[proofplan] We prove both parts by a Vitali covering argument. For part (i), we cover $A$ with balls on which the density ratio $\nu(B)/\mu(B)$ is at most $\alpha + \varepsilon$, extract a disjoint subcover via the Vitali covering theorem for Radon measures, sum the local estimates, and pass to the limit using outer regularity of $\mu$. Part (ii) is symmetric, with the inequality reversed and the covering theorem applied with respect to $\mu$ rather than $\nu$. [/proofplan] [step:Reduce to the case of finite total mass] Since $\mu$ and $\nu$ are [page:1008|Radon measures] on $\mathbb{R}^n$, they are inner regular with respect to compact sets. For each $m \in \mathbb{N}$, the restricted measures $\mu_m := \mu\lfloor_{\overline{B}(0,m)}$ and $\nu_m := \nu\lfloor_{\overline{B}(0,m)}$ have finite total mass. The derivative $D_\mu \nu(x)$ depends only on the behaviour of $\mu$ and $\nu$ in a neighbourhood of $x$, so $D_{\mu_m} \nu_m(x) = D_\mu \nu(x)$ for every $x \in \overline{B}(0, m)$. Applying the finite-mass result to $A_m := A \cap \overline{B}(0, m)$ gives $\nu(A_m) \leq \alpha \, \mu(A_m)$ (respectively $\geq$). Since $A_m \nearrow A$, continuity of measure from below yields $\nu(A) = \lim_{m \to \infty} \nu(A_m) \leq \alpha \lim_{m \to \infty} \mu(A_m) = \alpha \, \mu(A)$, and similarly for part (ii). We therefore assume without loss of generality that $\mu(\mathbb{R}^n) < \infty$ and $\nu(\mathbb{R}^n) < \infty$. [guided] Why do we reduce to finite total mass? The Vitali covering argument in the subsequent steps produces disjoint balls whose $\mu$-measures we sum. If $\mu(\mathbb{R}^n) = \infty$, controlling this sum requires care. By restricting to $\overline{B}(0, m)$, the restricted measures $\mu_m := \mu\lfloor_{\overline{B}(0,m)}$ and $\nu_m := \nu\lfloor_{\overline{B}(0,m)}$ are finite Radon measures. Since $D_\mu \nu(x)$ is determined by the values of $\mu$ and $\nu$ on balls centred at $x$ of arbitrarily small radius, and all such balls eventually lie in $\overline{B}(0, m)$ for $m$ large enough, we have $D_{\mu_m} \nu_m(x) = D_\mu \nu(x)$ for every $x \in \overline{B}(0, m)$. Applying the finite-mass result to $A_m := A \cap \overline{B}(0, m)$ gives $\nu(A_m) \leq \alpha \, \mu(A_m)$ for part (i). The sets $A_m$ increase to $A$, so by continuity of measure from below (which holds for any measure): \begin{align*} \nu(A) = \lim_{m \to \infty} \nu(A_m) \leq \alpha \lim_{m \to \infty} \mu(A_m) = \alpha \, \mu(A). \end{align*} The same argument applies to part (ii) with the inequality reversed. [/guided] [/step] [step:Cover $A$ by balls with controlled density ratio and extract a disjoint subcover] Fix $\varepsilon > 0$ and let $U \supset A$ be an open set. For each $a \in A$, the condition $D_\mu \nu(a) \leq \alpha$ means \begin{align*} \limsup_{r \to 0} \frac{\nu(B(a, r))}{\mu(B(a, r))} \leq \alpha. \end{align*} In particular, for every $\varepsilon > 0$ there exist arbitrarily small $r > 0$ with $\nu(B(a, r)) \leq (\alpha + \varepsilon) \mu(B(a, r))$ and $B(a, r) \subset U$. Define the family \begin{align*} \mathcal{F} := \{ B(a, r) \subset U \mid a \in A,\; \nu(B(a, r)) \leq (\alpha + \varepsilon) \mu(B(a, r)) \}. \end{align*} The family $\mathcal{F}$ is a Vitali cover of $A$: every point of $A$ is the centre of members of $\mathcal{F}$ of arbitrarily small radius. By the [Vitali covering theorem for Radon measures](/theorems/???) there exists a countable pairwise disjoint subfamily $\mathcal{G} \subset \mathcal{F}$ such that \begin{align*} \nu\!