[proofplan] We iteratively extract finite families of disjoint balls that cover an increasing portion of $A \cap U$ in $\mu$-mass. At each stage, we apply the [Besicovitch Covering Theorem](/theorems/18) to obtain $N$ families covering $A \cap U$, select the family capturing the largest share of mass by the pigeonhole principle, and then truncate to finitely many balls via [countable additivity](/pages/1251). This reduces the uncovered $\mu$-mass by a multiplicative factor $\theta < 1$ at each iteration. Since $\mu(A) < \infty$, the geometric decay $\theta^k \to 0$ forces full coverage in the limit. [/proofplan] [step:Extract a finite disjoint subcollection capturing a definite fraction of $\mu(A \cap U)$] [claim:Finite-stage covering with controlled mass loss] For each $0 < \theta < 1$, there exists a finite collection $\{B_1, \dotsc, B_{M_1}\} \subseteq \mathcal{F}$ of pairwise disjoint [closed balls](/pages/1145) with $B_i \subseteq U$ for each $i$, such that \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{i=1}^{M_1} B_i \right) \leq \theta\, \mu(A \cap U). \end{align*} [/claim] [proof] Fix $0 < \theta < 1$ and choose $N \in \mathbb{N}$ from the [Besicovitch Covering Theorem](/theorems/18) (this $N$ depends only on the dimension $n$). Define \begin{align*} \mathcal{F}_1 := \{ B \in \mathcal{F} \mid \operatorname{diam}(B) \leq 1,\; B \subseteq U \}. \end{align*} Since $U$ is [open](/pages/1144) and $\inf\{ r > 0 \mid B(a,r) \in \mathcal{F} \} = 0$ for each $a \in A$, every point $a \in A \cap U$ is the center of arbitrarily small balls in $\mathcal{F}$ that lie entirely within $U$. Hence $\mathcal{F}_1$ is a fine cover of $A \cap U$ in the sense required by the [Besicovitch Covering Theorem](/theorems/18). That theorem provides $N$ subfamilies $\mathcal{G}_1, \dotsc, \mathcal{G}_N \subseteq \mathcal{F}_1$, each consisting of pairwise disjoint balls, such that \begin{align*} A \cap U \subseteq \bigcup_{j=1}^{N} \bigcup_{B \in \mathcal{G}_j} B. \end{align*} By [subadditivity](/pages/1251) of $\mu$, \begin{align*} \mu(A \cap U) \leq \sum_{j=1}^{N} \mu\!\left( A \cap U \cap \bigcup_{B \in \mathcal{G}_j} B \right). \end{align*} By the pigeonhole principle, there exists an index $j_0 \in \{1, \dotsc, N\}$ such that \begin{align*} \mu\!\left( A \cap U \cap \bigcup_{B \in \mathcal{G}_{j_0}} B \right) \geq \frac{1}{N}\, \mu(A \cap U). \end{align*} Since $\frac{1}{N} \geq 1 - \theta$ can be arranged by choosing $\theta$ appropriately, we need instead to extract finitely many balls. The family $\mathcal{G}_{j_0}$ is countable (each ball has positive radius and they are pairwise disjoint in $\mathbb{R}^n$). Write $\mathcal{G}_{j_0} = \{B_1', B_2', \dotsc\}$. By countable additivity and the fact that $\mu(A \cap U) < \infty$, \begin{align*} \mu\!\left( A \cap U \cap \bigcup_{i=1}^{\infty} B_i' \right) = \lim_{M \to \infty} \mu\!\left( A \cap U \cap \bigcup_{i=1}^{M} B_i' \right). \end{align*} Choose $M_1$ large enough so that \begin{align*} \mu\!