[proofplan] The embedding $W^{1,1} \hookrightarrow BV$ is immediate: an $L^1$ gradient is a special case of a measure-valued derivative. The characterisation of $W^{1,1}$ as the absolutely continuous part of $BV$ follows from the Radon--Nikodym theorem: $Du \ll \mathcal{L}^n$ if and only if $Du$ has a density in $L^1$, which is precisely the weak derivative. [/proofplan] [step:Show the isometric embedding $W^{1,1}(U) \hookrightarrow BV(U)$] If $u \in W^{1,1}(U)$, the weak derivative $\nabla u \in L^1(U; \mathbb{R}^n)$ satisfies: \begin{align*} \int_U u\,\partial_i\phi \, d\mathcal{L}^n = -\int_U \phi\,(\nabla u)_i \, d\mathcal{L}^n \quad \text{for all } \phi \in C_c^\infty(U). \end{align*} This means the distributional derivative $Du$ is the measure $\nabla u \, d\mathcal{L}^n$, which is absolutely continuous with respect to $\mathcal{L}^n$. Its total variation is: \begin{align*} |Du|(U) = \int_U |\nabla u| \, d\mathcal{L}^n = \|\nabla u\|_{L^1}. \end{align*} Therefore $\|u\|_{BV} = \|u\|_{L^1} + \|\nabla u\|_{L^1} = \|u\|_{W^{1,1}}$. [/step] [step:Show $u \in BV(U)$ belongs to $W^{1,1}(U)$ if and only if $Du \ll \mathcal{L}^n$] The forward direction is the previous step: $u \in W^{1,1}$ implies $Du = \nabla u \, d\mathcal{L}^n \ll \mathcal{L}^n$. For the converse: if $Du \ll \mathcal{L}^n$, the Radon--Nikodym theorem gives a density $g \in L^1(U; \mathbb{R}^n)$ with $Du = g \, d\mathcal{L}^n$. Then for all $\phi \in C_c^\infty(U)$: \begin{align*} \int_U u\,\partial_i\phi \, d\mathcal{L}^n = -\int_U \phi\,g_i \, d\mathcal{L}^n, \end{align*} which is the definition of $g_i$ being the weak derivative $\partial_i u$. Since $g \in L^1$, $u \in W^{1,1}(U)$. [/step]