\Bigl(A \setminus \bigcup_{B \in \mathcal{G}} B\Bigr) = 0. \end{align*} [guided] We want to turn the pointwise bound $D_\mu \nu(a) \leq \alpha$ into the global estimate $\nu(A) \leq \alpha \mu(A)$. The strategy is to cover $A$ by balls on which we have local control of the density ratio, then sum. Fix $\varepsilon > 0$ and an open set $U \supset A$. Since $D_\mu \nu(a) \leq \alpha$ for each $a \in A$, the definition of the upper derivative as $\limsup_{r \to 0} \nu(B(a,r))/\mu(B(a,r)) \leq \alpha$ guarantees that there exist balls $B(a, r) \subset U$ of arbitrarily small radius satisfying \begin{align*} \nu(B(a, r)) \leq (\alpha + \varepsilon) \mu(B(a, r)). \end{align*} We collect all such balls into the family \begin{align*} \mathcal{F} := \{ B(a, r) \subset U \mid a \in A,\; \nu(B(a, r)) \leq (\alpha + \varepsilon) \mu(B(a, r)) \}. \end{align*} This is a Vitali cover of $A$: each point of $A$ lies in members of $\mathcal{F}$ of arbitrarily small diameter. Why does a disjoint subcover suffice? The [Vitali covering theorem for Radon measures](/theorems/???) extracts a countable disjoint subfamily $\mathcal{G} \subset \mathcal{F}$ satisfying $\nu(A \setminus \bigcup \mathcal{G}) = 0$. This means the disjoint balls capture all the $\nu$-mass of $A$, even though they do not literally cover $A$ as a set. The key hypothesis that makes the Vitali theorem applicable is that $\nu$ is a Radon measure (hence Borel regular and locally finite) and that $\mathcal{F}$ is a fine cover of $A$ — both conditions are satisfied by construction. [/guided] [/step] [step:Sum the local estimates and pass to the limit via outer regularity] Using the disjoint subfamily $\mathcal{G}$ from the previous step: \begin{align*} \nu(A) &\leq \nu\!\Bigl(\bigcup_{B \in \mathcal{G}} B\Bigr) \leq \sum_{B \in \mathcal{G}} \nu(B) \leq (\alpha + \varepsilon) \sum_{B \in \mathcal{G}} \mu(B) \leq (\alpha + \varepsilon) \, \mu(U). \end{align*} The first inequality uses $\nu(A \setminus \bigcup \mathcal{G}) = 0$, the second uses countable sub-additivity of $\nu$, the third uses the defining property of $\mathcal{F}$, and the fourth uses disjointness of $\mathcal{G}$ together with $\bigcup \mathcal{G} \subset U$. Since $\mu$ is a Radon measure, it is [page:1251|outer regular]: $\mu(A) = \inf\{\mu(U) \mid U \supset A,\, U \text{ open}\}$. Taking the infimum over all open $U \supset A$: \begin{align*} \nu(A) \leq (\alpha + \varepsilon) \, \mu(A). \end{align*} Since $\varepsilon > 0$ was arbitrary, sending $\varepsilon \to 0$ gives $\nu(A) \leq \alpha \, \mu(A)$, completing the proof of part (i). [guided] We now aggregate the local density bounds into a global estimate. Since $\nu(A \setminus \bigcup \mathcal{G}) = 0$, the $\nu$-mass of $A$ is concentrated on the disjoint balls: \begin{align*} \nu(A) \leq \nu\!\Bigl(\bigcup_{B \in \mathcal{G}} B\Bigr). \end{align*} By countable sub-additivity of $\nu$ (which holds for any [page:1251|measure]) and the defining property of each $B \in \mathcal{G} \subset \mathcal{F}$: \begin{align*} \nu\!\Bigl(\bigcup_{B \in \mathcal{G}} B\Bigr) \leq \sum_{B \in \mathcal{G}} \nu(B) \leq (\alpha + \varepsilon) \sum_{B \in \mathcal{G}} \mu(B). \end{align*} Because the balls in $\mathcal{G}$ are pairwise disjoint and all contained in $U$, the sum $\sum_{B \in \mathcal{G}} \mu(B) = \mu(\bigcup \mathcal{G}) \leq \mu(U)$. Combining: \begin{align*} \nu(A) \leq (\alpha + \varepsilon) \, \mu(U). \end{align*} The open set $U \supset A$ was arbitrary. Since $\mu$ is a Radon measure on $\mathbb{R}^n$, it is outer regular, meaning $\mu(A) = \inf\{\mu(U) \mid U \supset A,\, U \text{ open}\}$. Taking the infimum over all such $U$ yields $\nu(A) \leq (\alpha + \varepsilon) \mu(A)$. Since $\varepsilon > 0$ was arbitrary, sending $\varepsilon \to 0$ gives \begin{align*} \nu(A) \leq \alpha \, \mu(A), \end{align*} which is part (i). [/guided] [/step] [step:Reverse the argument for the lower bound by applying the covering theorem with respect to $\mu$] Fix $\varepsilon > 