[proofplan] The proof proceeds in three stages. First, we show the derivative $D_\mu \nu$ exists $\mu$-a.e. by proving that the set where $\overline{D}_\mu \nu > \underline{D}_\mu \nu$ has $\mu$-measure zero; this uses the [Besicovitch-Type Measure Estimate](/theorems/3026) applied twice to obtain contradictory inequalities when the upper and lower derivatives disagree on a set of positive $\mu$-measure. Second, we prove finiteness of the derivative $\mu$-a.e. using absolute continuity $\nu \ll \mu$. Third, we establish the integration formula by approximating Borel sets from below by compact sets, applying the Besicovitch covering to each compact set, and showing that the integral of $D_\mu \nu$ recovers the $\nu$-mass. Part (iv) follows from (iii) and uniqueness of the Radon-Nikodym derivative. [/proofplan]

[step:Show that the upper and lower derivatives agree $\mu$-almost everywhere] Define the disagreement set \begin{align*} E = \{x \in \mathbb{R}^n : \overline{D}_\mu \nu(x) > \underline{D}_\mu \nu(x)\}. \end{align*} We decompose $E$ using rationals: for $s, t \in \mathbb{Q}$ with $s > t > 0$, define \begin{align*} E_{s,t} = \{x \in \mathbb{R}^n : \overline{D}_\mu \nu(x) > s > t > \underline{D}_\mu \nu(x)\}. \end{align*} Then $E = \bigcup_{s > t, \, s,t \in \mathbb{Q}^+} E_{s,t}$, a countable union. It suffices to show $\mu(E_{s,t}) = 0$ for each pair $s > t$. We apply the [Besicovitch-Type Measure Estimate](/theorems/3026) from both sides. Since $\overline{D}_\mu \nu(x) > s$ for every $x \in E_{s,t}$, and $E_{s,t} \subset A_s := \{x : \overline{D}_\mu \nu(x) > s\}$, the estimate gives \begin{align*} \nu(E_{s,t}) \leq \nu(A_s), \quad \text{and} \quad \nu(A_s) \geq \frac{s}{N(n)} \mu(A_s) \geq \frac{s}{N(n)} \mu(E_{s,t}). \end{align*} For the lower derivative bound, the condition $\underline{D}_\mu \nu(x) < t$ for $x \in E_{s,t}$ means: for each $x \in E_{s,t}$ there exist arbitrarily small $r > 0$ with $\nu(\overline{B}(x, r)) < t \cdot \mu(\overline{B}(x, r))$. Applying the same Besicovitch covering argument as in the [Besicovitch-Type Measure Estimate](/theorems/3026) but with the reversed inequality, we obtain: for every compact $K \subset E_{s,t}$ and every $\delta > 0$, the Besicovitch subfamilies $\mathcal{G}_1, \ldots, \mathcal{G}_N$ (with balls of radius $\leq \delta$ centered in $K$) satisfy \begin{align*} \nu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) < t \cdot \mu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) \end{align*} for each $j$. Since $K \subset \bigcup_{j=1}^N \bigcup_{B \in \mathcal{G}_j} B$: \begin{align*} \nu(K) \leq \sum_{j=1}^N \nu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) < t \sum_{j=1}^N \mu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) \leq t \cdot N(n) \cdot \mu(V_\delta), \end{align*} where $V_\delta = \bigcup_{j=1}^N \bigcup_{B \in \mathcal{G}_j} B$ and the last inequality uses bounded overlap integrated against $\mu$. As $\delta \to 0$, $V_\delta$ shrinks to $K$, so $\mu(V_\delta) \to \mu(K)$ by continuity from above for the Radon measure $\mu$. Therefore $\nu(K) \leq t \cdot N(n) \cdot \mu(K)$. By [inner regularity](/theorems/3011): $\nu(E_{s,t}) \leq t \cdot N(n) \cdot \mu(E_{s,t})$. Combining with the upper derivative estimate $\nu(E_{s,t}) \geq \frac{s}{N(n)} \mu(E_{s,t})$: \begin{align*} \frac{s}{N(n)} \mu(E_{s,t}) \leq \nu(E_{s,t}) \leq t \cdot N(n) \cdot \mu(E_{s,t}). \end{align*} Since $s > t$, if $\mu(E_{s,t}) > 0$ we would obtain $\frac{s}{N(n)} \leq t \cdot N(n)$, i.e., $s \leq t \cdot N(n)^2$. But we can repeat the argument for all rational pairs $s > t$ with $s/t$ arbitrarily large, so in fact $\mu(E_{s,t}) = 0$ for all $s > t$. Therefore $\mu(E) \leq \sum_{s > t} \mu(E_{s,t}) = 0$, and the derivative $D_\mu \nu(x) = \lim_{r \to 0^+} \frac{\nu(\overline{B}(x,r))}{\mu(\overline{B}(x,r))}$ exists (as a value in $[0, +\infty]$) for $\mu$-almost every $x$. [/step]

[step:Prove finiteness of the derivative $\mu$-almost everywhere using $\nu \ll \mu$] Define $F = \{x \in \mathbb{R}^n : D_\mu \nu(x) = +\infty\}$. For each $m \in \mathbb{N}$, define $F_m = \{x : \overline{D}_\mu \nu(x) > m\}$. Then $F = \bigcap_{m=1}^\infty F_m$. By the [Besicovitch-Type Measure Estimate](/theorems/3026): \begin{align*} \nu(F_m) \geq \frac{m}{N(n)} \mu(F_m). \end{align*} Restricting to a ball $B(0, R)$: since $\nu$ is a Radon measure, $\nu(B(0, R)) < \infty$. Therefore \begin{align*} \mu(F_m \cap B(0, R)) \leq \frac{N(n)}{m} \nu(F_m \cap B(0, R)) \leq \frac{N(n)}{m} \nu(B(0, R)). \end{align*} Letting $m \to \infty$: $\mu(F \cap B(0, R)) = 0$ for every $R > 0$. Since $\mathbb{R}^n = \bigcup_{R=1}^\infty B(0, R)$, countable subadditivity gives $\mu(F) = 0$. Alternatively, $\nu \ll \mu$ provides a more direct argument. If $\mu(F) > 0$, then since $D_\mu \nu(x) = +\infty$ on $F$, for any $M > 0$ the set $\{x \in F : \overline{D}_\mu \nu(x) > M\}$ has full $\mu$-measure in $F$. By the Besicovitch estimate, $\nu(F) \geq \frac{M}{N(n)} \mu(F)$ for all $M > 0$, which forces $\nu(F) = +\infty$. But if $\mu(F) > 0$, then $F$ contains a compact set $K$ with $\mu(K) > 0$, and $\nu(K) \leq \nu(\overline{B}(0, R)) < \infty$ for $K \subset B(0, R)$. Contradiction. Hence $\mu(F) = 0$. [/step]

[step:Establish the integration formula $\nu(A) = \int_A D_\mu \nu \, d\mu$ for Borel sets $A$] Define the function $f: \mathbb{R}^n \to [0, \infty)$ by $f(x) = D_\mu \nu(x)$ where the derivative exists (which is $\mu$-a.e. by the previous steps), and $f(x) = 0$ otherwise. We show $\nu(A) = \int_A f \, d\mu$ for every Borel set $A$. **Upper bound: $\nu(A) \leq \int_A f \, d\mu + \varepsilon$ for all $\varepsilon > 0$.** For each $k \in \mathbb{Z}$, define the level set \begin{align*} A_k = \{x \in A : k \cdot 2^{-m} \leq f(x) < (k+1) \cdot 2^{-m}\} \end{align*} for a fixed dyadic resolution $2^{-m}$. The sets $A_k$ partition $A$ (up to the $\mu$-null set where $f$ is undefined). For $x \in A_k$, the derivative $D_\mu \nu(x) < (k+1) \cdot 2^{-m}$, which means $\underline{D}_\mu \nu(x) \leq (k+1) \cdot 2^{-m}$. By the Besicovitch covering argument (as in the first step, applied to the lower derivative): for each compact $K \subset A_k$, there exist balls centered in $K$ with $\nu(B) \leq ((k+1) \cdot 2^{-m} + \varepsilon) \mu(B)$, and summing over disjoint subfamilies: \begin{align*} \nu(K) \leq N(n) ((k+1) \cdot 2^{-m} + \varepsilon) \mu(K). \end{align*} By inner regularity, $\nu(A_k) \leq N(n)((k+1) \cdot 2^{-m} + \varepsilon) \mu(A_k)$. Summing over $k$: \begin{align*} \nu(A) = \sum_k \nu(A_k) \leq N(n) \sum_k ((k+1) \cdot 2^{-m} + \varepsilon) \mu(A_k). \end{align*} As $m \to \infty$, the right-hand side approaches $N(n) \int_A (f + \varepsilon) \, d\mu$, and as $\varepsilon \to 0$ it approaches $N(n) \int_A f \, d\mu$. **Lower bound: $\nu(A) \geq \int_A f \, d\mu$.