[proofplan] We use an iterative greedy extraction based on [Vitali's Covering Lemma](/theorems/15). Fix a contraction factor $\theta \in (1 - 5^{-n}, 1)$ and first assume $\mathcal{L}^n(U) < \infty$. At each stage, apply [Vitali's Covering Lemma](/theorems/15) to extract a finite disjoint subfamily of closed balls of diameter less than $\delta$ whose union captures a definite fraction of the remaining uncovered measure. The $5^n$-volume-scaling bound ensures that the uncovered measure contracts by at least the factor $\theta$ per stage, yielding geometric decay to zero. For the case $\mathcal{L}^n(U) = \infty$, decompose $U$ into bounded annuli of finite measure and apply the finite-measure argument to each. [/proofplan] [step:Fix the contraction factor and reduce to the finite-measure case] Fix $\theta \in (1 - 5^{-n}, 1)$. We first prove the theorem under the assumption $\mathcal{L}^n(U) < \infty$; the case $\mathcal{L}^n(U) = \infty$ is handled in the final step. [guided] The proof proceeds by iterative greedy extraction, and the constant $\theta$ governs the rate at which the uncovered measure decreases at each stage. We need to choose $\theta$ carefully so that each application of [Vitali's Covering Lemma](/theorems/15) guarantees a strict contraction of the leftover measure. Fix $\theta \in (1 - 5^{-n}, 1)$. Why this specific interval? [Vitali's Covering Lemma](/theorems/15) produces a disjoint subfamily whose $5$-fold dilations cover the target set. The volume-scaling property of [Lebesgue measure](/page/Lebesgue%20Integral) gives $\mathcal{L}^n(\widehat{B}) = 5^n \mathcal{L}^n(B)$ for each ball $B$ and its dilation $\widehat{B}$. This means the disjoint balls themselves capture at least a $5^{-n}$-fraction of the total measure, so the uncovered portion is at most $(1 - 5^{-n})$ of the original. Any $\theta$ strictly greater than $1 - 5^{-n}$ therefore satisfies $(1 - 5^{-n}) < \theta$, ensuring that the uncovered measure after each extraction stage is strictly less than $\theta$ times the measure before extraction — a genuine geometric contraction. The range $\theta \in (1 - 5^{-n}, 1)$ is thus the natural choice: the lower bound $1 - 5^{-n}$ is the worst-case fraction of uncovered measure guaranteed by the Vitali extraction, and any $\theta$ above this threshold yields a strict contraction factor at each stage. The upper bound $\theta < 1$ is needed so that $\theta^k \to 0$. We assume throughout the next three steps that $\mathcal{L}^n(U) < \infty$. This assumption is essential: the geometric bound $\theta^k \mathcal{L}^n(U) \to 0$ requires $\mathcal{L}^n(U)$ to be finite. Additionally, continuity of measure from above (used to pass from countably many to finitely many balls) likewise requires finite measure. The case $\mathcal{L}^n(U) = \infty$ is handled in the final step by decomposing $U$ into countably many bounded open annuli $U_m = \{ x \in U \mid m < |x| < m + 1 \}$, each of finite measure, and applying the finite-measure argument to each piece separately. The resulting disjoint families are then combined into a single countable disjoint family covering $U$ up to measure zero. [/guided] [/step] [step:Apply Vitali's Covering Lemma to extract the first disjoint subfamily] Define the collection of admissible balls: \begin{align*} \mathcal{F}_1 = \{ B \subset U \mid \text{$B$ is a closed ball with } \operatorname{diam} B < \delta \}. \end{align*} Since $U$ is [open](/page/Open%20Set), every point of $U$ is the centre of a closed ball of arbitrarily small radius contained in $U$, so $\mathcal{F}_1$ is a Vitali cover of $U$. By [Vitali's Covering Lemma](/theorems/15), there exists a countable pairwise disjoint subfamily $\mathcal{G}_1 \subset \mathcal{F}_1$ such that \begin{align*} U \subset \bigcup_{B \in \mathcal{G}_1} \widehat{B}, \end{align*} where $\widehat{B}$ denotes the closed ball concentric with $B$ having radius $5 \operatorname{rad}(B)$. By the scaling property of [Lebesgue measure](/page/Lebesgue%20Integral), $\mathcal{L}^n(\widehat{B}) = 5^n \mathcal{L}^n(B)$. Using the covering property and disjointness of $\mathcal{G}_1$: \begin{align*} \mathcal{L}^n(U) &\le \sum_{B \in \mathcal{G}_1} \mathcal{L}^n(\widehat{B}) = 5^n \sum_{B \in \mathcal{G}_1} \mathcal{L}^n(B) = 5^n \, \mathcal{L}^n\!\Bigl(\bigcup_{B \in \mathcal{G}_1} B\Bigr). \end{align*} Rearranging gives $\mathcal{L}^n(\bigcup_{B \in \mathcal{G}_1} B) \ge 5^{-n} \mathcal{L}^n(U)$, hence: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{B \in \mathcal{G}_1} B\Bigr) \le (1 - 5^{-n}) \mathcal{L}^n(U) < \theta \, \mathcal{L}^n(U). \end{align*} Since $\mathcal{G}_1$ is countable, $\mathcal{L}^n(U) < \infty$, and the series $\sum_{B \in \mathcal{G}_1} \mathcal{L}^n(B)$ converges, continuity of measure from above yields finitely many balls $B_1, \dotsc, B_{M_1} \in \mathcal{G}_1$ with \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{i=1}^{M_1} B_i\Bigr) \le \theta \, \mathcal{L}^n(U). \end{align*} [guided] We seek a disjoint family of small closed balls that covers nearly all of $U$. The idea is a greedy extraction: pick as many disjoint small balls as possible, then measure what remains. Define $\mathcal{F}_1 = \{ B \subset U \mid \text{$B$ is a closed ball with } \operatorname{diam} B < \delta \}$. Why is this a Vitali cover of $U$? Because $U$ is [open](/page/Open%20Set): for any $x \in U$ there exists $r > 0$ with $\overline{B}(x, r) \subset U$, and we can choose $r$ arbitrarily small (in particular $2r < \delta$). So every point of $U$ lies in arbitrarily small members of $\mathcal{F}_1$. [Vitali's Covering Lemma](/theorems/15) produces a countable pairwise disjoint subfamily $\mathcal{G}_1 \subset \mathcal{F}_1$ whose $5$-fold dilations cover $U$: $U \subset \bigcup_{B \in \mathcal{G}_1} \widehat{B}$. How much of $U$ is captured? Using the scaling property $\mathcal{L}^n(\widehat{B}) = 5^n \mathcal{L}^n(B)$ and the covering property: \begin{align*} \mathcal{L}^n(U) \le \sum_{B \in \mathcal{G}_1} \mathcal{L}^n(\widehat{B}) = 5^n \sum_{B \in \mathcal{G}_1} \mathcal{L}^n(B) = 5^n \, \mathcal{L}^n\!\Bigl(\bigcup_{B \in \mathcal{G}_1} B\Bigr), \end{align*} where the last equality holds because the balls are pairwise disjoint. Rearranging: $\mathcal{L}^n(\bigcup \mathcal{G}_1) \ge 5^{-n} \mathcal{L}^n(U)$, so: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{B \in \mathcal{G}_1} B\Bigr) \le (1 - 5^{-n}) \mathcal{L}^n(U) < \theta \, \mathcal{L}^n(U). \end{align*} Why pass to a finite sub-collection? For the iteration step, we need the remainder to be open (so that [Vitali's Covering Lemma](/theorems/15) applies again). Removing finitely many closed sets from an open set yields an open set. Since $\sum_{B \in \mathcal{G}_1} \mathcal{L}^n(B) \le \mathcal{L}^n(U) < \infty$, the tail of the series tends to zero. By continuity of measure from above, finitely many balls $B_1, \dotsc, B_{M_1} \in \mathcal{G}_1$ satisfy: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{i=1}^{M_1} B_i\Bigr) \le \theta \, \mathcal{L}^n(U). \end{align*} [/guided] [/step] [step:Iterate the Vitali extraction on the remainder] Define the open set $U_2 = U \setminus \bigcup_{i=1}^{M_1} B_i$. The collection $\mathcal{F}_2 = \{ B \subset U_2 \mid \text{$B$ closed ball, } \operatorname{diam} B < \delta \}$ is a Vitali cover of $U_2$. Applying [Vitali's Covering Lemma](/theorems/15) and passing to a finite sub-collection produces finitely many pairwise disjoint balls $B_{M_1+1}, \dotsc, B_{M_2} \subset U_2$ with \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{i=1}^{M_2} B_i\Bigr) \le \theta \, \mathcal{L}^n(U_2) \le \theta^2 \, \mathcal{L}^n(U). \end{align*} [guided] We repeat the extraction on the uncovered portion. Define $U_2 = U \setminus \bigcup_{i=1}^{M_1} B_i$. This is open because it is the difference of the open set $U$ and the finite union of closed balls $B_1, \dotsc, B_{M_1}$. Form the Vitali cover $\mathcal{F}_2 = \{ B \subset U_2 \mid \text{$B$ closed ball, } \operatorname{diam} B < \delta \}$ and apply [Vitali's Covering Lemma](/theorems/15) to $U_2$. Passing to a finite sub-collection (as in the previous step) gives pairwise disjoint balls $B_{M_1+1}, \dotsc, B_{M_2} \subset U_2$ with: \begin{align*} \mathcal{L}^n\!\Bigl(U_2 \setminus \bigcup_{i=M_1+1}^{M_2} B_i\Bigr) \le \theta \, \mathcal{L}^n(U_2). \end{align*} Since $U \setminus \bigcup_{i=1}^{M_2} B_i \subset U_2 \setminus \bigcup_{i=M_1+1}^{M_2} B_i$ and $\mathcal{L}^n(U_2) \le \theta \, \mathcal{L}^n(U)$ from the previous step: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{i=1}^{M_2} B_i\Bigr) \le \theta \, \mathcal{L}^n(U_2) \le \theta^2 \, \mathcal{L}^n(U). \end{align*} Each iteration multiplies the uncovered measure by at most $\theta < 1$, producing geometric decay. [/guided] [/step] [step:Conclude by geometric convergence of the leftover measure] Continuing inductively, after $k$ stages we obtain indices $M_1 < M_2 < \dotsb < M_k$ and pairwise disjoint closed balls $B_1, \dotsc, B_{M_k} \subset U$ with $\operatorname{diam} B_i < \delta$ and \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{i=1}^{M_k} B_i\Bigr) \le \theta^k \, \mathcal{L}^n(U). \end{align*} Since $\theta^k \to 0$, define $G = \{ B_i \}_{i=1}^{\infty}$. The sets $R_k := U \setminus \bigcup_{i=1}^{M_k} B_i$ form a decreasing sequence with $\bigcap_{k=1}^{\infty} R_k = U \setminus \bigcup_{B \in G} B$. Since $\mathcal{L}^n(R_1) \le \theta \, \mathcal{L}^n(U) < \infty$, continuity of measure from above gives: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{B \in G} B\Bigr) = \lim_{k \to \infty} \mathcal{L}^n(R_k) = 0. \end{align*} Every $B \in G$ satisfies $\operatorname{diam} B < \delta$ by construction, and the balls are pairwise disjoint because at each stage the new balls lie in the remainder, disjoint from all previously selected balls. [guided] After $k$ stages we have pairwise disjoint closed balls $B_1, \dotsc, B_{M_k} \subset U$ with $\operatorname{diam} B_i < \delta$ and the inductive bound: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{i=1}^{M_k} B_i\Bigr) \le \theta^k \, \mathcal{L}^n(U). \end{align*} Since $0 < \theta < 1$, this is geometric decay: $\theta^k \to 0$. Define $G = \{ B_i \}_{i=1}^{\infty}$ and $R_k = U \setminus \bigcup_{i=1}^{M_k} B_i$. The sets $R_k$ are nested: $R_1 \supset R_2 \supset \dotsb$, and $\bigcap_{k} R_k = U \setminus \bigcup_{B \in G} B$. To pass to the limit, we need continuity of measure from above, which requires $\mathcal{L}^n(R_1) < \infty$. We have $\mathcal{L}^n(R_1) \le \theta \, \mathcal{L}^n(U) < \infty$ (using the finite-measure assumption). Therefore: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{B \in G} B\Bigr) = \lim_{k \to \infty} \mathcal{L}^n(R_k) \le \lim_{k \to \infty} \theta^k \, \mathcal{L}^n(U) = 0. \end{align*} Why are all