[proofplan] The forward direction ($u \in BV \Rightarrow$ supremum $= |Du|(U)$) uses the definition of the distributional derivative and a Cauchy--Schwarz-type bound for measures, with the supremum achieved by aligning the test field with the Radon--Nikodym derivative of $Du$. The reverse direction (finite supremum $\Rightarrow u \in BV$) uses the Riesz representation theorem to extract the measure $Du$ from the bounded linear functional. [/proofplan] [step:Show $\int_U u\,\operatorname{div}\Phi \, d\mathcal{L}^n \leq |Du|(U)$ for $u \in BV$ and $|\Phi| \leq 1$] Since $u \in BV(U)$, the distributional derivative $Du$ is a finite $\mathbb{R}^n$-valued Radon measure with $\int_U u\,\partial_i\phi \, d\mathcal{L}^n = -\int_U \phi \, d(D_i u)$ for all $\phi \in C_c^\infty(U)$. Summing over $i$ with $\Phi = (\phi_1, \ldots, \phi_n)$: \begin{align*} \int_U u\,\operatorname{div}\Phi \, d\mathcal{L}^n = -\int_U \Phi \cdot dDu. \end{align*} Taking absolute values and using $|\Phi| \leq 1$: \begin{align*} \left|\int_U \Phi \cdot dDu\right| \leq \int_U |\Phi| \, d|Du| \leq |Du|(U). \end{align*} [/step] [step:Show the supremum equals $|Du|(U)$ by approximating the Radon--Nikodym derivative] The Radon--Nikodym theorem gives $Du = \sigma\,|Du|$ where $\sigma: U \to \mathbb{S}^{n-1}$ is $|Du|$-measurable with $|\sigma| = 1$ $|Du|$-a.e. For $\varepsilon > 0$, approximate $-\sigma$ in $L^1(|Du|)$ by $\Phi_k \in C_c^\infty(U; \mathbb{R}^n)$ with $|\Phi_k| \leq 1$ (using mollification and truncation). Then: \begin{align*} \int_U u\,\operatorname{div}\Phi_k \, d\mathcal{L}^n = -\int_U \Phi_k \cdot \sigma \, d|Du| \to \int_U |\sigma|^2 \, d|Du| = |Du|(U). \end{align*} Therefore the supremum equals $|Du|(U)$. [/step] [step:Prove the reverse direction: finite supremum implies $u \in BV$] Suppose $\sup\{\int_U u\,\operatorname{div}\Phi : |\Phi| \leq 1\} =: M < \infty$. For each $i \in \{1, \ldots, n\}$, the map $\phi \mapsto -\int_U u\,\partial_i\phi \, d\mathcal{L}^n$ defines a bounded linear functional on $C_c^\infty(U)$ (bounded by $M$ when $\|\phi\|_\infty \leq 1$). By the Riesz representation theorem for Radon measures, this functional extends to a finite signed Radon measure $\mu_i$ on $U$ with $|\mu_i|(U) \leq M$. Setting $Du = (\mu_1, \ldots, \mu_n)$, we have $u \in BV(U)$ with $|Du|(U) \leq M$. The reverse inequality $|Du|(U) \geq M$ follows from the previous step. Therefore $|Du|(U) = M$. [/step]