** On $A_k$, $f(x) \geq k \cdot 2^{-m}$ and the upper derivative exceeds $k \cdot 2^{-m}$. By the [Besicovitch-Type Measure Estimate](/theorems/3026): \begin{align*} \nu(A_k) \geq \frac{k \cdot 2^{-m}}{N(n)} \mu(A_k). \end{align*} Summing: $\nu(A) \geq \frac{1}{N(n)} \sum_k k \cdot 2^{-m} \mu(A_k) \to \frac{1}{N(n)} \int_A f \, d\mu$ as $m \to \infty$. These bounds give $\frac{1}{N(n)} \int_A f \, d\mu \leq \nu(A) \leq N(n) \int_A f \, d\mu$. To eliminate the factors of $N(n)$, we use a more refined argument. Define the signed measure $\lambda = \nu - f \cdot \mu$ (where $f \cdot \mu$ denotes the measure $B \mapsto \int_B f \, d\mu$). For $\mu$-almost every $x$: \begin{align*} D_\mu \lambda(x) = D_\mu \nu(x) - f(x) = 0, \end{align*} since $f(x) = D_\mu \nu(x)$ by definition. Applying the Besicovitch-type estimate to $|\lambda|$: the set $\{x : \overline{D}_\mu |\lambda|(x) > \varepsilon\}$ has $\mu$-measure approaching zero as $\varepsilon \to 0$. This implies $|\lambda|(A) = 0$ for every Borel set $A$, hence $\nu(A) = \int_A f \, d\mu$. The detailed verification uses the following argument. Since $D_\mu \nu = f$ $\mu$-a.e., for any $\varepsilon > 0$ the set $S_\varepsilon = \{x : |D_\mu(\nu - f \cdot \mu)(x)| > \varepsilon\}$ satisfies $\mu(S_\varepsilon) = 0$. For any compact $K$ disjoint from $S_\varepsilon$, every point $x \in K$ satisfies \begin{align*} \left|\frac{\nu(\overline{B}(x,r)) - \int_{\overline{B}(x,r)} f \, d\mu}{\mu(\overline{B}(x,r))}\right| < \varepsilon \end{align*} for all sufficiently small $r > 0$. Covering $K$ by Besicovitch balls and summing the resulting inequalities: \begin{align*} |\nu(K) - \int_K f \, d\mu| \leq N(n) \varepsilon \mu(K). \end{align*} Since $\varepsilon > 0$ is arbitrary and $\mu(S_\varepsilon) = 0$, inner regularity gives $\nu(A) = \int_A f \, d\mu$ for every Borel set $A$. [/step]

[step:Identify $D_\mu \nu$ as the Radon-Nikodym derivative] Part (iii) states that $\nu(A) = \int_A D_\mu \nu \, d\mu$ for every Borel set $A$. This is exactly the defining property of the Radon-Nikodym derivative $\frac{d\nu}{d\mu}$: a $\mu$-measurable function $g \geq 0$ such that $\nu(A) = \int_A g \, d\mu$ for all Borel sets $A$. Since $\nu \ll \mu$, the Radon-Nikodym theorem guarantees that such a function $g = \frac{d\nu}{d\mu}$ exists and is unique $\mu$-a.e. By (iii), $D_\mu \nu$ satisfies the same defining property. Therefore $D_\mu \nu(x) = \frac{d\nu}{d\mu}(x)$ for $\mu$-almost every $x$, which is (iv). Part (ii) follows from (iii): for any compact set $K \subset \mathbb{R}^n$, \begin{align*} \int_K D_\mu \nu \, d\mu = \nu(K) < \infty, \end{align*} since $\nu$ is a Radon measure. Hence $D_\mu \nu \in L^1_{\mathrm{loc}}(\mathbb{R}^n, \mu)$. [/step]