**Proof plan.** The necessity of the curl-free condition follows directly from the symmetry of mixed [distributional](/page/Distribution) [derivatives](/page/Derivative). The sufficiency uses the Poincaré lemma for distributions: on a simply connected domain, every curl-free vector-valued distribution is a gradient. The key ingredient is the regularisation of the measure $\mu$ by [mollification](/page/Standard%20Mollifier), which reduces the problem to the smooth Poincaré lemma. **Step 1: Necessity — curl-free is necessary.** [claim: Mixed Partials Commute] If $\mu = Du$ for some $u \in BV(U)$, then $\partial_i \mu_j = \partial_j \mu_i$ in $\mathcal{D}'(U)$ for all $i, j$. [/claim] [proof] For $\phi \in C_c^\infty(U)$: $\langle \partial_i \mu_j, \phi \rangle = -\int_U \partial_i\phi \, d\mu_j = -\int_U \partial_i\phi \, d(\partial_j u) = \int_U \partial_i\partial_j\phi \cdot u \, d\mathcal{L}^n$. By symmetry of the smooth partial derivatives $\partial_i\partial_j\phi = \partial_j\partial_i\phi$, interchanging $i$ and $j$ gives the same expression. [/proof] **Step 2: Sufficiency — curl-free implies gradient (simply connected case).** [claim: Distributional Poincaré Lemma] If $U$ is simply connected and $\mu$ is a curl-free $\mathbb{R}^n$-valued finite Radon measure, then $\mu = Du$ for some $u \in BV(U)$. [/claim] [proof] Mollify: $\mu_\varepsilon := \mu * \rho_\varepsilon$ is a smooth $\mathbb{R}^n$-valued function on $U_\varepsilon := \{x \in U : \operatorname{dist}(x, \partial U) > \varepsilon\}$. The curl-free condition $\partial_i\mu_j = \partial_j\mu_i$ in $\mathcal{D}'$ passes to the mollification: $\partial_i(\mu_\varepsilon)_j = (\partial_i\mu_j) * \rho_\varepsilon = (\partial_j\mu_i) * \rho_\varepsilon = \partial_j(\mu_\varepsilon)_i$. So $\mu_\varepsilon$ is a smooth curl-free vector field on $U_\varepsilon$. Since $U$ is simply connected and $U_\varepsilon$ is connected for small $\varepsilon$, the classical Poincaré lemma gives $\mu_\varepsilon = \nabla u_\varepsilon$ for some $u_\varepsilon \in C^\infty(U_\varepsilon)$. The [functions](/page/Function) $u_\varepsilon$ satisfy $\|\nabla u_\varepsilon\|_{L^1(V)} = \|\mu_\varepsilon\|_{L^1(V)} \leq |\mu|(U)$ for $V \Subset U$. After adjusting by constants (fixing $u_\varepsilon$ at a basepoint), the Poincaré inequality gives a uniform $L^1$ bound. By the [Compactness Theorem](/theorems/596), a subsequence converges in $L^1(V)$ to some $u \in BV(V)$. By the [Lower Semicontinuity](/theorems/597), $Du = \mu|_V$. A diagonal argument over $V \nearrow U$ gives $u \in BV(U)$ with $Du = \mu$. [/proof]