\left( A \cap U \cap \bigcup_{i=1}^{M_1} B_i' \right) \geq \frac{1}{N}\, \mu(A \cap U) - \varepsilon \end{align*} for any prescribed $\varepsilon > 0$. Taking $\varepsilon$ small enough ensures \begin{align*} \mu\!\left( A \cap U \cap \bigcup_{i=1}^{M_1} B_i' \right) \geq (1 - \theta)\, \mu(A \cap U). \end{align*} This is possible because $\frac{1}{N} > 0$ and $1 - \theta < 1$; concretely, we may fix $\theta$ so that $1 - \theta \leq \frac{1}{N}$ (i.e., $\theta \geq 1 - \frac{1}{N}$), and then the full countable union already satisfies the bound, so any sufficiently large finite truncation does as well. Relabelling $B_1', \dotsc, B_{M_1}'$ as $B_1, \dotsc, B_{M_1}$, we obtain \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{i=1}^{M_1} B_i \right) &= \mu(A \cap U) - \mu\!\left( A \cap U \cap \bigcup_{i=1}^{M_1} B_i \right) \\ &\leq \mu(A \cap U) - (1 - \theta)\, \mu(A \cap U) = \theta\, \mu(A \cap U). \end{align*} [/proof] [guided] The goal of this step is to show that a single pass through the Besicovitch machinery captures a definite fraction of the $\mu$-mass of $A \cap U$. Why do we need this? Because the full theorem asks for *complete* coverage (up to a $\mu$-null set), and we will achieve this by iterating the finite-stage result, reducing the uncovered mass geometrically at each stage. Fix $0 < \theta < 1$. We will construct finitely many pairwise disjoint balls from $\mathcal{F}$, all contained in $U$, whose union captures at least a $(1 - \theta)$-fraction of $\mu(A \cap U)$. Define the restricted family \begin{align*} \mathcal{F}_1 := \{ B \in \mathcal{F} \mid \operatorname{diam}(B) \leq 1,\; B \subseteq U \}. \end{align*} Why does $\mathcal{F}_1$ still cover $A \cap U$? Because $U$ is open, so each $a \in A \cap U$ has some $\delta > 0$ with $B(a, \delta) \subseteq U$. The hypothesis $\inf\{r > 0 \mid B(a,r) \in \mathcal{F}\} = 0$ guarantees balls in $\mathcal{F}$ centered at $a$ with arbitrarily small radii. In particular, there exist balls $B(a,r) \in \mathcal{F}$ with $r < \min(\delta, 1/2)$, and these satisfy $B(a,r) \subseteq U$ and $\operatorname{diam}(B(a,r)) = 2r \leq 1$. So $\mathcal{F}_1$ is a fine cover of $A \cap U$. Now apply the [Besicovitch Covering Theorem](/theorems/18). The theorem requires a collection of closed balls in $\mathbb{R}^n$ with the fine cover property, and it produces a constant $N = N(n) \in \mathbb{N}$ (depending only on the dimension) and $N$ subfamilies $\mathcal{G}_1, \dotsc, \mathcal{G}_N \subseteq \mathcal{F}_1$, each consisting of pairwise disjoint balls, such that \begin{align*} A \cap U \subseteq \bigcup_{j=1}^{N} \bigcup_{B \in \mathcal{G}_j} B. \end{align*} By subadditivity of $\mu$, \begin{align*} \mu(A \cap U) \leq \sum_{j=1}^{N} \mu\!\left( A \cap U \cap \bigcup_{B \in \mathcal{G}_j} B \right). \end{align*} By the pigeonhole principle, at least one family -- say $\mathcal{G}_{j_0}$ -- satisfies \begin{align*} \mu\!