[proofplan] The strategy is an iterative greedy extraction using Vitali's covering lemma. We first treat the case $\mathcal{L}^n(U) < \infty$. At each stage, we apply Vitali's covering lemma to select a finite disjoint subfamily of small closed balls whose union captures a definite fraction of the remaining uncovered measure. The $5^n$-volume-scaling property of the Vitali lemma guarantees that the uncovered set shrinks by a factor $\theta \in (1 - 5^{-n}, 1)$ at each stage, so the leftover measure decays geometrically to zero. For the case $\mathcal{L}^n(U) = \infty$, we decompose $U$ into bounded annuli of finite measure and apply the finite-measure argument to each. [/proofplan] [step:Fix the contraction factor $\theta$ and reduce to the finite-measure case] Fix $\theta \in (1 - 5^{-n}, 1)$. We first prove the theorem under the additional assumption $\mathcal{L}^n(U) < \infty$. The case $\mathcal{L}^n(U) = \infty$ is handled in the final step by decomposing $U$ into bounded pieces of finite measure. [guided] Fix $\theta \in (1 - 5^{-n}, 1)$. We prove the theorem under the assumption $\mathcal{L}^n(U) < \infty$; the general case is handled in the final step by decomposing $U$ into bounded pieces. Why this specific range for $\theta$? The Vitali covering lemma guarantees that a disjoint subfamily of balls captures at least a $5^{-n}$-fraction of the total measure. After one extraction, the uncovered set has measure at most \begin{align*} \mathcal{L}^n(U_{\text{remaining}}) \le (1 - 5^{-n}) \, \mathcal{L}^n(U). \end{align*} Since $\theta > 1 - 5^{-n}$, the contraction factor $\theta$ strictly exceeds the worst-case leftover fraction. This means after $k$ extraction stages, the remaining measure satisfies $\mathcal{L}^n(U_k) < \theta^k \, \mathcal{L}^n(U) \to 0$ as $k \to \infty$ — geometric decay is guaranteed. The finite-measure assumption $\mathcal{L}^n(U) < \infty$ is needed to truncate each [countable](/page/1215) disjoint family to a finite sub-family while keeping the measure loss below $\theta$. Without finite measure, the tail of the countable family could carry infinite measure, and no finite truncation would suffice. The infinite-measure case is recovered in the final step by writing $U = \bigcup_{j=1}^{\infty} U_j$ where each $U_j$ is bounded (hence finite-measure), applying the finite-measure result to each piece, and taking the union of the resulting countable families. [/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 an [open set](/page/1144), every point $x \in U$ is the centre of a closed ball of arbitrarily small positive radius contained in $U$, so $\mathcal{F}_1$ is a Vitali cover of $U$. By [Vitali's covering lemma](/page/1298), there exists a countable 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 $5 \operatorname{rad}(B)$. Since the balls in $\mathcal{G}_1$ are pairwise disjoint and $\mathcal{L}^n(\widehat{B}) = 5^n \mathcal{L}^n(B)$ by the scaling property of [Lebesgue measure](/page/1251), we estimate: \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*} The last equality uses disjointness. Rearranging: \begin{align*} \mathcal{L}^n\!\Bigl(\bigcup_{B \in \mathcal{G}_1} B\Bigr) \ge 5^{-n} \mathcal{L}^n(U). \end{align*} Therefore: \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 and $\mathcal{L}^n(U) < \infty$, the continuity of measure from above implies that there exist finitely many balls $B_1, \dotsc, B_{M_1} \in \mathcal{G}_1$ satisfying \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 need a family of balls that covers $U$ up to measure zero. The idea is to use a greedy selection: pick as many disjoint small balls as possible, then measure what is left over. 