A Lebesgue measure condition Suppose that $\{B(x_j,r_j)\}_{j=1}^{n}$ is a finite collection of balls in $\mathbb{R}^d$. Show that there is a subcollection $\{B(x_{j_k},r_{j_k})\}_{k=1}^{l}$ of pairwise disjoint balls such that $$m\left(\bigcup_{k=1}^{l}B(x_{j_k},r_{j_k})\right) \geq 3^{-d} m\left(\bigcup_{j=1}^{n}B(x_{j},r_{j})\right)$$
where $m$ denotes Lebesgue measure.
I want to constrain this problem by attempting to not trivialize it by applying a covering lemma.
Attempted proof: Suppose that $\{B(x_j,r_j)\}_{j=1}^{n}$ is a finite collection of balls in $\mathbb{R}^d$ such that $$\{B(x_j,r_j)\}_{j=1}^{n} \rightarrow \{B(x,r)\} \ \ \text{as} \ \ n\rightarrow \infty$$ Then there exists a subsequence of balls $\{B(x_{j_k},r_{j_k})\}_{j=1}^{n}$ such that $$\{B(x_{j_k},r_{j_k})\}_{j=1}^{n}\rightarrow \{B(x,r)\} \ \ \text{as} \ \ n,k\rightarrow \infty $$ Let $\delta = 3^{d}$, then choose an $x_{j_k},r_{j_k}$ such that $$|x_{j_k} - r_{j_k}| \leq \frac{(b - a)}{\delta}$$ where the subsequence $\{B(x_{j_k},r_{j_k})\}$ lies in the interval $[a,b]$. Then let $$A = \bigcup_{1}^{l}B(x_{j_k},r_{j_k})$$ Since $\{B(x_{j_k},r_{j_k})\}$ is convergent and bounded, taking the Lebesgue measure it follows that $$m\left(\bigcup_{k=1}^{l}B(x_{j_k},r_{j_k})\right) \geq 3^{-d} m\left(\bigcup_{k=1}^{l}B(x_{j},r_{j})\right)$$
I am not sure if we can have balls just on some interval $[a,b]$ but I figured they could be on some place in $\mathbb{R}^d$. I am not sure if I can just conclude with that if I know it is convergent and bounded. Any suggestions is greatly appreciated.
 A: The $3$ does have some geometric meaning here. One can solve this exercise in the context of usual $3$-dimensional spheres or even $2$-dimension circles, since one only needs elementary geometric thoughts and no methods from measure theory.
It goes like this: Denote the balls by $B_1, \dotsc, B_n$.
In the first step, pick a ball $B_{j_1}$ with maximal size.
In the $k$-th step, pick a ball $B_{j_k}$ with maximal size among these, which are disjoint to $B_{j_1} \cup \dotsc \cup B_{j_{k-1}}$.
The process stops at some point, not later than at the $n$-th step. Nevertheless one can show that the desired inequality holds, when the proess stops.
The key is the following observation: Say the process stopped after $s$ steps, i.e. we have obtained the disjoint collection $B_{j_1}, \dotsc, B_{j_s}$.
By construction, any ball $B_i$ meets some $B_{j_k}$ (since the process has stopped) and we have (if we assume $B_{j_k}$ maximal among these, who meet $B_i$) $r_{j_k} \geq r_i$ (if not, we could have picked $B_i$ in the $k$-th step instead of $B_{j_k}$). Hence we get the punchline: $B_i$ is contained in $B(x_{j_k},3r_{j_k})$, note the three.
Now you can see, where the $3^d$ comes into play, because we have $m(B(x_{j_k},3r_{j_k}))=3^d \cdot m(B(x_{j_k},r_{j_k}))$.