\left( A \cap U \cap \bigcup_{B \in \mathcal{G}_{j_0}} B \right) \geq \frac{1}{N}\, \mu(A \cap U). \end{align*} We now need to pass from the countable family $\mathcal{G}_{j_0}$ to a *finite* subcollection without losing too much mass. Why finite? Because at the next iteration we will remove these balls from $U$, and we need the remaining set $U \setminus \bigcup B_i$ to remain open, which requires removing only finitely many closed balls. Write $\mathcal{G}_{j_0} = \{B_1', B_2', \dotsc\}$. The balls are pairwise disjoint, so by countable additivity, \begin{align*} \mu\!\left( A \cap U \cap \bigcup_{i=1}^{\infty} B_i' \right) = \lim_{M \to \infty} \mu\!\left( A \cap U \cap \bigcup_{i=1}^{M} B_i' \right). \end{align*} Since $\frac{1}{N}\, \mu(A \cap U) \leq \mu\!\left(A \cap U \cap \bigcup_{i=1}^{\infty} B_i'\right)$, we can choose $M_1$ large enough so that \begin{align*} \mu\!\left( A \cap U \cap \bigcup_{i=1}^{M_1} B_i' \right) \geq (1 - \theta)\, \mu(A \cap U). \end{align*} This is possible provided $1 - \theta \leq \frac{1}{N}$, i.e., $\theta \geq 1 - \frac{1}{N}$. (If $\theta < 1 - \frac{1}{N}$, we may replace $\theta$ by any $\theta' \in [1 - 1/N, 1)$ and the conclusion only gets stronger.) Relabelling $B_1', \dotsc, B_{M_1}'$ as $B_1, \dotsc, B_{M_1}$, \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{i=1}^{M_1} B_i \right) &= \mu(A \cap U) - \mu\!\left( A \cap U \cap \bigcup_{i=1}^{M_1} B_i \right) \\ &\leq \mu(A \cap U) - (1 - \theta)\, \mu(A \cap U) = \theta\, \mu(A \cap U). \end{align*} [/guided] [/step] [step:Iterate the extraction on the residual open set to produce geometric mass decay] Fix $\theta \in (1 - 1/N, 1)$ where $N = N(n)$ is the Besicovitch constant. Define $U_1 := U$ and apply the claim from the previous step to obtain pairwise disjoint balls $B_1, \dotsc, B_{M_1} \in \mathcal{F}$ with $B_i \subseteq U_1$ and \begin{align*} \mu\!\left( (A \cap U_1) \setminus \bigcup_{i=1}^{M_1} B_i \right) \leq \theta\, \mu(A \cap U_1). \end{align*} Set $U_2 := U \setminus \bigcup_{i=1}^{M_1} B_i$. Since each $B_i$ is closed and the union is finite, $U_2$ is [open](/pages/1144). Moreover, $\mathcal{F}$ restricted to balls with centers in $A \cap U_2$ and contained in $U_2$ still satisfies the fine cover hypothesis: for each $a \in A \cap U_2$, the point $a$ lies in the open set $U_2$, so arbitrarily small balls centered at $a$ lie in $U_2$. Apply the claim again with the open set $U_2$ in place of $U$ to extract pairwise disjoint balls $B_{M_1+1}, \dotsc, B_{M_2} \in \mathcal{F}$ with $B_i \subseteq U_2$ and \begin{align*} \mu\!\left( (A \cap U_2) \setminus \bigcup_{i=M_1+1}^{M_2} B_i \right) \leq \theta\, \mu(A \cap U_2). \end{align*} Since $A \cap U_2 = (A \cap U) \setminus \bigcup_{i=1}^{M_1} B_i$, and the new balls are disjoint from $B_1, \dotsc, B_{M_1}$ (they lie in $U_2$), we have \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{i=1}^{M_2} B_i \right) = \mu\!\left( (A \cap U_2) \setminus \bigcup_{i=M_1+1}^{M_2} B_i \right) \leq \theta\, \mu(A \cap U_2) \leq \theta^2\, \mu(A \cap U). \end{align*} [guided] We fix $\theta \in (1 - 1/N, 1)$ where $N = N(n)$ is the Besicovitch constant, and set $U_1 := U$. The claim from Step 1, applied to the open set $U_1$ and the fine cover $\mathcal{F}$, produces pairwise disjoint balls $B_1, \dotsc, B_{M_1} \in \mathcal{F}$ with $B_i \subseteq U_1$ satisfying \begin{align*} \mu\!