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*} Why is $\mathcal{F}_1$ a Vitali cover of $U$? Because $U$ is open: for every $x \in U$, there exists $r > 0$ with $\overline{B}(x, r) \subset U$, and we can choose $r$ as small as we like (in particular, $2r < \delta$). So every point of $U$ is contained in arbitrarily small members of $\mathcal{F}_1$. By Vitali's covering lemma (applied to the Vitali cover $\mathcal{F}_1$ of $U$), there exists a countable pairwise disjoint subfamily $\mathcal{G}_1 \subset \mathcal{F}_1$ such that $U \subset \bigcup_{B \in \mathcal{G}_1} \widehat{B}$, where $\widehat{B}$ is the closed ball concentric with $B$ of radius $5 \operatorname{rad}(B)$. How much of $U$ do the selected balls cover? By the scaling property of Lebesgue measure in $\mathbb{R}^n$, $\mathcal{L}^n(\widehat{B}) = 5^n \mathcal{L}^n(B)$. Using the covering property and disjointness: \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 in $\mathcal{G}_1$ are pairwise disjoint. Rearranging gives $\mathcal{L}^n(\bigcup_{B \in \mathcal{G}_1} B) \ge 5^{-n} \mathcal{L}^n(U)$, so the uncovered portion satisfies: \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*} using the choice $\theta > 1 - 5^{-n}$. Why pass to a finite sub-collection? The countable disjoint family $\mathcal{G}_1$ covers all but a $\theta$-fraction, but for the inductive step we need the remainder $U \setminus \bigcup B_i$ to be open (so that the Vitali lemma can be applied again). Removing finitely many closed balls from the open set $U$ yields an open set. Since $\mathcal{L}^n(U) < \infty$ and $\mathcal{L}^n(\bigcup_{B \in \mathcal{G}_1} B) = \sum_{B \in \mathcal{G}_1} \mathcal{L}^n(B)$ is a convergent series, the tail tends to zero. By continuity of measure from above, there exist 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] [/step] [step:Iterate the Vitali extraction on the remainder] Define the open remainder: \begin{align*} U_2 = U \setminus \bigcup_{i=1}^{M_1} B_i. \end{align*} Since $U$ is open and $B_1, \dotsc, B_{M_1}$ are finitely many closed sets, $U_2$ is open. 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 the same Vitali extraction as in the previous step to $U_2$ yields finitely many pairwise disjoint closed 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 now repeat the extraction on whatever measure is left. Define the open remainder: \begin{align*} U_2 = U \setminus \bigcup_{i=1}^{M_1} B_i. \end{align*} This set is open because it is the difference of the open set $U$ and the finite union of closed sets $B_1, \dotsc, B_{M_1}$. We form the Vitali cover $\mathcal{F}_2 = \{ B \subset U_2 \mid \text{$B$ closed ball, } \operatorname{diam} B < \delta \}$ of $U_2$. Applying Vitali's covering lemma to $U_2$ with cover $\mathcal{F}_2$ and then passing to a finite sub-collection (exactly as in the previous step) produces finitely many pairwise disjoint closed balls $B_{M_1+1}, \dotsc, B_{M_2} \subset U_2$ satisfying: \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 $U \setminus \bigcup_{i=1}^{M_2} B_i \subset U_2 \setminus \bigcup_{i=M_1+1}^{M_2} B_i$, we have: \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 \cdot \theta \, \mathcal{L}^n(U) = \theta^2 \, \mathcal{L}^n(U), \end{align*} where the second inequality uses $\mathcal{L}^n(U_2) \le \theta \, \mathcal{L}^n(U)$ from the previous step. The key point is that each iteration multiplies the uncovered measure by at most $\theta < 1$. [/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 $B_1, B_2, \dotsc, B_{M_k} \subset U$, each with $\operatorname{diam} B_i < \delta$, such that \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 $0 < \theta < 1$, we have $\theta^k \to 0$ as $k \to \infty$. Define $G = \{ B_i \}_{i=1}^{\infty}$. The sets $U \setminus \bigcup_{i=1}^{M_k} B_i$ are nested (decreasing in $k$), so by [continuity of measure from above](/page/1251) (valid because $\mathcal{L}^n(U) < \infty$): \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{B \in G} B\Bigr) = \lim_{k \to \infty} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{i=1}^{M_k} B_i\Bigr) = 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 are chosen inside the remainder, disjoint from all previously selected balls. [guided] After $k$ stages of the Vitali extraction, we have accumulated pairwise disjoint closed balls $B_1, \dotsc, B_{M_k} \subset U$ with $\operatorname{diam} B_i < \delta$ for each $i$, and the inductive estimate: \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*} This is geometric decay: since $\theta \in (0, 1)$, we have $\theta^k \to 0$ as $k \to \infty$. Define $G = \{ B_i \}_{i=1}^{\infty} = \bigcup_{k=1}^{\infty} \{ B_1, \dotsc, B_{M_k} \}$. To pass from the finite-stage estimate to the full family, observe that the sets $R_k := U \setminus \bigcup_{i=1}^{M_k} B_i$ form a decreasing sequence: $R_1 \supset R_2 \supset \dotsb$, and $U \setminus \bigcup_{B \in G} B = \bigcap_{k=1}^{\infty} R_k$. Since $\mathcal{L}^n(R_1) \le \theta \, \mathcal{L}^n(U) < \infty$, we may apply continuity of measure from above to conclude: \begin{align*} \mathcal{L}^n\!