[proofplan] The proof has three stages. **Stage 1** (Steps 1-4) identifies the multi-valued subdifferential $\partial f$ with the single-valued classical gradient $\nabla f$ on the full-measure set $\Omega_2$ of Alexandrov twice-differentiability points, and produces a Kirszbraun-extended Lipschitz representative $\widetilde{\nabla f}$ on small balls together with the identification of its derivative with the Alexandrov Hessian. **Stage 2** (Steps 5-6) computes the Lebesgue density $\lim_{r \to 0^+} \mu_f(B(x_0, r))/\mathcal{L}^n(B(x_0, r)) = \det D^2 f(x_0)$ at every $x_0 \in \Omega_2$ that is also a Lebesgue point of $\det D^2 f$: Case 1 uses the area formula in the non-degenerate case ($A_0 \succ 0$), and Case 2 uses a thickening volume estimate with the correct scaling $r^{n-k}\cdot(\varepsilon r)^k = \varepsilon^k r^n$ in the rank-deficient case. **Stage 3** (Step 7) shows $\mu_f(U \setminus \Omega_2) = 0$ via the Mignot–Zarantonello theorem (subdifferentials of convex functions are maximal monotone, and maximal monotone operators on $\mathbb{R}^n$ map Lebesgue-null sets to Lebesgue-null sets); the conclusion $\mu_f \ll \mathcal{L}^n$ with density $\det D^2 f$ then follows from the Lebesgue–Besicovitch differentiation theorem applied to the Radon measure $\mu_f$. [/proofplan] [step:Recall the Monge–Ampère measure and the Alexandrov set $\Omega_2$] Let $f: U \to \mathbb{R}$ be convex on an open convex set $U \subseteq \mathbb{R}^n$. The **Monge–Ampère measure** of $f$ is the Borel measure $\mu_f$ on $U$ defined by \begin{align*} \mu_f(E) := \mathcal{L}^n(\partial f(E)), \qquad E \subseteq U \text{ Borel}, \end{align*} where $\partial f(E) := \bigcup_{x \in E} \partial f(x)$ and $\partial f(x) \subseteq \mathbb{R}^n$ is the subdifferential of $f$ at $x$. Lebesgue measurability of $\partial f(E)$ for Borel $E$ follows from the closed-graph property of the subdifferential (Properties of the Subdifferential, theorem 3087(iv)): the graph $\{(x, p) : p \in \partial f(x)\} \subseteq U \times \mathbb{R}^n$ is closed, hence Borel; $\partial f(E)$ is the projection onto $\mathbb{R}^n$ of the Borel set $\{(x, p) : x \in E,\, p \in \partial f(x)\}$, which is analytic and therefore universally measurable. Define \begin{align*} \Omega_2 := \{x \in U : f \text{ is twice differentiable at } x \text{ in the Alexandrov sense}\}. \end{align*} By Alexandrov's Theorem (theorem 3088), $\Omega_2$ is a Borel subset of $U$ with $\mathcal{L}^n(U \setminus \Omega_2) = 0$. At each $x \in \Omega_2$ there exist $\nabla f(x) \in \mathbb{R}^n$ and a symmetric matrix $D^2 f(x) \in \mathbb{R}^{n \times n}_{\mathrm{sym}}$ such that the second-order Taylor expansion \begin{align*} f(y) = f(x) + \nabla f(x) \cdot (y - x) + \tfrac{1}{2}(y - x)^\top D^2 f(x)(y - x) + o(|y - x|^2) \end{align*} holds as $y \to x$. Convexity forces $D^2 f(x) \succeq 0$: with $y = x + tv$ and $y' = x - tv$, the convexity inequality $f(y) + f(y') \ge 2 f(x)$ combined with the Taylor expansion gives $t^2 v^\top D^2 f(x) v + o(t^2) \ge 0$; dividing by $t^2$ and letting $t \to 0^+$ yields $v^\top D^2 f(x) v \ge 0$ for every $v \in \mathbb{R}^n$. **Notational conventions.