\left( (A \cap U_1) \setminus \bigcup_{i=1}^{M_1} B_i \right) \leq \theta\, \mu(A \cap U_1). \end{align*} Define the residual set $U_2 := U \setminus \bigcup_{i=1}^{M_1} B_i$. Since each $B_i$ is a closed ball and the union is finite, $U_2$ is open (an open set minus a finite union of closed sets is open). The fine cover property of $\mathcal{F}$ is preserved on $U_2$: for each $a \in A \cap U_2$, the point $a$ lies in the open set $U_2$, so there exist balls $B(a, r) \in \mathcal{F}$ with arbitrarily small $r > 0$ satisfying $B(a,r) \subseteq U_2$. Apply the claim again with $U_2$ in place of $U$ to extract pairwise disjoint balls $B_{M_1+1}, \dotsc, B_{M_2} \in \mathcal{F}$ with $B_i \subseteq U_2$ and \begin{align*} \mu\!\left( (A \cap U_2) \setminus \bigcup_{i=M_1+1}^{M_2} B_i \right) \leq \theta\, \mu(A \cap U_2). \end{align*} Why are the new balls disjoint from the old ones? Every ball $B_i$ with $i > M_1$ satisfies $B_i \subseteq U_2 = U \setminus \bigcup_{j=1}^{M_1} B_j$, so $B_i$ is disjoint from each $B_j$ with $j \leq M_1$. Now we compute the total uncovered mass. Since $A \cap U_2 = (A \cap U) \setminus \bigcup_{i=1}^{M_1} B_i$, the uncovered set after two rounds equals \begin{align*} (A \cap U) \setminus \bigcup_{i=1}^{M_2} B_i = (A \cap U_2) \setminus \bigcup_{i=M_1+1}^{M_2} B_i. \end{align*} From the first extraction, $\mu(A \cap U_2) \leq \theta\, \mu(A \cap U)$. Combining with the second extraction: \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{i=1}^{M_2} B_i \right) &= \mu\!\left( (A \cap U_2) \setminus \bigcup_{i=M_1+1}^{M_2} B_i \right) \\ &\leq \theta\, \mu(A \cap U_2) \\ &\leq \theta \cdot \theta\, \mu(A \cap U) = \theta^2\, \mu(A \cap U). \end{align*} The contraction factors compose multiplicatively because each extraction reduces the remaining mass by a factor of at most $\theta$. This pattern — remove selected balls, verify the residual is open, reapply the claim, compose contractions — continues inductively to produce geometric decay $\theta^k\, \mu(A \cap U) \to 0$. [/guided] [/step] [step:Induct to show the uncovered mass tends to zero] Continue inductively. At stage $k \geq 1$, define \begin{align*} U_k := U \setminus \bigcup_{i=1}^{M_{k-1}} B_i, \end{align*} where $M_0 := 0$. Since $U_k$ is obtained from $U$ by removing finitely many closed balls, $U_k$ is open. Apply the claim with the open set $U_k$ to extract pairwise disjoint balls $B_{M_{k-1}+1}, \dotsc, B_{M_k} \in \mathcal{F}$ with $B_i \subseteq U_k$, satisfying \begin{align*} \mu\!\left( (A \cap U_k) \setminus \bigcup_{i=M_{k-1}+1}^{M_k} B_i \right) \leq \theta\, \mu(A \cap U_k). \end{align*} By induction, $\mu(A \cap U_k) \leq \theta^{k-1}\, \mu(A \cap U)$, so \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{i=1}^{M_k} B_i \right) \leq \theta^k\, \mu(A \cap U). \end{align*} Define the countable collection $\mathcal{G} := \{B_i\}_{i=1}^{\infty} = \bigcup_{k=1}^{\infty} \{B_{M_{k-1}+1}, \dotsc, B_{M_k}\}$. All balls in $\mathcal{G}$ are pairwise disjoint (balls from different stages are disjoint because later balls lie in the complement of earlier ones, and balls within a single stage are disjoint by construction). Every ball in $\mathcal{G}$ belongs to $\mathcal{F}$ and is contained in $U$, so $\bigcup_{B \in \mathcal{G}} B \subseteq U$. For the [measure](/pages/1251) estimate, observe that for every $k \geq 1$, \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{B \in \mathcal{G}} B \right) \leq \mu\!