[proofplan] We iteratively apply the [Vitali Covering Theorem](/theorems/3020). At each stage, the uncovered portion of $A$ admits a Vitali cover by balls disjoint from all previously selected balls. The Vitali theorem extracts a disjoint subcollection whose five-fold dilates cover the uncovered set. Since the uncovered set $A_k$ is contained in the union of dilated balls minus the union of undilated balls, and the undilated balls are disjoint, the uncovered measure satisfies $\mathcal{L}^n(A_k) \leq (1 - 5^{-n}) \mathcal{L}^n(A_{k-1})$. After $k$ stages the uncovered measure is at most $(1 - 5^{-n})^k \mathcal{L}^n(A)$, and we choose $k$ so that this is less than $\varepsilon$. [/proofplan] [step:Reduce to the case where $A$ is bounded and the balls have bounded diameter] Since $\mathcal{L}^n(A) < \infty$, for any $\delta > 0$ there exists $m \in \mathbb{N}$ with $\mathcal{L}^n(A \setminus \overline{B}(0, m)) < \delta$. It therefore suffices to prove the result for $A' := A \cap \overline{B}(0, m)$ with tolerance $\varepsilon - \delta$ (provided $\delta < \varepsilon$), then combine. So assume $A$ is bounded. Choose a bounded open set $U \supset A$ with $\mathcal{L}^n(U) < \infty$. Since $\mathcal{V}$ is a Vitali cover, for every $x \in A$ and every $\eta > 0$ there exists $B \in \mathcal{V}$ with $x \in B$ and $\operatorname{diam}(B) < \eta$. Restricting $\mathcal{V}$ to those balls contained in $U$ preserves the Vitali cover property and ensures $\sup\{\operatorname{diam}(B) : B \in \mathcal{V}\} < \infty$. [guided] The reduction to bounded $A$ uses only $\mathcal{L}^n(A) < \infty$: the tails $A \setminus \overline{B}(0, m)$ have measure tending to zero by continuity of measure from above. Once $A$ is bounded, we embed it in a bounded open set $U$ so that all selected balls lie inside a region of finite volume. Why restrict to balls inside $U$? We need the diameter bound $\sup \operatorname{diam}(B) < \infty$ to invoke the [Vitali Covering Theorem](/theorems/3020), and we need the balls to lie in a set of finite measure to ensure the selected disjoint balls form a countable collection. The Vitali cover property is preserved because for each $x \in A$ and $\eta > 0$, there exists $B \in \mathcal{V}$ with $x \in B$ and $\operatorname{diam}(B) < \eta$. Choosing $\eta < \operatorname{dist}(x, \mathbb{R}^n \setminus U)$ ensures $B \subset U$. [/guided] [/step] [step:Set up the iteration: apply the Vitali Covering Theorem at each stage to the uncovered set] Set $A_0 := A$, $\mathcal{V}_0 := \mathcal{V}$, and $\mathcal{D}_0 := \varnothing$. We define sequences inductively. At stage $k \geq 1$: **Selection.** Apply the [Vitali Covering Theorem](/theorems/3020) to $\mathcal{V}_{k-1}$: since $\sup\{\operatorname{diam}(B) : B \in \mathcal{V}_{k-1}\} < \infty$ (all balls lie in $U$), there exists a countable pairwise disjoint subcollection $\mathcal{C}_k \subset \mathcal{V}_{k-1}$ with \begin{align*} \bigcup_{B \in \mathcal{V}_{k-1}} B \;\subseteq\; \bigcup_{B \in \mathcal{C}_k} 5B. \end{align*} Since $A_{k-1} \subset \bigcup_{B \in \mathcal{V}_{k-1}} B$ ($\mathcal{V}_{k-1}$ covers $A_{k-1}$), we have $A_{k-1} \subset \bigcup_{B \in \mathcal{C}_k} 5B$. **Update.** Set $\mathcal{D}_k := \mathcal{D}_{k-1} \cup \mathcal{C}_k$, $A_k := A \setminus \bigcup_{B \in \mathcal{D}_k} B$, and $\mathcal{V}_k := \{B \in \mathcal{V} : B \cap \bigcup_{B' \in \mathcal{D}_k} B' = \varnothing\}$. **$\mathcal{V}_k$ is a Vitali cover of $A_k$.** For each $x \in A_k$, the set $\bigcup_{B' \in \mathcal{D}_k} B'$ is closed (it is a locally finite union of closed balls in the bounded set $U$) and does not contain $x$, so $\operatorname{dist}(x, \bigcup_{B' \in \mathcal{D}_k} B') > 0$. Since $\mathcal{V}$ is a Vitali cover of $A \supset A_k$, there exist balls in $\mathcal{V}$ of arbitrarily small diameter containing $x$. Any such ball with diameter less than this distance lies in $\mathcal{V}_k$. **$\mathcal{D}_k$ is pairwise disjoint.