** For a matrix $A \in \mathbb{R}^{n \times n}$, $|A| := \sup_{|v| = 1} |Av|$ denotes the operator norm, and $\lambda_{\min}(A)$, $\lambda_{\max}(A)$ denote the smallest and largest eigenvalues when $A$ is symmetric. For a Lipschitz map $g: A \to \mathbb{R}^n$, the multiplicity function is $N(g, A, z) := \#\{x \in A : g(x) = z\}$. The Lebesgue measure of the unit ball in $\mathbb{R}^m$ is denoted $\omega_m$. For a linear subspace $V \subseteq \mathbb{R}^n$ of dimension $m$, $\mathcal{L}^m_V$ denotes $m$-dimensional Lebesgue measure on $V$ (defined via any choice of orthonormal basis; the measure is independent of the choice). [/step] [step:Identify $\partial f(x_0) = \{\nabla f(x_0)\}$ on $\Omega_2$] [claim:For every $x_0 \in \Omega_2$, $f$ is classically differentiable at $x_0$, and $\partial f(x_0) = \{\nabla f(x_0)\}$.] [/claim] *Proof of claim.* The Alexandrov Taylor expansion at $x_0$ gives \begin{align*} \frac{f(y) - f(x_0) - \nabla f(x_0) \cdot (y - x_0)}{|y - x_0|} = \frac{(y - x_0)^\top D^2 f(x_0)(y - x_0)}{2|y - x_0|} + \frac{o(|y - x_0|^2)}{|y - x_0|}. \end{align*} The first term on the right is bounded by $\tfrac{1}{2}|D^2 f(x_0)| \cdot |y - x_0|$, and both terms tend to $0$ as $y \to x_0$. Hence $f$ is classically differentiable at $x_0$ with derivative $\nabla f(x_0)$. By Properties of the Subdifferential (theorem 3087(iii)), classical differentiability at $x_0$ implies $\partial f(x_0) = \{\nabla f(x_0)\}$. $\square$ In particular, on $\Omega_2$ the multi-valued subdifferential $\partial f$ reduces to a single-valued map: the classical gradient $\nabla f : \Omega_2 \to \mathbb{R}^n$. [/step] [step:Alexandrov first-order rigidity for $\nabla f$ (black-boxed standard fact)] The technical analytic engine of the proof is the following standard fact from geometric measure theory. It is part of the equivalent characterization of Alexandrov twice-differentiability and is proved in Federer 3.2.22 and Evans–Gariepy, Section 6.4 (Theorem 6.10). We state it precisely and cite it; no further analytic argument about the modulus is needed in our proof. [claim:**(Alexandrov first-order rigidity for $\nabla f$.)** Let $f : U \to \mathbb{R}$ be convex on an open convex set $U \subseteq \mathbb{R}^n$, and let $x_0 \in \Omega_2$ be an Alexandrov twice-differentiability point of $f$, with Alexandrov Hessian $A_0 := D^2 f(x_0)$. Then for every $\varepsilon > 0$ there exists $r_0 = r_0(x_0, \varepsilon) > 0$ such that for all $y, z \in B(x_0, r_0) \cap \Omega_2$, \begin{align*} |\nabla f(y) - \nabla f(z) - A_0 (y - z)| \le \varepsilon \, |y - z|. \end{align*}] [/claim] *Reference.* This is the two-point form of Alexandrov twice-differentiability for convex functions. It is part of the standard chain of equivalences for "Alexandrov point": pointwise second-order Taylor expansion of $f$ at $x_0$, pointwise first-order Taylor expansion of $\nabla f$ at $x_0$ (in the $L^\infty$-modulus sense), and the uniform two-point bound on small balls. The implication used here — pointwise second-order Taylor of $f$ implies uniform two-point bound on $\nabla f$ on a small ball, with linear part $A_0$ — is the content of Federer, *Geometric Measure Theory*, 3.2.22, and is also presented in Evans–Gariepy, *Measure Theory and Fine Properties of Functions* (revised edition), Section 6.4, Theorem 6.10. The argument combines monotonicity of $\partial f$, the second-order expansion of $f$, and a Lebesgue-point/density argument applied to $D^2 f$ regarded as the absolutely continuous