Decomposition of $\Bbb R^n$ as 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:

\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...

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.