0$ and let $U \supset A$ be open. For each $a \in A$, the condition $D_\mu \nu(a) \geq \alpha$ ensures there exist arbitrarily small $r > 0$ with $\nu(B(a, r)) \geq (\alpha - \varepsilon) \mu(B(a, r))$ and $B(a, r) \subset U$. Define \begin{align*} \mathcal{F}' := \{ B(a, r) \subset U \mid a \in A,\; \nu(B(a, r)) \geq (\alpha - \varepsilon) \mu(B(a, r)) \}. \end{align*} By the [Vitali covering theorem for Radon measures](/theorems/???) applied with respect to $\mu$, there exists a countable pairwise disjoint subfamily $\mathcal{G}' \subset \mathcal{F}'$ such that \begin{align*} \mu\!\Bigl(A \setminus \bigcup_{B \in \mathcal{G}'} B\Bigr) = 0. \end{align*} Since $\mu(A \setminus \bigcup \mathcal{G}') = 0$ and the balls are pairwise disjoint, $\mu(A) = \mu(\bigcup \mathcal{G}') = \sum_{B \in \mathcal{G}'} \mu(B)$. Since $\bigcup \mathcal{G}' \subset U$, monotonicity and disjointness give: \begin{align*} \nu(U) \geq \nu\!\Bigl(\bigcup_{B \in \mathcal{G}'} B\Bigr) = \sum_{B \in \mathcal{G}'} \nu(B) \geq (\alpha - \varepsilon) \sum_{B \in \mathcal{G}'} \mu(B) = (\alpha - \varepsilon) \, \mu(A). \end{align*} Since $\nu$ is a Radon measure, it is [page:1251|outer regular]: $\nu(A) = \inf\{\nu(U) \mid U \supset A,\, U \text{ open}\}$. Taking the infimum over all open $U \supset A$: \begin{align*} \nu(A) \geq (\alpha - \varepsilon) \, \mu(A). \end{align*} Sending $\varepsilon \to 0$ gives $\nu(A) \geq \alpha \, \mu(A)$, completing the proof of part (ii). [guided] The lower bound argument mirrors the upper bound, but with a crucial difference in which measure the covering theorem is applied to. Fix $\varepsilon > 0$ and an open $U \supset A$. For each $a \in A$, since $D_\mu \nu(a) \geq \alpha$, there exist balls $B(a, r) \subset U$ of arbitrarily small radius with \begin{align*} \nu(B(a, r)) \geq (\alpha - \varepsilon) \mu(B(a, r)). \end{align*} The collection $\mathcal{F}'$ of such balls is a Vitali cover of $A$. We define \begin{align*} \mathcal{F}' := \{ B(a, r) \subset U \mid a \in A,\; \nu(B(a, r)) \geq (\alpha - \varepsilon) \mu(B(a, r)) \}. \end{align*} Why do we apply the Vitali covering theorem with respect to $\mu$ here, rather than $\nu$? Because the inequality we seek is a lower bound on $\nu(A)$ in terms of $\mu(A)$. We need the disjoint subcover to capture all the $\mu$-mass of $A$, so we require $\mu(A \setminus \bigcup \mathcal{G}') = 0$. The [Vitali covering theorem for Radon measures](/theorems/???) can be applied with respect to any Radon measure; here we apply it with $\mu$. The extracted disjoint subfamily $\mathcal{G}' \subset \mathcal{F}'$ satisfies $\mu(A \setminus \bigcup \mathcal{G}') = 0$. Since the balls are pairwise disjoint: \begin{align*} \mu(A) = \mu\!\Bigl(\bigcup_{B \in \mathcal{G}'} B\Bigr) = \sum_{B \in \mathcal{G}'} \mu(B). \end{align*} Now, can we conclude $\nu(A) \geq \sum \nu(B)$ directly? No — the balls in $\mathcal{G}'$ are centred at points of $A$ but extend beyond $A$, so $\bigcup \mathcal{G}' \not\subseteq A$ in general and monotonicity does not give $\nu(A) \geq \nu(\bigcup \mathcal{G}')$. Instead, we use the open set $U$. Since $\bigcup \mathcal{G}' \subset U$, monotonicity and disjointness give: \begin{align*} \nu(U) \geq \nu\!\Bigl(\bigcup_{B \in \mathcal{G}'} B\Bigr) = \sum_{B \in \mathcal{G}'} \nu(B) \geq (\alpha - \varepsilon) \sum_{B \in \mathcal{G}'} \mu(B) = (\alpha - \varepsilon) \, \mu(A). \end{align*} Since $\nu$ is a Radon measure, it is [page:1251|outer regular]: $\nu(A) = \inf\{\nu(U) \mid U \supset A,\, U \text{ open}\}$. Taking the infimum over all open $U \supset A$ gives $\nu(A) \geq (\alpha - \varepsilon) \mu(A)$. Sending $\varepsilon \to 0$ yields \begin{align*} \nu(A) \geq \alpha \, \mu(A), \end{align*} which is part (ii). [/guided] [/step]