Covering Lemma (Folland Lemma 3.15) Lemma 3.15 from Folland's Real Analysis: Let $\mathcal{C}$ be a collection of open balls in $\mathbb{R}^n$, and let $U = \cup_{B \in \mathcal{C}}B$. If $c < m(U)$, there exist disjoint $B_1,\ldots,B_k \in \mathcal{C}$ such that $\sum_1^k m(B_j) > 3^{-n}c.$
I've gone through the proof and looked through Vitali's covering lemma, but I am heaving a hard time understanding where the factor 3 comes from. Why does the radius of $B_J^*$ have to be three times that of $B_j$ (i.e., why is there a $3^{-n}$ in the inequality)?
 A: Let me quote the proof in Folland to refer to it and so that people have it.

If $c < m(U)$, by Theorem $2.40$ there is a compact $K\subset U$ with $m(K) > 
c$, and finitely many of the balls in $C$ - say, $A_1$, . . . , $A_m$ - cover $K$. Let $B_1$ be the largest of the $A_j$'s (that is, choose $B_1$ to have maximal radius), let $B_2$ be the largest of the $A_j$'s that are disjoint from $B_1$ , $B_3$ the largest of the $A_j$'s that are disjoint from $B_1$ and $B_2$, and so on until the list of $A_j$'s is exhausted. According to this construction, if $A_i$ is not one of the $B_j$'s, there is a $j$ such that $A_i\cap B_j \neq\emptyset$, and if $j$ is the smallest integer with this property, the radius of $A_i$ is at most that of $B_j$. Hence $A_i\subset B_j^*$, 
  where $B_j^*$ is the ball concentric with $B_j$ whose radius is three times that of $B_j$. But then $K\subset\bigcup_{i=1}^{k} B_j^{*}$, so
$$c < m(K) < \sum_{1}^{k}m(B_j) = 3^n \sum_{1}^{k} m(B_j). $$

We have that $B_j$ and $A_i$ intersect. The worst case scenario is that they are almost exterior tangent to each other. The radius of $A_i$ is $\leq$ the radius of $B_j$. The worst case scenario is if the radius of $A_i$ is almost that of $B_j$. How much you  would have to expand $B_j$ to cover $A_i$ (i.e. to be interior tangent)? Radius of $B_j$ plus diameter of $A_i$, which is at worst $2$ times the radius of $B_j$. Then the $1+2$ is the $3$ that they are saying.