** Each $\mathcal{C}_k \subset \mathcal{V}_{k-1}$ consists of balls disjoint from $\bigcup_{B' \in \mathcal{D}_{k-1}} B'$ and pairwise disjoint among themselves. [/step] [step:Establish the geometric decay $\mathcal{L}^n(A_k) \leq (1 - 5^{-n})^k \mathcal{L}^n(A)$] Define $\sigma := 1 - 5^{-n} \in (0,1)$. We prove $\mathcal{L}^n(A_k) \leq \sigma \cdot \mathcal{L}^n(A_{k-1})$ for each $k \geq 1$. From the previous step, $A_{k-1} \subset \bigcup_{B \in \mathcal{C}_k} 5B$. The set $A_k = A_{k-1} \setminus \bigcup_{B \in \mathcal{C}_k} B$ satisfies $A_k \subset A_{k-1}$ and $A_k \cap \bigcup_{B \in \mathcal{C}_k} B = \varnothing$. Combined: \begin{align*} A_k \subset \left(\bigcup_{B \in \mathcal{C}_k} 5B\right) \setminus \left(\bigcup_{B \in \mathcal{C}_k} B\right). \end{align*} Since $\bigcup_{B \in \mathcal{C}_k} B \subset \bigcup_{B \in \mathcal{C}_k} 5B$ and the balls in $\mathcal{C}_k$ are pairwise disjoint: \begin{align*} \mathcal{L}^n(A_k) &\leq \mathcal{L}^n\!\left(\bigcup_{B \in \mathcal{C}_k} 5B\right) - \mathcal{L}^n\!\left(\bigcup_{B \in \mathcal{C}_k} B\right) \leq \sum_{B \in \mathcal{C}_k} \mathcal{L}^n(5B) - \sum_{B \in \mathcal{C}_k} \mathcal{L}^n(B) \\ &= (5^n - 1) \sum_{B \in \mathcal{C}_k} \mathcal{L}^n(B), \end{align*} where the second inequality uses subadditivity for the first sum and disjointness for the second. From $A_{k-1} \subset \bigcup_{B \in \mathcal{C}_k} 5B$: \begin{align*} \mathcal{L}^n(A_{k-1}) \leq \sum_{B \in \mathcal{C}_k} \mathcal{L}^n(5B) = 5^n \sum_{B \in \mathcal{C}_k} \mathcal{L}^n(B). \end{align*} Dividing the first estimate by the second (both sides are positive when $\mathcal{L}^n(A_{k-1}) > 0$; if $\mathcal{L}^n(A_{k-1}) = 0$ the result is immediate): \begin{align*} \frac{\mathcal{L}^n(A_k)}{\mathcal{L}^n(A_{k-1})} \leq \frac{(5^n - 1)\sum \mathcal{L}^n(B)}{5^n \sum \mathcal{L}^n(B)} = \frac{5^n - 1}{5^n} = 1 - 5^{-n} = \sigma. \end{align*} By induction, $\mathcal{L}^n(A_k) \leq \sigma^k \mathcal{L}^n(A_0) = (1 - 5^{-n})^k \mathcal{L}^n(A)$. [guided] The geometric decay is powered by two estimates that use the same quantity $\sum \mathcal{L}^n(B)$ in complementary ways. The covering estimate from the Vitali theorem gives a lower bound on $\sum \mathcal{L}^n(B)$ in terms of $\mathcal{L}^n(A_{k-1})$. The containment $A_k \subset (\bigcup 5B) \setminus (\bigcup B)$ gives an upper bound on $\mathcal{L}^n(A_k)$ in terms of $\sum \mathcal{L}^n(B)$. Taking the ratio cancels $\sum \mathcal{L}^n(B)$ and yields the contraction factor $\sigma = (5^n - 1)/5^n$. Geometrically, the factor $5^{-n}$ represents the fraction of volume that the undilated ball occupies within its five-fold dilate: $\mathcal{L}^n(B)/\mathcal{L}^n(5B) = 5^{-n}$. At each stage, the undilated balls "capture" at least this fraction of the volume covered by the dilates, so the uncovered remainder $A_k$ loses at least a fraction $5^{-n}$ of the previous uncovered measure. Note that this argument uses the scaling property $\mathcal{L}^n(5B) = 5^n \mathcal{L}^n(B)$ specific to Lebesgue measure. For a non-doubling measure, this geometric decay would fail, and the Vitali covering lemma does not hold in general without some doubling condition on the measure. [/guided] [/step] [step:Choose $k$ large enough and conclude] Since $\sigma = 1 - 5^{-n} \in (0, 1)$, the sequence $\sigma^k \to 0$ as $k \to \infty$. Choose $k$ large enough that $(1 - 5^{-n})^k \mathcal{L}^n(A) < \varepsilon$. The combined collection $\mathcal{D}_k = \bigcup_{i=1}^k \mathcal{C}_i$ is a countable pairwise disjoint subcollection of $\mathcal{V}$, and \begin{align*} \mathcal{L}^n\!\left(A \setminus \bigcup_{B \in \mathcal{D}_k} B\right) = \mathcal{L}^n(A_k) \leq (1 - 5^{-n})^k \mathcal{L}^n(A) < \varepsilon. \end{align*} This completes the proof. [/step]