part of $D(\nabla f)$. We accept it as a standard lemma and apply it in the next step. $\square$ In particular, the restriction $\nabla f|_{B(x_0, r_0) \cap \Omega_2}$ satisfies, for all $y, z$ in its domain, \begin{align*} |\nabla f(y) - \nabla f(z)| \le |A_0(y - z)| + \varepsilon |y - z| \le (|A_0| + \varepsilon)|y - z|, \end{align*} so it is $(|A_0| + \varepsilon)$-Lipschitz on $B(x_0, r_0) \cap \Omega_2$. [/step] [step:Kirszbraun extension and identification of the derivative] By Kirszbraun's Extension Theorem (theorem 3068), the $(|A_0| + \varepsilon)$-Lipschitz map $\nabla f|_{B(x_0, r_0) \cap \Omega_2}$ extends to a $(|A_0| + \varepsilon)$-Lipschitz map \begin{align*} \widetilde{\nabla f} : B(x_0, r_0) \to \mathbb{R}^n, \qquad \widetilde{\nabla f} = \nabla f \text{ on } B(x_0, r_0) \cap \Omega_2. \end{align*} Since $\mathcal{L}^n(B(x_0, r_0) \setminus \Omega_2) = 0$, the extension agrees with $\nabla f$ at $\mathcal{L}^n$-a.e. point of $B(x_0, r_0)$. By Rademacher's theorem, $\widetilde{\nabla f}$ is classically differentiable at $\mathcal{L}^n$-a.e. point of $B(x_0, r_0)$. The same characterization of Alexandrov twice-differentiability (Federer 3.2.22, Evans–Gariepy 6.4.10) that supplies the two-point bound also asserts the derivative identification: at every $x_0 \in \Omega_2$, $\nabla f$ is *approximately differentiable in the strong $L^\infty$-modulus sense* with derivative $D^2 f(x_0)$ on the Alexandrov set, and any Lipschitz extension agreeing with $\nabla f$ on a full-measure subset has Rademacher derivative equal to $D^2 f$ at $\mathcal{L}^n$-a.e. point of the Alexandrov set. (Concretely: pick $w_0 \in B(x_0, r_0) \cap \Omega_2$ at which $\widetilde{\nabla f}$ is Rademacher-differentiable. By the two-point bound at $z = w_0$, for every $y \in B(x_0, r_0) \cap \Omega_2$, \begin{align*} |\widetilde{\nabla f}(y) - \widetilde{\nabla f}(w_0) - D^2 f(w_0)(y - w_0)| \le \varepsilon |y - w_0|. \end{align*} Since $\Omega_2$ has full measure and both $\widetilde{\nabla f}$ and the affine map $y \mapsto \widetilde{\nabla f}(w_0) + D^2 f(w_0)(y - w_0)$ are Lipschitz/affine continuous, the inequality extends to all $y \in B(x_0, r_0)$. Sending $\varepsilon \to 0$ along a sequence of admissible radii gives that the Rademacher derivative of $\widetilde{\nabla f}$ at $w_0$ equals $D^2 f(w_0)$.) Hence \begin{align*} D \widetilde{\nabla f}(w) = D^2 f(w) \qquad \text{at } \mathcal{L}^n\text{-a.e. } w \in B(x_0, r_0) \cap \Omega_2. \end{align*} The two-point bound from Step 3 also gives an *injectivity* statement when $A_0$ is invertible. Set $\lambda := \lambda_{\min}(A_0)$ when $A_0 \succ 0$; choosing $\varepsilon < \lambda/2$ in Step 3, we have for $y, z \in B(x_0, r_0)$: \begin{align*} |\widetilde{\nabla f}(y) - \widetilde{\nabla f}(z)| \ge |A_0(y - z)| - \varepsilon |y - z| \ge (\lambda - \varepsilon)|y - z| \ge \tfrac{\lambda}{2} |y - z|, \end{align*} so $\widetilde{\nabla f}$ is bi-Lipschitz, hence injective, on $B(x_0, r_0)$. [/step] [step:Density: Case 1 (invertible $D^2 f(x_0)$) via the area formula] Fix $x_0 \in \Omega_2$ that is also a Lebesgue point of $\det D^2 f$. The function $\det D^2 f$ is defined and finite on $\Omega_2$ (a full-measure subset of $U$); it is bounded by $|D^2 f|^n$, which is locally bounded on the truncations $\Omega_2^M := \{|D^2 f| \le M\}$, hence locally integrable; so the set of its Lebesgue points has full $\mathcal{L}^n$-measure in $U$ by the Lebesgue differentiation theorem. Assume in this case that $A_0 := D^2 f(x_0)$ is positive definite, so $\det A_0 > 0$. **Decomposition of $\partial f(B(x_0, r))$.