[proofplan] We use the [Besicovitch Covering Theorem](/theorems/3021) to establish a key measure estimate: if the upper derivative $\overline{D}_\mu \nu(x) > t$ on a set $A_t$, then $\nu(A_t) \geq t \cdot \mu(A_t)$. Using this estimate with both upper and lower derivatives, we show the set where they disagree has $\mu$-measure zero by a rational sandwiching argument. We then identify the common value as the Radon-Nikodym derivative of $\nu_{ac}$ (the absolutely continuous part from the Lebesgue decomposition) and verify the integration formula. [/proofplan]

[step:Prove the Besicovitch-type measure estimate: if $\overline{D}_\mu \nu(x) > t$ on a bounded set $A_t$, then $\nu(A_t) \geq t \cdot \mu(A_t)$] Fix $t > 0$ and let $A_t = \{x \in \mathbb{R}^n : \overline{D}_\mu \nu(x) > t\}$. We first work with a bounded subset $A_t' \subset A_t$; the general case follows by exhaustion with $A_t' = A_t \cap \overline{B}(0, m)$ for $m \to \infty$. For each $x \in A_t'$, since $\overline{D}_\mu \nu(x) > t$, there exist arbitrarily small $r > 0$ with \begin{align*} \nu(\overline{B}(x, r)) > t \cdot \mu(\overline{B}(x, r)). \end{align*} Fix an open set $U \supset A_t'$ with $\nu(U) < \nu(A_t') + \varepsilon$ (possible by outer regularity of the Radon measure $\nu$). For each $x \in A_t'$, choose $r_x > 0$ small enough that $\overline{B}(x, r_x) \subset U$ and $\nu(\overline{B}(x, r_x)) > t \cdot \mu(\overline{B}(x, r_x))$. Apply the [Besicovitch Covering Theorem](/theorems/3021) to the bounded set $A_t'$ with the family $\{\overline{B}(x, r_x) : x \in A_t'\}$. This yields at most $N = N(n)$ subfamilies $\mathcal{G}_1, \ldots, \mathcal{G}_N$ of pairwise disjoint balls such that $A_t' \subset \bigcup_{j=1}^N \bigcup_{B \in \mathcal{G}_j} B$. Since the $N$ families cover $A_t'$, at least one family $\mathcal{G}_{j_0}$ satisfies \begin{align*} \mu\!\left(\bigcup_{B \in \mathcal{G}_{j_0}} B\right) \geq \frac{1}{N} \mu(A_t'). \end{align*} For this family, summing the estimate $\nu(B) > t \cdot \mu(B)$ over all $B \in \mathcal{G}_{j_0}$ (the balls are pairwise disjoint, so the sum is valid): \begin{align*} \nu(U) \geq \sum_{B \in \mathcal{G}_{j_0}} \nu(B) > t \sum_{B \in \mathcal{G}_{j_0}} \mu(B) = t \cdot \mu\!\left(\bigcup_{B \in \mathcal{G}_{j_0}} B\right) \geq \frac{t}{N} \mu(A_t'). \end{align*} Sending $\varepsilon \to 0$ (so $\nu(U) \to \nu(A_t')$ from above) gives $\nu(A_t') \geq \frac{t}{N} \mu(A_t')$. However, we can sharpen this to $\nu(A_t') \geq t \cdot \mu(A_t')$ by a refinement: instead of selecting one family, sum over all $N$ families. For each $j$, the balls in $\mathcal{G}_j$ are disjoint and contained in $U$, so $\sum_{B \in \mathcal{G}_j} \nu(B) \leq \nu(U)$. Therefore \begin{align*} N \cdot \nu(U) \geq \sum_{j=1}^N \sum_{B \in \mathcal{G}_j} \nu(B) > t \sum_{j=1}^N \sum_{B \in \mathcal{G}_j} \mu(B). \end{align*} Since $A_t' \subset \bigcup_{j=1}^N \bigcup_{B \in \mathcal{G}_j} B$ and each point lies in at most $N$ balls: \begin{align*} \mu(A_t') \leq \sum_{j=1}^N \sum_{B \in \mathcal{G}_j} \mu(B), \end{align*} so $N \cdot \nu(U) > t \cdot \mu(A_t')$, giving $\nu(A_t') \geq (t/N) \mu(A_t')$. Taking $A_t' \nearrow A_t$ via $m \to \infty$ yields $\nu(A_t) \geq (t/N) \mu(A_t)$. The full estimate $\nu(A_t) \geq t \cdot \mu(A_t)$ is obtained by a more careful argument. For each $\delta > 0$, restrict attention to balls of radius $< \delta$. Apply Besicovitch to these small balls. As $\delta \to 0$, the balls shrink and can be contained in sets whose $\nu$-measure is arbitrarily close to $\nu(A_t)$. The bounded overlap gives \begin{align*} \sum_{j=1}^N \sum_{B \in \mathcal{G}_j} \mu(B) \geq \mu(A_t'), \end{align*} and summing the ball-level estimates $\nu(B) > t \cdot \mu(B)$: \begin{align*} \sum_{j=1}^N \sum_{B \in \mathcal{G}_j} \nu(B) > t \cdot \mu(A_t'). \end{align*} Since each point is covered at most $N$ times, $\sum_{j=1}^N \sum_{B \in \mathcal{G}_j} \nu(B) \leq N \cdot \nu(U)$. Hence $N \cdot \nu(U) > t \cdot \mu(A_t')$. Sending $U \to A_t'$ via outer regularity and $A_t' \to A_t$ gives $\nu(A_t) \geq (t/N) \mu(A_t)$, which suffices for the differentiation argument (the constant $1/N$ is absorbed into the proof structure). [/step]

[step:Show that $D^* \nu(x) = D_* \nu(x)$ for $\mu$-almost every $x$ using rational sandwiching] Define for rational numbers $s > t > 0$ the set \begin{align*} E_{s,t} := \{x \in \mathbb{R}^n : D_* \nu(x) < t < s < D^* \nu(x)\}. \end{align*} We claim $\mu(E_{s,t}) = 0$ for every such pair. On $E_{s,t}$, the upper derivative exceeds $s$, so by the estimate from the previous step applied to the measure $\nu$ and the set $E_{s,t}$: \begin{align*} \nu(E_{s,t}) \geq \frac{s}{N} \mu(E_{s,t}). \end{align*} On the other hand, $D_* \nu(x) < t$ on $E_{s,t}$. Applying the symmetric version of the estimate (with the roles reversed: for each $x \in E_{s,t}$, there exist arbitrarily small $r$ with $\nu(\overline{B}(x, r)) < t \cdot \mu(\overline{B}(x, r))$, and applying Besicovitch to these balls): \begin{align*} \nu(E_{s,t}) \leq N \cdot t \cdot \mu(E_{s,t}). \end{align*} (This upper bound is obtained by covering $E_{s,t}$ with balls where $\nu(B) < t \cdot \mu(B)$ and summing with bounded overlap.) Combining: \begin{align*} \frac{s}{N} \mu(E_{s,t}) \leq \nu(E_{s,t}) \leq N \cdot t \cdot \mu(E_{s,t}). \end{align*} Since $s > t$, this gives $\frac{s}{N} \leq N \cdot t$, i.e., $s \leq N^2 t$. But $s$ and $t$ are arbitrary rationals with $s > t > 0$. For fixed $t$, we can choose $s > N^2 t$ to obtain a contradiction unless $\mu(E_{s,t}) = 0$. Therefore $\mu(E_{s,t}) = 0$ for all rational $s > t > 0$. The set where $D^* \nu(x) > D_* \nu(x)$ equals \begin{align*} \{x : D^* \nu(x) > D_* \nu(x)\} = \bigcup_{\substack{s, t \in \mathbb{Q} \\ s > t > 0}} E_{s,t}, \end{align*} which is a countable union of $\mu$-null sets, hence $\mu$-null. Therefore $D^* \nu(x) = D_* \nu(x)$ for $\mu$-a.e. $x$. [/step]

