \mathrm{dist}  (x,K) < 10r\}} g 
$$
The numbers 1/2 and 10 are not important, but to do this it is natural to try to find a collection of balls as described. In specific cases I have constructed this collection myself but now I wonder if there is a general lemma that I happen not to have heard of.

REPLY [6 votes]: Let $r$ be less than half the distance from $K$ to the complement of $B_1(0)$. Start with a ball of radius $r$ centered at each point of $K$. By https://en.wikipedia.org/wiki/Besicovitch_covering_theorem there is a subcover of $K$ by balls that is a union of $C_n$ collections $A_i$, where each $A_i$ consists of pairwise disjoint balls of radius $r$. For each collection $A_i$ of disjoint balls, the balls of twice the radius can overlap at most with multiplicity $3^n$ because a ball of radius $3r$ around the point of overlap can contain at most $3^n$ disjoint balls of radius $r$. So overall we conclude that $N(n) \le 3^n C_n$.
PS Here is a self-contained argument that avoids using the Besicovitch covering lemma: 
Given $r$, let $Z=\{z_j\}_{j=1}^m$ be a maximal subset of $K$ such that the   open balls $\{B(z,r/2): z \in Z\}$ are pairwise disjoint. Then the open balls 
$\{B(z,r): z \in Z\}$ form a cover of $K$ (If $y \in K$ was not covered, that would contradict the maximality of $Z$). Finally, the collection  of balls $S=\{B(z,2r): z \in Z\}$ has overlap multiplicity at most $5^n$: 
Indeed, if a point $w$ is in the intersection of $L$   balls from $S$, then $B(w,5r/2)$ contains  $L$ disjoint balls of radius $r/2$; comparing volumes gives $L \le 5^n$.