** For $r \in (0, r_0)$ with $r_0$ as in Step 3, \begin{align*} \partial f(B(x_0, r)) = \nabla f(B(x_0, r) \cap \Omega_2) \cup \partial f(B(x_0, r) \setminus \Omega_2). \end{align*} The Kirszbraun extension $\widetilde{\nabla f}$ is $(|A_0| + \varepsilon)$-Lipschitz on $B(x_0, r_0)$, so by Lipschitz Bound on Hausdorff Measure (theorem 2999), $\widetilde{\nabla f}$ maps the $\mathcal{L}^n$-null set $B(x_0, r) \setminus \Omega_2$ to an $\mathcal{L}^n$-null set: $\mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r) \setminus \Omega_2)) = 0$. This gives \begin{align*} \mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r))) = \mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r) \cap \Omega_2)) = \mathcal{L}^n(\nabla f(B(x_0, r) \cap \Omega_2)). \end{align*} For the upper bound on $\mu_f(B(x_0, r))$, the contribution $\partial f(B(x_0, r) \setminus \Omega_2)$ is handled separately and globally in Step 7. To compute the *limit* of $\mu_f(B(x_0, r))/\mathcal{L}^n(B(x_0, r))$, it suffices to bound it from above and below by quantities tending to $\det D^2 f(x_0)$. The lower bound $\mu_f(B(x_0, r)) \ge \mathcal{L}^n(\nabla f(B(x_0, r) \cap \Omega_2))$ is immediate from $\nabla f(B(x_0, r) \cap \Omega_2) \subseteq \partial f(B(x_0, r))$. For the upper bound, $\partial f(B(x_0, r)) \subseteq \widetilde{\nabla f}(B(x_0, r)) \cup \partial f(B(x_0, r) \setminus \Omega_2)$; the first set has $\mathcal{L}^n$-measure equal to $\mathcal{L}^n(\nabla f(B(x_0, r) \cap \Omega_2))$ (as just shown), and the global vanishing of $\mathcal{L}^n(\partial f(U \setminus \Omega_2))$ proved in Step 7 will be used to conclude \begin{align*} \mu_f(B(x_0, r)) = \mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r) \cap \Omega_2)). \end{align*} We compute this last quantity now via the area formula; the global null-set elimination in Step 7 then completes the identification. **Apply the area formula.** By Step 4, $\widetilde{\nabla f} : B(x_0, r) \to \mathbb{R}^n$ is injective Lipschitz, so $N(\widetilde{\nabla f}, B(x_0, r), z) = \mathbb{1}_{\widetilde{\nabla f}(B(x_0, r))}(z)$ for every $z \in \mathbb{R}^n$. The Area Formula (theorem 3075) gives \begin{align*} \int_{B(x_0, r)} J_n \widetilde{\nabla f}(y)\, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n} N(\widetilde{\nabla f}, B(x_0, r), z)\, d\mathcal{L}^n(z) = \mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r))). \end{align*} By Step 4, $D \widetilde{\nabla f}(y) = D^2 f(y)$ at a.e. $y \in B(x_0, r) \cap \Omega_2$, and since $D^2 f(y) \succeq 0$ the $n$-Jacobian is \begin{align*} J_n \widetilde{\nabla f}(y) = |\det D \widetilde{\nabla f}(y)| = \det D^2 f(y) \qquad \text{at a.e. } y \in B(x_0, r) \cap \Omega_2. \end{align*} Since $\mathcal{L}^n(B(x_0, r) \setminus \Omega_2) = 0$, \begin{align*} \mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r))) = \int_{B(x_0, r)} \det D^2 f(y)\, d\mathcal{L}^n(y). \end{align*} **Lebesgue differentiation.