$R^n$ can be decomposed into a union of countable disjoint closed balls and a null set

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?

Solutions Collecting From Web of "$R^n$ can be decomposed into a union of countable disjoint closed balls and a null set"

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:

\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}

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…

it’s overkill but you can use Vitali-Lebesgue covering theorem (they cover subsets of finite measure, so e.g. decompose $\mathbb{R}^n$ to cubes by hyperplanes and then cover the interior of each of those cubes)

Divide $\mathbf R^n$ into an integer mesh $\mathcal M_0$, that is divide the space into cubes with integer vertices. Now fill the cubes with closed balls of diameter $\frac12$, call this mesh of balls $\mathcal C_0$. Now we can get from $\mathcal M_0$ an infinite sequence of meshes $\mathcal M_k = 2^{-2k} \mathcal M_0$ by bisecting the sides in $2^{2k}$ parts. So for each cube in $\mathcal M_k$ we get $2^{2n}$ cubes in $\mathcal M_{k + 1}$ and each cube in $\mathcal M_k$ has side length $2^{-k-1}$, so they have diameter $\sqrt{n} 2^{-k-1}$.

From these meshes we get meshes of closed balls $\mathcal C_k$ just like we have obtained $\mathcal C_0$.

Now define $\Omega_k = \{Q \in \mathcal C_n : n = 1,\ldots, k – 1\}$, these are all the previous balls. So the “cover” of the space is now

$$\mathcal F = \bigcup_k \{Q \in \mathcal C_k : Q \cap \Omega_k = \emptyset\}.$$

Note that $\mathcal F^c = \bigcap_k \{Q \in \mathcal C_k : Q \cap \Omega_k \neq \emptyset\}$.

Try #2. The idea is to fill up the mesh with balls where there is horizontally and vertically the space of one diameters between them, then we split up the mesh-sides again in $4$ pieces, fill those up again, then make sure you only select the disjoint ones.

Here is an idea without all of the details.

Fix some sphere packing with positive density $d$ so that all spheres have radius $1$.

A positive proportion $c$ of the complement of any collection of spheres of radius at least $1$ is of distance at least $\epsilon$ away from the spheres.

Rescale the sphere packing to be spheres of radius $\epsilon$ to cover at least $dc$ of the complement.

Iterate, so at step $n$ you use spheres of radius $\epsilon^n$ to cover at least $dc$ times what you haven’t covered before.

This is an idea which I cannot see if it ends up working or not. Maybe someone can?

Consider the set $\mathcal S$ of all families of closed balls in $\mathbb R^n$ which are pairwise disjoint. Ordered by inclusion, this poset satisfies the hypothesis in Zorn’s lemma, so there exist maximal elements $S\in\mathcal S$.
One could hope for $S$ to be a candidate…

Now: is $\mathbb R^n\setminus\bigcup_{B\in S}B$ a null set? I don’t see how to prove this…

Notice, though, that every decomposition like those gylns wants does give a maximal element in $\mathcal S$.

Later: This idea does not work: Mike has explicitely constructed a counterexample in the comments below.

Is there a way to fix this? I mean: can one select a smaller set $\mathcal S$ such that the union of its maximal elements have null complement?