TITLE: Decomposition of $\Bbb R^n$ as union of countable disjoint closed balls and a null set
QUESTION [14 upvotes]: This is a problem in Frank Jones's Lebesgue integration on Euclidean space (p.57), 

$$\mathbb{R}^n = N \cup \bigcup_{k=1}^\infty \overline{B}_k$$
where $\lambda(N)=0$, and the closed balls are disjoint.

could any one give some hints?

REPLY [11 votes]: Fix some dimension $d \geq 1$. It suffices to prove that the subspace $X = [0,1)^d \subset \mathbf{R}^d$ is the union of a disjoint family of closed balls and a null set (with respect to Lebesgue measure $\lambda$ on $X$). Let's call any subset of $X$ with the form $\prod_{i=1}^d[x_i,x_i + s)$ where $s>0$ a box. A disjoint union of finitely many boxes (resp. closed balls) will be called a square (resp. round) set. Using a grid construction, it is not difficult to see that every open subset of $X$ is the disjoint union of countably many boxes. Thus follows:
Lemma 1: If $U$ is an open subset of $X$ and $\epsilon > 0$ there is a square set $S \subset U$ with $\lambda(U) - \lambda(S) < \epsilon$.
Let $a \in (0,1)$ be a constant (depending on $d$)  such that every box contains a closed ball of $a$ times the measure. Choosing a ball for every box in a square set gives:
Lemma 2: For every square set $S \subset X$ there is a round set $R \subset S$ such that $\lambda(R) = a \lambda(S)$.
Choose $\epsilon > 0$ so that $(1-a) + \epsilon < 1$. We can now construct the desired family of disjoint balls. We will construct recursively for each $n=1,2,\ldots$, a round set $R_n$ such that $\lambda(X - R_n) \leq (1-a)^n + (1-a+\epsilon)^n$.
Since $X$ is square, there is a round set $R_1$ (just a ball actually) with $\lambda(R_1) = a \lambda(X) = a$ whence $\lambda(X - R_1) = 1-a \leq (1-a) + (1-a+\epsilon)$.
Now suppose that $R_n$ given for $n \geq 1$ and that $\lambda(X - R_n) \leq (1-a)^n + (1-a+\epsilon)^n$ holds. Since $X-R_n$ is open, lemma 1 gives us a square set $S$ disjoint from $R_n$ with $\lambda(R_n) - \lambda(S) \leq \epsilon^{n+1}$. Then, by lemma 2, there is a round set $R \subset S$ with $\lambda(R) = a \lambda(S)$. Putting $R_{n+1} = R_n \sqcup R$ we see that:
\begin{align*}
\lambda(X-R_{n + 1}) &= \lambda(X - R_n) - \lambda(R)\\
&= [ \lambda(X-R_n) - \lambda(S)] + (1-a) \lambda(S)\\
&\leq [ \lambda(X-R_n) - \lambda(S)] + (1-a) \lambda(X-R_n)\\
&\leq \epsilon^{n+1} + (1-a)[(1-a)^n + (1-a+\epsilon)^n]\\
&= [\epsilon^{n+1} - \epsilon (1-a+\epsilon)^n] + (1-a)^{n+1} + (1-a+\epsilon)^{n+1}\\
&\leq (1-a)^{n+1} + (1-a+ \epsilon)^{n+1}
\end{align*}
and the bound is established. It is clear from our construction that $\bigcup_{n=1}^\infty R_n$ is a disjoint union of closed balls and the bound shows that its complement in $X$ is a null set so we are done.
Hopefully this doesn't have too many mistakes and is somewhat readable. I wasn't expecting the analysis portion of the problem to be as finicky as it turned out to be when I started typing this up...