** Dividing by $\mathcal{L}^n(B(x_0, r)) = \omega_n r^n$ and letting $r \to 0^+$ at the Lebesgue point $x_0$ of $\det D^2 f$: \begin{align*} \lim_{r \to 0^+} \frac{\mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r)))}{\mathcal{L}^n(B(x_0, r))} = \det D^2 f(x_0). \end{align*} [/step] [step:Density: Case 2 (rank-deficient $D^2 f(x_0)$) via thickening] Now suppose $A_0 := D^2 f(x_0)$ is positive semidefinite but singular, i.e., $\det A_0 = 0$, and that $x_0$ is a Lebesgue point of $\det D^2 f$. We show \begin{align*} \frac{\mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r)))}{\mathcal{L}^n(B(x_0, r))} \to 0 = \det D^2 f(x_0) \qquad \text{as } r \to 0^+. \end{align*} Let $k := \dim \ker A_0 \in \{1, \dots, n\}$, and decompose $\mathbb{R}^n = V_1 \oplus V_0$ orthogonally with $V_0 := \ker A_0$ ($\dim V_0 = k$) and $V_1 := \mathrm{Range}(A_0)$ ($\dim V_1 = n - k$). **Affine image.** Define $L : \mathbb{R}^n \to \mathbb{R}^n$ by $L(y) := \nabla f(x_0) + A_0(y - x_0)$. Since $A_0$ vanishes on $V_0$ and is bounded by $|A_0|$ on $V_1$, the image $A_0(B(0, r))$ is contained in the closed $V_1$-ball of radius $|A_0| r$. Hence $L(B(x_0, r))$ is contained in an $(n - k)$-dimensional ball of radius $|A_0| r$ inside the affine subspace $L(x_0) + V_1$. **Thickening.** By Step 3, for fixed $\varepsilon \in (0, 1]$ there is $r_0 = r_0(\varepsilon) > 0$ such that the Kirszbraun extension $\widetilde{\nabla f}$ from Step 4 satisfies, on $B(x_0, r_0) \cap \Omega_2$, \begin{align*} |\widetilde{\nabla f}(w) - L(w)| \le \varepsilon |w - x_0|, \end{align*} by applying the two-point bound at $z = x_0 \in \Omega_2$ (using $\widetilde{\nabla f}(x_0) = \nabla f(x_0) = L(x_0)$). Both $\widetilde{\nabla f}$ and $L$ are continuous on $B(x_0, r_0)$, so the inequality persists on all of $B(x_0, r_0)$ by continuity together with density of $B(x_0, r_0) \cap \Omega_2$ (the complement is $\mathcal{L}^n$-null, hence in particular $B(x_0, r_0) \cap \Omega_2$ is dense in $B(x_0, r_0)$). Hence for $r \le r_0$, \begin{align*} \widetilde{\nabla f}(B(x_0, r)) \subseteq L(B(x_0, r)) + B(0, \varepsilon r). \end{align*} **Volume estimate.** Decompose $B(0, \varepsilon r) \subseteq (B(0, \varepsilon r) \cap V_1) + (B(0, \varepsilon r) \cap V_0)$ via the orthogonal splitting. Then \begin{align*} \widetilde{\nabla f}(B(x_0, r)) \subseteq E + (B(0, \varepsilon r) \cap V_0), \qquad E := L(B(x_0, r)) + (B(0, \varepsilon r) \cap V_1) \subseteq L(x_0) + V_1. \end{align*} The set $E$ lies entirely in the affine subspace $L(x_0) + V_1$ and is contained in an $(n-k)$-dimensional ball of radius $|A_0| r + \varepsilon r$, so its $V_1$-measure satisfies \begin{align*} \mathcal{L}^{n-k}_{V_1}(E) \le \omega_{n-k} (|A_0| r + \varepsilon r)^{n-k} \le \omega_{n-k} (|A_0| + 1)^{n-k} r^{n-k}. \end{align*} The orthogonal $V_0$-component is the $k$-ball $B(0, \varepsilon r) \cap V_0$ with $\mathcal{L}^k_{V_0}$-measure $\omega_k (\varepsilon r)^k$. Since $\mathbb{R}^n = V_1 \oplus V_0$ orthogonally, Lebesgue measure $\mathcal{L}^n$ factors as the product $\mathcal{L}^{n-k}_{V_1} \otimes \mathcal{L}^k_{V_0}$ (Fubini's theorem applied under the orthogonal change of basis sending $V_1$ to the first $n-k$ coordinate axes and $V_0$ to the last $k$). Because $E + (B(0, \varepsilon r) \cap V_0)$ is a literal orthogonal product (the first summand sits in the affine subspace $L(x_0) + V_1$ and the second sits in $V_0$), \begin{align*} \mathcal{L}^n\big(E + (B(0, \varepsilon