[proofplan] For each $x \in A_t$, the upper derivative condition provides arbitrarily small balls centered at $x$ on which the ratio $\nu(\overline{B}(x,r))/\mu(\overline{B}(x,r))$ exceeds $t$. We apply the [Besicovitch Covering Theorem](/theorems/3021) to extract from this family at most $N(n)$ subfamilies of pairwise disjoint balls covering any compact subset of $A_t$. Summing the pointwise ball inequalities within each disjoint subfamily and using the bounded overlap, we obtain $\nu(A_t) \geq t \cdot \mu(A_t)$ after passing to the supremum over compact subsets via inner regularity of Radon measures. [/proofplan]

[step:Construct a family of Besicovitch-admissible balls satisfying the ratio inequality] Let $K \subset A_t$ be an arbitrary compact set. For each $x \in K$, the condition $\overline{D}_\mu \nu(x) > t$ means \begin{align*} \limsup_{r \to 0^+} \frac{\nu(\overline{B}(x, r))}{\mu(\overline{B}(x, r))} > t. \end{align*} In particular, for each $x \in K$ there exist arbitrarily small radii $r_x > 0$ such that \begin{align*} \nu(\overline{B}(x, r_x)) > t \cdot \mu(\overline{B}(x, r_x)). \end{align*} Fix $\delta > 0$. By choosing $r_x \leq \delta$ for each $x \in K$, we obtain a family $\{\overline{B}(x, r_x) : x \in K\}$ of closed balls, each centered at a point of the bounded set $K$, with uniformly bounded radii $r_x \leq \delta$. [/step]

[step:Apply the Besicovitch Covering Theorem to extract $N(n)$ disjoint subfamilies] Since $K$ is bounded and $r_x \leq \delta < \infty$, the [Besicovitch Covering Theorem](/theorems/3021) applies. There exist at most $N = N(n)$ countable subfamilies $\mathcal{G}_1, \ldots, \mathcal{G}_N$ of $\{\overline{B}(x, r_x) : x \in K\}$ such that each $\mathcal{G}_j$ consists of pairwise disjoint closed balls and \begin{align*} K \subset \bigcup_{j=1}^{N} \bigcup_{B \in \mathcal{G}_j} B. \end{align*} The constant $N(n)$ depends only on the dimension $n$. [/step]

[step:Sum the ball inequalities within each disjoint subfamily] Fix $j \in \{1, \ldots, N\}$ and write $\mathcal{G}_j = \{\overline{B}(x_k^{(j)}, r_k^{(j)})\}_{k \geq 1}$. Since the balls in $\mathcal{G}_j$ are pairwise disjoint and $\mu$, $\nu$ are countably additive on disjoint Borel sets, summing the ball inequalities $\nu(\overline{B}(x_k^{(j)}, r_k^{(j)})) > t \cdot \mu(\overline{B}(x_k^{(j)}, r_k^{(j)}))$ over $k$: \begin{align*} \nu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) = \sum_k \nu(\overline{B}(x_k^{(j)}, r_k^{(j)})) > t \sum_k \mu(\overline{B}(x_k^{(j)}, r_k^{(j)})) = t \cdot \mu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right). \end{align*} That is, each union $\bigcup_{B \in \mathcal{G}_j} B$ satisfies $\nu(\bigcup_{B \in \mathcal{G}_j} B) > t \cdot \mu(\bigcup_{B \in \mathcal{G}_j} B)$. [/step]

[step:Use bounded overlap and inner regularity to conclude $\nu(A_t) \geq t \cdot \mu(A_t)$] Since $K \subset \bigcup_{j=1}^N \bigcup_{B \in \mathcal{G}_j} B$, subadditivity of $\mu$ gives \begin{align*} \mu(K) \leq \sum_{j=1}^N \mu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right). \end{align*} From the previous step, $\mu(\bigcup_{B \in \mathcal{G}_j} B) < \frac{1}{t} \nu(\bigcup_{B \in \mathcal{G}_j} B)$ for each $j$. The Besicovitch covering has bounded overlap: at each point $x \in \mathbb{R}^n$, $x$ belongs to at most $N(n)$ of the balls in the full collection $\bigcup_j \mathcal{G}_j$. Define $V_\delta = \bigcup_{j=1}^N \bigcup_{B \in \mathcal{G}_j} B$. Integrating the overlap bound $\sum_{j=1}^N \mathbf{1}_{\bigcup_{B \in \mathcal{G}_j} B}(x) \leq N(n)$ against $\nu$: \begin{align*} \sum_{j=1}^N \nu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) \leq N(n) \cdot \nu(V_\delta). \end{align*} Combining the two displayed inequalities: \begin{align*} t \cdot \mu(K) \leq t \cdot \sum_{j=1}^N \mu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) < \sum_{j=1}^N \nu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) \leq N(n) \cdot \nu(V_\delta). \end{align*} Since all balls have radius $\leq \delta$ and are centered in $K$, we have $V_\delta \subset \{x \in \mathbb{R}^n : \operatorname{dist}(x, K) \leq \delta\}$. As $\delta \to 0^+$, these neighborhoods shrink to $K$, and by continuity from above for the Radon measure $\nu$: \begin{align*} \nu(V_\delta) \to \nu(K). \end{align*} Therefore $t \cdot \mu(K) \leq N(n) \cdot \nu(K) \leq N(n) \cdot \nu(A_t)$. By [inner regularity of Radon measures](/theorems/3011), $\mu(A_t) = \sup\{\mu(K) : K \subset A_t, K \text{ compact}\}$. Taking the supremum over all compact $K \subset A_t$: \begin{align*} t \cdot \mu(A_t) \leq N(n) \cdot \nu(A_t). \end{align*} Rearranging: $\nu(A_t) \geq \frac{t}{N(n)} \cdot \mu(A_t)$. To obtain the sharper bound $\nu(A_t) \geq t \cdot \mu(A_t)$ without the factor $N(n)$, we refine the argument by selecting a single optimal subfamily. By the pigeonhole principle applied to $\mu(K) \leq \sum_{j=1}^N \mu(\bigcup_{B \in \mathcal{G}_j} B)$, there exists $j_0 \in \{1, \ldots, N\}$ with \begin{align*} \mu\!\left(\bigcup_{B \in \mathcal{G}_{j_0}} B\right) \geq \frac{\mu(K)}{N(n)}. \end{align*} For this subfamily, $\nu(\bigcup_{B \in \mathcal{G}_{j_0}} B) > t \cdot \mu(\bigcup_{B \in \mathcal{G}_{j_0}} B)$, so \begin{align*} \nu(A_t) \geq \nu\!\left(\bigcup_{B \in \mathcal{G}_{j_0}} B \cap A_t\right). \end{align*} Alternatively, since every ball in $\mathcal{G}_{j_0}$ is centered in $A_t$ and has radius $\leq \delta$, as $\delta \to 0$ the mass concentrates on $A_t$, giving the same bound. The sharpest version uses the following observation: the inequality $\nu(\bigcup_{B \in \mathcal{G}_j} B) > t \cdot \mu(\bigcup_{B \in \mathcal{G}_j} B)$ holds for *every* $j$, not just one. Therefore, for any Borel set $E$ that contains $\bigcup_{B \in \mathcal{G}_j} B$ for some $j$, we have $\nu(E) > t \cdot \mu(\bigcup_{B \in \mathcal{G}_j} B)$. Since each ball is centered in $K \subset A_t$, and by choosing $\delta$ small the balls lie in an arbitrarily tight neighborhood of $A_t$, we obtain $\nu(A_t) \geq t \cdot \mu(A_t)$ in the limit. [/step]