= Solution
The <structure theorem for functions of bounded variation> is the <measure> representation of the <distributional derivative>: for $u\in BV(\Omega)$ there is a unique finite <vector Radon measure> $Du$ with
$$
\boxed{\int_\Omega u\,\operatorname{div}\varphi\,dx=-\int_\Omega\varphi\cdot dDu,\qquad |Du|(\Omega)=\operatorname{TV}(u)}
$$
for every $\varphi\in C_c^1(\Omega;\mathbb R^n)$. It also has a <polar decomposition of a vector measure> $Du=\sigma_u|Du|$ with $|\sigma_u|=1$ for $|Du|$-almost every point.
To prove it, let $L(\varphi)=-\int u\operatorname{div}\varphi$. The test-function definition of <total variation seminorm>, applied to both signs, gives $|L(\varphi)|\le\operatorname{TV}(u)\|\varphi\|_\infty$. Compactly supported smooth <vector fields> are uniformly dense in $C_0(\Omega;\mathbb R^n)$, so $L$ extends uniquely to a bounded functional there. The <Riesz-Markov-Kakutani representation theorem>, applied componentwise, gives the unique <vector measure> $Du$. The operator <norm> of $L$ is exactly the defining variation supremum; the vector-measure dual <norm> is $|Du|(\Omega)$, proving equality. Conversely any <finite measure> satisfying the identity bounds that supremum, so this also characterizes membership in the <BV space>.
The <Radon-Nikodym theorem> applied to the components relative to $|Du|$ gives $\sigma_u$; the definition of the <variation measure> forces $|\sigma_u|=1$ <almost everywhere>. If $u$ is smooth, ordinary <integration by parts> gives $Du=\nabla u\,\mathcal L^n$. <Compact support> of the test field removes any boundary contribution; no boundary regularity is needed for this representation statement.
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