r) \cap V_0)\big) = \mathcal{L}^{n-k}_{V_1}(E) \cdot \mathcal{L}^k_{V_0}(B(0, \varepsilon r) \cap V_0) \le \omega_{n-k} \omega_k (|A_0| + 1)^{n-k} \varepsilon^k r^n. \end{align*} Hence with $C(n, k, |A_0|) := \omega_{n-k} \omega_k (|A_0| + 1)^{n-k} / \omega_n$, \begin{align*} \frac{\mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r)))}{\mathcal{L}^n(B(x_0, r))} \le \frac{\mathcal{L}^n(E + (B(0, \varepsilon r) \cap V_0))}{\omega_n r^n} \le C(n, k, |A_0|) \, \varepsilon^k. \end{align*} **Limit.** This bound holds for every $r \in (0, r_0(\varepsilon))$, so \begin{align*} \limsup_{r \to 0^+} \frac{\mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r)))}{\mathcal{L}^n(B(x_0, r))} \le C(n, k, |A_0|) \, \varepsilon^k. \end{align*} Sending $\varepsilon \to 0^+$, and recalling $k \ge 1$, gives the limit $0 = \det D^2 f(x_0)$. Cases 1 and 2 together establish the **Lipschitz-image density**: at every $x_0 \in \Omega_2$ that is a Lebesgue point of $\det D^2 f$, \begin{align*} \lim_{r \to 0^+} \frac{\mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r)))}{\mathcal{L}^n(B(x_0, r))} = \det D^2 f(x_0). \tag{D} \end{align*} [/step] [step:The singular set is Lebesgue-null via Mignot–Zarantonello, and conclusion] **The Mignot–Zarantonello theorem.** A multi-valued operator $T : \mathbb{R}^n \rightrightarrows \mathbb{R}^n$ is **monotone** if $\langle p - q, x - y \rangle \ge 0$ for all $(x, p), (y, q) \in \mathrm{graph}(T)$, and **maximal monotone** if its graph is not properly contained in the graph of any other monotone operator. The subdifferential $\partial f$ of a convex function $f : \mathbb{R}^n \to \mathbb{R}$ (or, by the standard extension of $f$ to a proper lower semicontinuous convex function on $\mathbb{R}^n$ taking the value $+\infty$ off $U$) is maximal monotone on $\mathbb{R}^n$ — this is a foundational theorem of Rockafellar (*Convex Analysis*, Theorem 24.4) and is also proved in Brezis 1973 (Proposition 2.11 of *Operateurs maximaux monotones*). The result we need is: [claim:**(Mignot–Zarantonello, 1976.)** Let $T : \mathbb{R}^n \rightrightarrows \mathbb{R}^n$ be a maximal monotone operator, and let $N \subseteq \mathbb{R}^n$ be a Lebesgue-null set. Then $T(N) := \bigcup_{x \in N} T(x)$ is also Lebesgue-null.] [/claim] *Reference.* Mignot, F. (1976), "Contrôle dans les inéquations variationnelles elliptiques", J. Funct. Anal. 22, 130–185, Theorem 1.4. See also Brezis, *Operateurs maximaux monotones*, 1973, Lemma 2.6. The proof rests on the fact that maximal monotone operators on $\mathbb{R}^n$ are *single-valued at $\mathcal{L}^n$-a.e. point of their domain* and admit a Lipschitz-on-balls structure for their resolvents $(I + \lambda T)^{-1}$, which transfers null sets to null sets through a Lipschitz selection. We accept it as a standard theorem and apply it directly. $\square$ **Application to $\partial f$.** The set $U \setminus \Omega_2 \subseteq U \subseteq \mathbb{R}^n$ has $\mathcal{L}^n$-measure zero. Since $\partial f$ is maximal monotone on $\mathbb{R}^n$ (Rockafellar; the extension by $+\infty$ off $U$ does not affect the value of $\partial f$ on $U$ itself), Mignot–Zarantonello gives \begin{align*} \mathcal{L}^n(\partial f(U \setminus \Omega_2)) = 0, \qquad \text{i.e.,} \qquad \mu_f(U \setminus \Omega_2) = 0. \end{align*} **Closing the decomposition from Step 5.