[proofplan] Completeness follows from completeness of $L^1$: a Cauchy sequence in $BV$ converges in $L^1$, and the [Lower Semicontinuity of Total Variation](/theorems/597) ensures the limit has finite total variation. Non-separability is shown by exhibiting an uncountable family of characteristic functions at mutual $BV$-distance bounded below. [/proofplan] [step:Prove completeness of $BV(U)$] Let $(u_k)$ be Cauchy in $(BV(U), \|\cdot\|_{BV})$. Since $\|u_j - u_k\|_{L^1} \leq \|u_j - u_k\|_{BV} \to 0$, the sequence is Cauchy in $L^1(U)$. By completeness of $L^1$, there exists $u \in L^1(U)$ with $u_k \to u$ in $L^1$. For any $\Phi \in C_c^\infty(U; \mathbb{R}^n)$ 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*} Since $(u_k)$ is Cauchy in $BV$, $\sup_k |Du_k|(U) < \infty$. Taking the supremum over $\Phi$ and applying the [Dual Characterisation](/theorems/591): $|Du|(U) \leq \liminf_k |Du_k|(U) < \infty$, so $u \in BV(U)$. For the norm convergence: $\|u_k - u\|_{BV} = \|u_k - u\|_{L^1} + |D(u_k - u)|(U)$. The first term tends to $0$. For the second, the [Lower Semicontinuity of Total Variation](/theorems/597) applied to the sequence $(u_k - u_j)_{j \geq k}$ gives: \begin{align*} |D(u_k - u)|(U) \leq \liminf_{j \to \infty} |D(u_k - u_j)|(U) \leq \sup_{j \geq k}\|u_k - u_j\|_{BV} \to 0 \end{align*} as $k \to \infty$ (by the Cauchy property). [/step] [step:Prove non-separability of $BV(U)$] For distinct points $x_1, x_2 \in U$ with $|x_1 - x_2|$ sufficiently small, the balls $B(x_1, r)$ and $B(x_2, r)$ (with $r$ small enough that both lie in $U$) satisfy: \begin{align*} \|\mathbb{1}_{B(x_1, r)} - \mathbb{1}_{B(x_2, r)}\|_{L^1} = \mathcal{L}^n(B(x_1, r) \triangle B(x_2, r)) > 0. \end{align*} The family $\{\mathbb{1}_{B(x, r) \cap U}\}$ for $x$ ranging over an uncountable dense subset of $U$ contains an uncountable subcollection at mutual $L^1$-distance bounded below by some $\delta > 0$. Since the $BV$ norm dominates the $L^1$ norm, these elements are at mutual $BV$-distance at least $\delta$. An uncountable collection at mutual distance $\geq \delta$ cannot be covered by countably many balls of radius $\delta/3$, so $BV(U)$ is not separable. [/step]