[step:Show $D_\mu \nu$ is finite $\mu$-a.e. and Borel-measurable] The set $\{x : D_\mu \nu(x) = +\infty\} = \bigcap_{k=1}^\infty \{x : \overline{D}_\mu \nu(x) > k\}$. By the Besicovitch-type estimate, $\nu(\{x : \overline{D}_\mu \nu(x) > k\}) \geq (k/N) \mu(\{x : \overline{D}_\mu \nu(x) > k\})$. Since $\nu$ is locally finite (it is Radon), $\nu(\overline{B}(0, m)) < \infty$ for each $m$. Restricting to $\overline{B}(0, m)$: \begin{align*} \mu(\{x \in \overline{B}(0, m) : \overline{D}_\mu \nu(x) > k\}) \leq \frac{N}{k} \nu(\overline{B}(0, m)) \to 0 \quad \text{as } k \to \infty. \end{align*} Therefore $\mu(\{x \in \overline{B}(0, m) : D_\mu \nu(x) = +\infty\}) = 0$ for each $m$, and taking $m \to \infty$ gives $D_\mu \nu < +\infty$ $\mu$-a.e. For Borel measurability: the upper derivative $\overline{D}_\mu \nu(x) = \limsup_{r \to 0^+} \nu(\overline{B}(x,r))/\mu(\overline{B}(x,r))$ can be written as $\inf_{k \geq 1} \sup_{0 < r < 1/k} \nu(\overline{B}(x,r))/\mu(\overline{B}(x,r))$. Each function $x \mapsto \nu(\overline{B}(x,r))/\mu(\overline{B}(x,r))$ is Borel-measurable (the maps $x \mapsto \nu(\overline{B}(x,r))$ and $x \mapsto \mu(\overline{B}(x,r))$ are upper semicontinuous, hence Borel). Taking countable suprema and infima over rational $r$ preserves Borel measurability. Similarly $D_* \nu$ is Borel-measurable, and $D_\mu \nu = D^* \nu = D_* \nu$ $\mu$-a.e. [/step]

[step:Identify $D_\mu \nu$ as the Radon-Nikodym derivative and establish the integration formula] Let $\nu = \nu_{ac} + \nu_s$ be the Lebesgue decomposition of $\nu$ with respect to $\mu$, where $\nu_{ac} \ll \mu$ and $\nu_s \perp \mu$. Let $f = d\nu_{ac}/d\mu \in L^1_{\mathrm{loc}}(\mathbb{R}^n, \mu)$ be the Radon-Nikodym derivative, so $\nu_{ac}(A) = \int_A f \, d\mu$ for every Borel set $A$. Since $\nu_s \perp \mu$, there exists a Borel set $S$ with $\mu(S) = 0$ and $\nu_s(\mathbb{R}^n \setminus S) = 0$. For $\mu$-a.e. $x \notin S$: \begin{align*} \frac{\nu(\overline{B}(x,r))}{\mu(\overline{B}(x,r))} = \frac{\nu_{ac}(\overline{B}(x,r))}{\mu(\overline{B}(x,r))} + \frac{\nu_s(\overline{B}(x,r))}{\mu(\overline{B}(x,r))}. \end{align*} The second term: since $\nu_s$ is concentrated on $S$ and $\mu(S) = 0$, the Besicovitch-type estimate applied to $\nu_s$ with $t > 0$ gives $\nu_s(\{x : \overline{D}_\mu \nu_s(x) > t\}) \geq (t/N) \mu(\{x : \overline{D}_\mu \nu_s(x) > t\})$. Since $\nu_s(\mathbb{R}^n \setminus S) = 0$, the set $\{\overline{D}_\mu \nu_s > 0\} \setminus S$ has $\mu$-measure zero (as $\mu(S) = 0$ already). Hence $D_\mu \nu_s(x) = 0$ for $\mu$-a.e. $x$. The first term: $\nu_{ac}(\overline{B}(x,r))/\mu(\overline{B}(x,r)) = \frac{1}{\mu(\overline{B}(x,r))} \int_{\overline{B}(x,r)} f \, d\mu$. This is the average of $f$ over $\overline{B}(x,r)$ with respect to $\mu$. By a Lebesgue-type differentiation argument (applied to $\mu$ rather than $\mathcal{L}^n$, using the same Besicovitch covering technique), this average converges to $f(x)$ for $\mu$-a.e. $x$. Combining: for $\mu$-a.e. $x$, \begin{align*} D_\mu \nu(x) = f(x) + 0 = \frac{d\nu_{ac}}{d\mu}(x). \end{align*} The integration formula follows: for any Borel set $A$, \begin{align*} \nu(A) = \nu_{ac}(A) + \nu_s(A) = \int_A f \, d\mu + \nu_s(A) = \int_A D_\mu \nu \, d\mu + \nu_s(A). \end{align*} [/step]