measure of the union taken from a finite subcollection

Question

Let $c>0$ be a constant and $\mathcal C$ be a collection of balls in $\mathbb R^d$ such that $m(\cup_{B\in \mathcal C}B)>c$. I wonder if we can find finitely many $B_i$'s from $\mathcal C$ such that $m(\cup_{i}B_i)>c$. $m$ is the Lebesgue measure.

Answer 1

Yes, if your balls are open balls. Since $\mathbb R^{d}$ is second countable we can express any union of open balls as a countable union, say $\cup_{n=1}^{\infty } B_n$. Since $m(\cup_{n=1}^{\infty } B_n)=lim_{n\to \infty }m(\cup_{k=1}^{n } B_k)$ we can find $n$ such that $m(\cup_{k=1}^{n } B_k)>c$