[proofplan] The [Dual Characterisation](/theorems/591) expresses $|Du|(U)$ as a supremum of integrals $\int_U u\,\operatorname{div}\Phi \, d\mathcal{L}^n$ over test fields $\Phi$ with $|\Phi| \leq 1$. Each such integral is continuous in $L^1$ convergence (since $\operatorname{div}\Phi \in L^\infty$). The supremum of continuous functions is lower semicontinuous, giving $|Du|(U) \leq \liminf_k |Du_k|(U)$. [/proofplan] [step:Express total variation as a supremum of $L^1$-continuous functionals] By the [Dual Characterisation of Total Variation](/theorems/591): \begin{align*} |Du|(U) = \sup\left\{\int_U u\,\operatorname{div}\Phi \, d\mathcal{L}^n : \Phi \in C_c^\infty(U; \mathbb{R}^n),\, |\Phi| \leq 1\right\}. \end{align*} For each fixed $\Phi$, the map $u \mapsto \int_U u\,\operatorname{div}\Phi \, d\mathcal{L}^n$ is continuous with respect to $L^1$ convergence: \begin{align*} \left|\int_U (u_k - u)\,\operatorname{div}\Phi \, d\mathcal{L}^n\right| \leq \|\operatorname{div}\Phi\|_{L^\infty}\,\|u_k - u\|_{L^1} \to 0. \end{align*} [/step] [step:Take the supremum to conclude lower semicontinuity] For any fixed $\Phi$ with $|\Phi| \leq 1$: \begin{align*} \int_U u\,\operatorname{div}\Phi \, d\mathcal{L}^n = \lim_{k \to \infty}\int_U u_k\,\operatorname{div}\Phi \, d\mathcal{L}^n \leq \liminf_{k \to \infty} |Du_k|(U), \end{align*} where the inequality uses the [Dual Characterisation](/theorems/591) applied to each $u_k$. Taking the supremum over all admissible $\Phi$: \begin{align*} |Du|(U) = \sup_\Phi \int_U u\,\operatorname{div}\Phi \, d\mathcal{L}^n \leq \liminf_{k \to \infty} |Du_k|(U). \end{align*} If the right side is finite, the [Dual Characterisation](/theorems/591) also gives $u \in BV(U)$. [/step]