[proofplan] Since $\nu = \nu_{ac} + \nu_s$ with $\nu_{ac} \ll \mu$ and $\nu_s \perp \mu$, we handle each component separately. For $\nu_{ac}$, the [Differentiation Theorem for Radon Measures](/theorems/3027) applies directly and gives existence, finiteness, and the integration formula. For $\nu_s$, we show $D_\mu \nu_s = 0$ $\mu$-a.e. by using the mutual singularity to find a Borel set $S$ with $\mu(S) = 0$ and $\nu_s(\mathbb{R}^n \setminus S) = 0$, then applying the [Besicovitch-Type Measure Estimate](/theorems/3026) to bound $\mu(\{D_\mu \nu_s > \varepsilon\})$ for each $\varepsilon > 0$. The statement $D_\mu \nu_s = +\infty$ $\nu_s$-a.e. follows by a symmetric argument using Besicovitch with the roles of $\mu$ and $\nu_s$ exchanged. The five parts then combine via linearity of differentiation. [/proofplan] [step:Apply the Differentiation Theorem to the absolutely continuous part $\nu_{ac}$] Since $\nu_{ac} \ll \mu$ and both $\mu$ and $\nu_{ac}$ are Radon measures on $\mathbb{R}^n$, the [Differentiation Theorem for Radon Measures](/theorems/3027) applies to the pair $(\mu, \nu_{ac})$. It yields: - $D_\mu \nu_{ac}(x)$ exists and is finite for $\mu$-almost every $x$. - $D_\mu \nu_{ac} \in L^1_{\mathrm{loc}}(\mathbb{R}^n, \mu)$. - For every Borel set $A$: $\nu_{ac}(A) = \int_A D_\mu \nu_{ac} \, d\mu$. - $D_\mu \nu_{ac}(x) = \frac{d\nu_{ac}}{d\mu}(x)$ for $\mu$-almost every $x$. [/step] [step:Show $D_\mu \nu_s(x) = 0$ for $\mu$-almost every $x$] Since $\nu_s \perp \mu$, there exists a Borel set $S \subset \mathbb{R}^n$ with \begin{align*} \mu(S) = 0 \quad \text{and} \quad \nu_s(\mathbb{R}^n \setminus S) = 0. \end{align*} Define, for $\varepsilon > 0$: \begin{align*} G_\varepsilon = \{x \in \mathbb{R}^n \setminus S : \overline{D}_\mu \nu_s(x) > \varepsilon\}. \end{align*} We show $\mu(G_\varepsilon) = 0$. By the [Besicovitch-Type Measure Estimate](/theorems/3026) applied to the measures $\mu$ and $\nu_s$: \begin{align*} \nu_s(G_\varepsilon) \geq \frac{\varepsilon}{N(n)} \mu(G_\varepsilon). \end{align*} Since $G_\varepsilon \subset \mathbb{R}^n \setminus S$ and $\nu_s(\mathbb{R}^n \setminus S) = 0$, we have $\nu_s(G_\varepsilon) = 0$. Therefore $\mu(G_\varepsilon) = 0$. The set $\{x \in \mathbb{R}^n \setminus S : \overline{D}_\mu \nu_s(x) > 0\} = \bigcup_{k=1}^\infty G_{1/k}$ is a countable union of $\mu$-null sets, hence $\mu$-null. Since $\mu(S) = 0$, we conclude \begin{align*} \overline{D}_\mu \nu_s(x) = 0 \quad \text{for } \mu\text{-almost every } x \in \mathbb{R}^n. \end{align*} In particular, $D_\mu \nu_s(x) = 0$ for $\mu$-almost every $x$ (since $0 \leq \underline{D}_\mu \nu_s(x) \leq \overline{D}_\mu \nu_s(x) = 0$ forces the limit to exist and equal zero). [/step] [step:Show $D_\mu \nu_s(x) = +\infty$ for $\nu_s$-almost every $x$] We use the same Borel set $S$ with $\mu(S) = 0$ and $\nu_s(\mathbb{R}^n \setminus S) = 0$. For $M > 0$, define \begin{align*} H_M = \{x \in S : \overline{D}_\mu \nu_s(x) \leq M\}. \end{align*} We show $\nu_s(H_M) = 0$. For each $x \in H_M$, the condition $\overline{D}_\mu \nu_s(x) \leq M$ means: for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $0 < r < \delta$, \begin{align*} \nu_s(\overline{B}(x, r)) \leq (M + \varepsilon) \mu(\overline{B}(x, r)). \end{align*} Actually, since $\overline{D}_\mu \nu_s(x) \leq M$, for all sufficiently small $r$ we have $\nu_s(\overline{B}(x,r)) \leq (M + 1) \mu(\overline{B}(x,r))$. Let $K \subset H_M$ be compact and $\delta > 0$. For each $x \in K$, choose $r_x \leq \delta$ with $\nu_s(\overline{B}(x, r_x)) \leq (M+1) \mu(\overline{B}(x, r_x))$. Apply the [Besicovitch Covering Theorem](/theorems/3021) to $K$: there exist subfamilies $\mathcal{G}_1, \ldots, \mathcal{G}_N$ covering $K$, each consisting of pairwise disjoint balls. For each $j$, summing over the disjoint balls in $\mathcal{G}_j$: \begin{align*} \nu_s\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) = \sum_B \nu_s(B) \leq (M+1) \sum_B \mu(B) = (M+1) \mu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right). \end{align*} Since $K \subset \bigcup_{j=1}^N \bigcup_{B \in \mathcal{G}_j} B$ and the overlap is at most $N(n)$: \begin{align*} \nu_s(K) \leq \sum_{j=1}^N \nu_s\!\left(\bigcup_{B \in \mathcal{G}_j} B\right) \leq (M+1) \sum_{j=1}^N \mu\!\left(\bigcup_{B \in \mathcal{G}_j} B\right). \end{align*} Let $V_\delta = \bigcup_{j=1}^N \bigcup_{B \in \mathcal{G}_j} B$. By bounded overlap, $\sum_{j=1}^N \mu(\bigcup_{B \in \mathcal{G}_j} B) \leq N(n) \mu(V_\delta)$. Since all balls have radius $\leq \delta$ and are centered in $K \subset S$, as $\delta \to 0$ the set $V_\delta$ shrinks toward $K \subset S$. By outer regularity, $\mu(V_\delta) \to \mu(K)$. But $K \subset S$ and $\mu(S) = 0$, so $\mu(K) = 0$. Therefore $\nu_s(K) = 0$. By [inner regularity](/theorems/3011) of $\nu_s$: $\nu_s(H_M) = \sup\{\nu_s(K) : K \subset H_M, K \text{ compact}\} = 0$. Since $\{x \in S : D_\mu \nu_s(x) < +\infty\} = \bigcup_{M=1}^\infty H_M$, countable subadditivity gives $\nu_s(\{x \in S : D_\mu \nu_s(x) < +\infty\}) = 0$. Since $\nu_s(\mathbb{R}^n \setminus S) = 0$, we conclude $D_\mu \nu_s(x) = +\infty$ for $\nu_s$-almost every $x$. [/step] [step:Combine the results to prove all five parts] **(i)** By linearity of differentiation, $D_\mu \nu = D_\mu \nu_{ac} + D_\mu \nu_s$ wherever both limits exist. The limit $D_\mu \nu_{ac}$ exists and is finite for $\mu$-a.e. $x$ (from Step 1). The limit $D_\mu \nu_s$ exists and equals $0$ for $\mu$-a.e. $x$ (from Step 2). Therefore $D_\mu \nu(x) = D_\mu \nu_{ac}(x) + 0 = D_\mu \nu_{ac}(x)$ exists and is finite for $\mu$-almost every $x$. **(ii)** From (i): $D_\mu \nu(x) = D_\mu \nu_{ac}(x) = \frac{d\nu_{ac}}{d\mu}(x)$ for $\mu$-almost every $x$, where the last equality is from the [Differentiation Theorem](/theorems/3027) applied to $\nu_{ac}$. **(iii)** Proved in Step 2: $D_\mu \nu_s(x) = 0$ for $\mu$-almost every $x$. **(iv)** Proved in Step 3: $D_\mu \nu_s(x) = +\infty$ for $\nu_s$-almost every $x$. **(v)** For every Borel set $A$: \begin{align*} \nu(A) = \nu_{ac}(A) + \nu_s(A) = \int_A D_\mu \nu_{ac} \, d\mu + \nu_s(A) = \int_A D_\mu \nu \, d\mu + \nu_s(A), \end{align*} where the second equality uses the integration formula from the [Differentiation Theorem](/theorems/3027) applied to $\nu_{ac}$, and the third uses $D_\mu \nu = D_\mu \nu_{ac}$ $\mu$-a.e. from (ii). [/step]