[proofplan] We use the Besicovitch differentiation machinery developed in Chapter 5. Define the set $S = \{x \in \mathbb{R}^n : D_\mu \nu(x) = +\infty\}$ and its complement. The [Differentiation Recovers the Lebesgue Decomposition](/theorems/3028) theorem provides the existence of $D_\mu \nu$ $\mu$-a.e. and shows that the singular part concentrates on $S$. We define $\nu_{ac}(A) = \int_A D_\mu \nu \, d\mu$ and $\nu_s = \nu - \nu_{ac}$, verify that $\nu_{ac} \ll \mu$ by construction and $\nu_s \perp \mu$ because $\nu_s$ is supported on the $\mu$-null set $S$, and prove uniqueness by showing any other decomposition must agree with this one $\mu$-a.e. [/proofplan] [step:Establish existence and properties of the derivative $D_\mu \nu$ via the Besicovitch differentiation theorem] By the [Differentiation Recovers the Lebesgue Decomposition](/theorems/3028), since $\mu$ and $\nu$ are Radon measures on $\mathbb{R}^n$, the derivative \begin{align*} D_\mu \nu(x) = \lim_{r \to 0^+} \frac{\nu(\overline{B}(x, r))}{\mu(\overline{B}(x, r))} \end{align*} exists and is finite for $\mu$-almost every $x \in \mathbb{R}^n$. Moreover, $D_\mu \nu$ is $\mu$-measurable and belongs to $L^1_{\mathrm{loc}}(\mathbb{R}^n, \mu)$. Define the set where the derivative is infinite: \begin{align*} S = \{x \in \mathbb{R}^n : D_\mu \nu(x) = +\infty\}. \end{align*} By the [Differentiation Recovers the Lebesgue Decomposition](/theorems/3028), $\mu(S) = 0$. [/step] [step:Define the absolutely continuous part $\nu_{ac}$ and the singular part $\nu_s$] Define the function $f : \mathbb{R}^n \to [0, \infty)$ by $f(x) = D_\mu \nu(x)$ for $x \notin S$ and $f(x) = 0$ for $x \in S$. Since $\mu(S) = 0$, the value of $f$ on $S$ does not affect integrals with respect to $\mu$. Define the set function \begin{align*} \nu_{ac}(A) = \int_A f \, d\mu \end{align*} for every Borel set $A \subset \mathbb{R}^n$. Since $f \geq 0$ and $f \in L^1_{\mathrm{loc}}(\mathbb{R}^n, \mu)$, the set function $\nu_{ac}$ is a well-defined Radon measure on $\mathbb{R}^n$ (it is Borel regular, locally finite because $f$ is locally $\mu$-integrable, and inner regular because $\mu$ is inner regular). By construction, $\nu_{ac} \ll \mu$: if $\mu(A) = 0$, then $\nu_{ac}(A) = \int_A f \, d\mu = 0$. Define $\nu_s = \nu - \nu_{ac}$. By the [Differentiation Recovers the Lebesgue Decomposition](/theorems/3028), for every Borel set $A \subset \mathbb{R}^n$, \begin{align*} \nu(A) = \int_A D_\mu \nu \, d\mu + \nu_s(A), \end{align*} where $\nu_s$ is a positive Radon measure. This gives $\nu = \nu_{ac} + \nu_s$. [/step] [step:Verify that $\nu_s \perp \mu$ by showing $\nu_s$ is supported on the $\mu$-null set $S$] We show that $\nu_s(\mathbb{R}^n \setminus S) = 0$. By the [Differentiation Recovers the Lebesgue Decomposition](/theorems/3028), $D_\mu \nu_s(x) = 0$ for $\mu$-a.e. $x$. Moreover, $D_\mu \nu_s(x) = +\infty$ for $\nu_s$-a.e. $x$. This