\Bigl(U \setminus \bigcup_{B \in G} B\Bigr) = \mathcal{L}^n\!\Bigl(\bigcap_{k=1}^{\infty} R_k\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 the balls in $G$ pairwise disjoint? At stage $j$, the new balls $B_{M_{j-1}+1}, \dotsc, B_{M_j}$ are chosen inside $U_j = U \setminus \bigcup_{i=1}^{M_{j-1}} B_i$, which is disjoint from all previously selected balls. Within each stage, the Vitali lemma produces a disjoint sub-collection. Hence all balls across all stages are pairwise disjoint. This completes the proof when $\mathcal{L}^n(U) < \infty$. [/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 (as the intersection of $U$ with the open annulus $\{ x \in \mathbb{R}^n \mid m < |x| < m+1 \}$) and satisfies $\mathcal{L}^n(U_m) < \infty$ since $U_m \subset \overline{B}(0, m+1)$. The sets $U_m$ are pairwise disjoint and $\mathcal{L}^n(U \setminus \bigcup_{m=0}^{\infty} U_m) = 0$ because $U \setminus \bigcup_{m=0}^{\infty} U_m \subset \bigcup_{m=0}^{\infty} \{ x \mid |x| = m \}$, which is a countable union of spheres, each having $\mathcal{L}^n$-measure zero. Apply the finite-measure argument (the preceding steps) to each $U_m$ with the same $\delta$ to obtain a countable pairwise disjoint family $G_m$ of closed balls in $U_m$ with $\operatorname{diam} B < \delta$ for every $B \in G_m$ and $\mathcal{L}^n(U_m \setminus \bigcup_{B \in G_m} B) = 0$. Since the annuli $U_m$ are pairwise disjoint, the families $G_m$ are pairwise disjoint. Set $G = \bigcup_{m=0}^{\infty} G_m$. Then $G$ is a countable family of pairwise disjoint closed balls contained in $U$ with $\operatorname{diam} B < \delta$ for every $B \in G$, and by countable 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} \mathcal{L}^n\!\Bigl(U_m \setminus \bigcup_{B \in G_m} B\Bigr) + 0 = 0. \end{align*} This completes the proof. [guided] What if $U$ has infinite Lebesgue measure? The finite-measure argument relied on $\mathcal{L}^n(U) < \infty$ in two places: when passing from a countable to a finite sub-collection (continuity of measure from above requires finite measure), and when bounding $\theta^k \mathcal{L}^n(U) \to 0$. To handle the infinite case, we decompose $U$ into pieces of finite measure. 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 because it is the intersection of $U$ (open) with the open annulus $\{ x \in \mathbb{R}^n \mid m < |x| < m + 1 \}$. Each $U_m$ is bounded (contained in $\overline{B}(0, m+1)$), hence $\mathcal{L}^n(U_m) < \infty$. The sets $U_m$ are pairwise disjoint (the annuli are nested shells). Does their union cover $U$ up to measure zero? The uncovered set is $U \setminus \bigcup_{m=0}^{\infty} U_m \subset \bigcup_{m=0}^{\infty} \{ x \mid |x| = m \}$. Each sphere $\{ x \in \mathbb{R}^n \mid |x| = m \}$ is an $(n-1)$-dimensional surface, so $\mathcal{L}^n(\{ |x| = m \}) = 0$. A countable union of measure-zero sets has measure zero, so $\mathcal{L}^n(U \setminus \bigcup_{m=0}^{\infty} U_m) = 0$. Apply the finite-measure result (the preceding steps) to each $U_m$ with the same diameter bound $\delta$, producing 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 pairwise disjoint, so are the families $G_m$. Define $G = \bigcup_{m=0}^{\infty} G_m$. Then $G$ is countable (a countable union of countable sets), pairwise disjoint, consists of closed balls in $U$ of diameter less than $\delta$, and by sub-additivity and countable additivity: \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 constructed family $G$ satisfies all requirements of the theorem, completing the proof. [/guided] [/step]