Open set in $\mathbb{R}^2$ can be decomposed into pairwise disjoint open balls and a null set

Question

Let $U\subset\mathbb{R}^2$ be an open set. Because $\mathbb{R}^2$ is $\sigma-$finite. I can assume $m(U)$ is finite.

I tried to mimic the proof of 5-time covering lemma. But I found that it is not so relevant to measure of $U$.

Answer 1

We may assume $U$ is bounded, as we can always write it as a countable union of open bounded sets and a null set. For each $x \in U$, consider an open ball $B_x$ centerd at $x$, of maximal radius $r_x$ such that $B_x \subset U$. Let $B'_x$ be the ball centered at $x$, of radius $r'_x= 2r_x$. Then the union of $$\bigcup_{x \in U} B'_x$$ covers $\overline{U}$, which is compact. Hence we can find a finite union of $B'_{x_n}$ covering $\overline{U}$. We may assume that when no $B'_{x_n}$ is included in another $B'_{x_m}$ as otherwise we can just remove this $B'_{x_n}$ from our union. Therefore, if we denote by $B''_x$ the ball centered at $x$ and of radius $r''_x=r'_x/4=r_x/2$, then the $B''_{x_n}$ are pairwise disjoint and included in $U$. But the measure of $$\bigcup_{n} B'_{x_n}$$ is larger than the measure of $U$ therefore the measure of $$\bigcup_{n} B''_{x_n}$$ is larger than a fourth of the measure of $U$. Now consider the open set $U_1$ obtained from $U$ by removing the closure of $\bigcup_{n} B''_{x_n}$, that is the union of a null set and disjoint balls. It has measure at most $3/4$ the measure of $U$. Do the same thing by replacing $U$ with $U_1$, and iterate.