means $\nu_s$ is concentrated on the set where $D_\mu \nu_s = +\infty$, which is contained in $S$ (the set where $D_\mu \nu = +\infty$, since $\nu_s \leq \nu$ implies $D_\mu \nu_s \leq D_\mu \nu$ wherever both are defined). More precisely, define $S^c = \mathbb{R}^n \setminus S = \{x : D_\mu \nu(x) < +\infty\}$. For $\nu_s$-a.e. $x$, the derivative $D_\mu \nu_s(x) = +\infty$, so any such $x$ must have $D_\mu \nu(x) = +\infty$ (since $\nu_s \leq \nu$ and thus $D_\mu \nu_s \leq D_\mu \nu$ where both limits exist). Therefore $\nu_s$-almost every $x$ belongs to $S$, which gives $\nu_s(S^c) = 0$, i.e., $\nu_s(\mathbb{R}^n \setminus S) = 0$. Since $\mu(S) = 0$ and $\nu_s(\mathbb{R}^n \setminus S) = 0$, the measures $\mu$ and $\nu_s$ are mutually singular: $\nu_s \perp \mu$. [/step] [step:Prove uniqueness of the decomposition] Suppose $\nu = \tilde{\nu}_{ac} + \tilde{\nu}_s$ is another decomposition with $\tilde{\nu}_{ac} \ll \mu$ and $\tilde{\nu}_s \perp \mu$. Since $\tilde{\nu}_s \perp \mu$, there exists a Borel set $\tilde{E}$ with $\mu(\tilde{E}) = 0$ and $\tilde{\nu}_s(\mathbb{R}^n \setminus \tilde{E}) = 0$. Then $\nu_{ac} - \tilde{\nu}_{ac} = \tilde{\nu}_s - \nu_s$. The left side is absolutely continuous with respect to $\mu$ (since both $\nu_{ac} \ll \mu$ and $\tilde{\nu}_{ac} \ll \mu$). The right side is singular with respect to $\mu$: it is supported on the $\mu$-null set $S \cup \tilde{E}$, since $\nu_s(\mathbb{R}^n \setminus S) = 0$ and $\tilde{\nu}_s(\mathbb{R}^n \setminus \tilde{E}) = 0$. A signed measure that is simultaneously absolutely continuous and singular with respect to $\mu$ must be the zero measure. To verify: let $\lambda = \nu_{ac} - \tilde{\nu}_{ac}$. Then $\lambda \ll \mu$ and $\lambda$ is supported on a $\mu$-null set $N = S \cup \tilde{E}$. For any Borel set $A$, $\lambda(A) = \lambda(A \cap N) + \lambda(A \setminus N) = \lambda(A \cap N)$ (since $\lambda$ is supported on $N$). But $\mu(A \cap N) \leq \mu(N) = 0$, and $\lambda \ll \mu$ gives $\lambda(A \cap N) = 0$. Hence $\lambda = 0$, so $\nu_{ac} = \tilde{\nu}_{ac}$ and $\nu_s = \tilde{\nu}_s$. [/step] [step:Identify the density function as the Besicovitch derivative] By definition, $\nu_{ac} = f \cdot \mu$ where $f = D_\mu \nu$ $\mu$-a.e. The [Differentiation Theorem for Radon Measures](/theorems/3027) applied to $\nu_{ac}$ and $\mu$ shows that for $\mu$-a.e. $x$, \begin{align*} D_\mu \nu_{ac}(x) = \lim_{r \to 0^+} \frac{\nu_{ac}(B(x,r))}{\mu(B(x,r))} = f(x). \end{align*} Since $D_\mu \nu_s(x) = 0$ for $\mu$-a.e. $x$ (established in the singularity step), the additivity of limits gives \begin{align*} D_\mu \nu(x) = D_\mu \nu_{ac}(x) + D_\mu \nu_s(x) = f(x) + 0 = f(x) \end{align*} for $\mu$-a.e. $x$. This confirms that the density of $\nu_{ac}$ with respect to $\mu$ is recovered by the pointwise differentiation formula $f(x) = \lim_{r \to 0} \nu(B(x,r))/\mu(B(x,r))$. [/step]