balls in $G$ pairwise disjoint? At stage $j$, the new balls $B_{M_{j-1}+1}, \dotsc, B_{M_j}$ are selected inside $U_j = U \setminus \bigcup_{i=1}^{M_{j-1}} B_i$, which is disjoint from all previously chosen balls. Within each stage, [Vitali's Covering Lemma](/theorems/15) produces a disjoint sub-collection. Hence all balls across all stages are pairwise disjoint. This completes the proof for $\mathcal{L}^n(U) < \infty$. [/guided] [/step] [step:Handle the case $\mathcal{L}^n(U) = \infty$ by decomposing into bounded annuli] If $\mathcal{L}^n(U) = \infty$, define the bounded open sets: \begin{align*} U_m = \{ x \in U \mid m < |x| < m + 1 \}, \quad m \in \mathbb{N} \cup \{0\}. \end{align*} Each $U_m$ is open (intersection of $U$ with the open annulus $\{ x \in \mathbb{R}^n \mid m < |x| < m + 1 \}$) and has $\mathcal{L}^n(U_m) < \infty$ since $U_m \subset \overline{B}(0, m+1)$. The annuli are pairwise disjoint, and $\mathcal{L}^n(U \setminus \bigcup_{m=0}^{\infty} U_m) = 0$ because $U \setminus \bigcup_{m=0}^{\infty} U_m \subset \bigcup_{m=0}^{\infty} \{ x \mid |x| = m \}$, a countable union of $(n-1)$-dimensional spheres, each with $\mathcal{L}^n$-measure zero. Apply the finite-measure argument to each $U_m$ with the same $\delta$ to obtain a disjoint family $G_m$ of closed balls in $U_m$ with $\operatorname{diam} B < \delta$ and $\mathcal{L}^n(U_m \setminus \bigcup_{B \in G_m} B) = 0$. Since the annuli are pairwise disjoint, so are the families $G_m$. Set $G = \bigcup_{m=0}^{\infty} G_m$. Then $G$ is countable, pairwise disjoint, and: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{B \in G} B\Bigr) &\le \sum_{m=0}^{\infty} \mathcal{L}^n\!\Bigl(U_m \setminus \bigcup_{B \in G_m} B\Bigr) + \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{m=0}^{\infty} U_m\Bigr) = 0. \end{align*} This completes the proof. [guided] When $\mathcal{L}^n(U) = \infty$, the finite-measure argument breaks down: continuity of measure from above requires a set of finite measure, and the bound $\theta^k \mathcal{L}^n(U)$ does not tend to zero. The remedy is to decompose $U$ into bounded pieces. Define $U_m = \{ x \in U \mid m < |x| < m + 1 \}$ for $m \in \mathbb{N} \cup \{0\}$. Each $U_m$ is open (the intersection of two open sets) and bounded (contained in $\overline{B}(0, m+1)$), so $\mathcal{L}^n(U_m) < \infty$. The annuli are pairwise disjoint by construction. Does their union cover $U$ up to measure zero? The uncovered set is $U \setminus \bigcup_m U_m \subset \bigcup_{m=0}^{\infty} \{ x \mid |x| = m \}$. Each sphere $\{ x \in \mathbb{R}^n \mid |x| = m \}$ has $\mathcal{L}^n$-measure zero (it is an $(n-1)$-dimensional manifold in $\mathbb{R}^n$), and a countable union of null sets is null. Apply the finite-measure result to each $U_m$ with diameter bound $\delta$, obtaining a countable pairwise disjoint family $G_m$ of closed balls in $U_m$ with $\operatorname{diam} B < \delta$ for all $B \in G_m$ and $\mathcal{L}^n(U_m \setminus \bigcup_{B \in G_m} B) = 0$. Since the annuli $U_m$ are disjoint, balls from different families $G_m$ cannot overlap. Set $G = \bigcup_{m=0}^{\infty} G_m$. This is a countable union of countable sets, hence countable. By sub-additivity of $\mathcal{L}^n$: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{B \in G} B\Bigr) &\le \mathcal{L}^n\!\Bigl(\bigcup_{m=0}^{\infty} \bigl(U_m \setminus \bigcup_{B \in G_m} B\bigr)\Bigr) + \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{m=0}^{\infty} U_m\Bigr) \\ &\le \sum_{m=0}^{\infty} 0 + 0 = 0. \end{align*} The family $G$ satisfies all requirements, completing the proof. [/guided] [/step]