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


1. $x\_j \in K$ and $B\_{r\_j}(x\_j) \subset B\_1(0)$ for each $j$;
2. $K \subset \bigcup\_j B\_{r\_j}(x\_j)$; and
3. 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,
$$
\sum\_j \mathbf{1}\_{B\_{2r\_j}(x\_j)} \leq N
$$




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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.



Can I cover a compact set by balls {B} such that {2B} has bounded overlap?
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 $L \le 5^n$. 