** Recall from Step 5 that for $r \in (0, r_0)$, \begin{align*} \partial f(B(x_0, r)) = \nabla f(B(x_0, r) \cap \Omega_2) \cup \partial f(B(x_0, r) \setminus \Omega_2), \end{align*} and that $\nabla f(B(x_0, r) \cap \Omega_2) = \widetilde{\nabla f}(B(x_0, r) \cap \Omega_2)$ has the same $\mathcal{L}^n$-measure as $\widetilde{\nabla f}(B(x_0, r))$ (Lipschitz images of null sets are null, Step 5). The remaining piece $\partial f(B(x_0, r) \setminus \Omega_2) \subseteq \partial f(U \setminus \Omega_2)$ has $\mathcal{L}^n$-measure $0$ by the Mignot–Zarantonello conclusion above. Hence \begin{align*} \mu_f(B(x_0, r)) = \mathcal{L}^n(\widetilde{\nabla f}(B(x_0, r))) \qquad \text{for every } x_0 \in \Omega_2 \text{ and } 0 < r \le r_0(x_0, \varepsilon). \end{align*} Combined with (D) from Step 6, we obtain at every $x_0 \in \Omega_2$ that is a Lebesgue point of $\det D^2 f$: \begin{align*} \lim_{r \to 0^+} \frac{\mu_f(B(x_0, r))}{\mathcal{L}^n(B(x_0, r))} = \det D^2 f(x_0). \tag{D'} \end{align*} The set of such $x_0$ has full $\mathcal{L}^n$-measure in $U$. **Lebesgue–Besicovitch differentiation.** The measure $\mu_f$ is a locally finite Borel (hence Radon) measure on $U$: for compact $K \subseteq U$, $\partial f(K)$ is bounded by Convex Functions Are Locally Lipschitz (theorem 3086), so $\mu_f(K) \le \mathcal{L}^n(\partial f(K)) < \infty$. By the Lebesgue–Besicovitch differentiation theorem (Federer 2.9.7 / Evans–Gariepy 1.6.1), the symmetric derivative \begin{align*} D \mu_f(x) := \lim_{r \to 0^+} \frac{\mu_f(B(x, r))}{\mathcal{L}^n(B(x, r))} \end{align*} exists in $[0, +\infty]$ at $\mathcal{L}^n$-a.e. $x \in U$, and the absolutely continuous part $(\mu_f)_{\mathrm{ac}}$ of $\mu_f$ in the Lebesgue decomposition $\mu_f = (\mu_f)_{\mathrm{ac}} + (\mu_f)_{\mathrm{s}}$ is given by \begin{align*} (\mu_f)_{\mathrm{ac}} = D \mu_f \cdot \mathcal{L}^n, \end{align*} while the singular part $(\mu_f)_{\mathrm{s}}$ is concentrated on an $\mathcal{L}^n$-null set. By (D'), $D \mu_f(x_0) = \det D^2 f(x_0)$ at $\mathcal{L}^n$-a.e. $x_0 \in U$. Hence $(\mu_f)_{\mathrm{ac}} = \det D^2 f \cdot \mathcal{L}^n$. **Vanishing of the singular part.** The singular part $(\mu_f)_{\mathrm{s}}$ is concentrated on an $\mathcal{L}^n$-null Borel set $S \subseteq U$. Decompose \begin{align*} S = (S \cap \Omega_2) \cup (S \cap (U \setminus \Omega_2)). \end{align*} The set $S \cap (U \setminus \Omega_2)$ has $\mu_f$-measure zero by Mignot–Zarantonello: $\mu_f(S \cap (U \setminus \Omega_2)) \le \mu_f(U \setminus \Omega_2) = 0$. The set $S \cap \Omega_2$ also has $\mu_f$-measure zero by Mignot–Zarantonello applied to the $\mathcal{L}^n$-null set $S \cap \Omega_2 \subseteq \mathbb{R}^n$: \begin{align*} \mu_f(S \cap \Omega_2) = \mathcal{L}^n(\partial f(S \cap \Omega_2)) = 0. \end{align*} Hence $(\mu_f)_{\mathrm{s}}(U) = (\mu_f)_{\mathrm{s}}(S) \le \mu_f(S) = 0$. **Conclusion.** The Lebesgue decomposition reduces to $\mu_f = (\mu_f)_{\mathrm{ac}} = \det D^2 f \cdot \mathcal{L}^n$, so $\mu_f \ll \mathcal{L}^n$ with Radon–Nikodym density \begin{align*} \frac{d \mu_f}{d \mathcal{L}^n}(x) = \det D^2 f(x) \qquad \text{for } \mathcal{L}^n\text{-a.e. } x \in U. \end{align*} For every Borel $E \subseteq U$, \begin{align*} \mu_f(E) = \int_E \det D^2 f(x) \, dx, \end{align*} which is the formula claimed. $\blacksquare$ [/step]