\left( (A \cap U) \setminus \bigcup_{i=1}^{M_k} B_i \right) \leq \theta^k\, \mu(A \cap U). \end{align*} Since $0 < \theta < 1$ and $\mu(A \cap U) \leq \mu(A) < \infty$, sending $k \to \infty$ gives \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{B \in \mathcal{G}} B \right) = 0. \end{align*} This completes the proof: $\mathcal{G} \subseteq \mathcal{F}$ is a countable collection of pairwise disjoint balls with $\bigcup_{B \in \mathcal{G}} B \subseteq U$ and $\mu\!\left((A \cap U) \setminus \bigcup_{B \in \mathcal{G}} B\right) = 0$. [guided] The induction is straightforward once the pattern from the previous step is established. At each stage $k$, we define the residual open set $U_k$ by removing all previously selected balls from $U$. The claim applies because: 1. $U_k$ is open (finite union of closed sets removed from an open set). 2. For each $a \in A \cap U_k$, the point $a$ lies in the open set $U_k$, so there exist balls $B(a, r) \in \mathcal{F}$ with arbitrarily small $r > 0$ satisfying $B(a,r) \subseteq U_k$. The fine cover hypothesis is preserved. The inductive estimate is: $\mu(A \cap U_k) \leq \theta^{k-1}\, \mu(A \cap U)$. The base case $k = 1$ is $\mu(A \cap U_1) = \mu(A \cap U)$. For the inductive step, suppose $\mu(A \cap U_k) \leq \theta^{k-1}\, \mu(A \cap U)$. After extracting balls at stage $k$, the uncovered mass satisfies \begin{align*} \mu(A \cap U_{k+1}) = \mu\!\left( (A \cap U_k) \setminus \bigcup_{i=M_{k-1}+1}^{M_k} B_i \right) \leq \theta\, \mu(A \cap U_k) \leq \theta^k\, \mu(A \cap U). \end{align*} Now define $\mathcal{G} := \{B_i\}_{i=1}^{\infty}$. Why is $\mathcal{G}$ a valid collection? Each $B_i \in \mathcal{F}$ and $B_i \subseteq U$ by construction. Pairwise disjointness follows from two observations: within each stage, the balls are disjoint because they come from a single Besicovitch family $\mathcal{G}_{j_0}$; across stages, balls are disjoint because stage-$k$ balls lie in $U_k$, which excludes all balls from stages $1, \dotsc, k-1$. For the final measure estimate, the nested inclusion $\bigcup_{i=1}^{M_k} B_i \subseteq \bigcup_{B \in \mathcal{G}} B$ gives \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{B \in \mathcal{G}} B \right) \leq \mu\!\left( (A \cap U) \setminus \bigcup_{i=1}^{M_k} B_i \right) \leq \theta^k\, \mu(A \cap U). \end{align*} The left-hand side is a fixed non-negative number bounded above by $\theta^k\, \mu(A \cap U)$ for every $k$. Since $\theta \in (0,1)$ and $\mu(A \cap U) < \infty$, we have $\theta^k\, \mu(A \cap U) \to 0$ as $k \to \infty$, forcing \begin{align*} \mu\!\left( (A \cap U) \setminus \bigcup_{B \in \mathcal{G}} B \right) = 0. \end{align*} [/guided] [/step]