= Solution
Take the <indicator function> $u(x_1,x_2)=\chi_{\{x_1>1/2\}}$ inside $(0,1)^2$. For a compactly supported test <vector field>, integration in $x_1$ and then $x_2$ gives
$$
\int_\Omega u\,\operatorname{div}\xi\,dx=-\int_0^1\xi_1(1/2,x_2)\,dx_2.
$$
The vertical <derivative> term integrates to zero. The absolute pairing is at most one, and test <vector fields> with $\xi_1=-1$ along all but arbitrarily short endpoint pieces approach one. Thus $|Du|(\Omega)=1$. More explicitly, the <distributional derivative> is $Du=e_1\mathcal H^1\!\restriction\{(1/2,x_2):0<x_2<1\}$, where $\mathcal H^1$ is <Hausdorff measure> along the segment. Also $\|u\|_1=1/2$.
A <Sobolev space> <function> in $W^{1,1}(\Omega)$ has each weak <derivative> represented by an $L^1$ density with respect to planar <Lebesgue measure>. The nonzero segment measure here is singular with respect to that measure, so this requirement fails. Equivalently, the <Sobolev fundamental theorem of calculus on lines> would require an absolutely <continuous> representative on almost every horizontal slice, whereas each slice has a jump. Therefore
$$
\boxed{u\in BV(\Omega)\setminus W^{1,1}(\Omega),\qquad\|u\|_{BV}=\tfrac32.}
$$
This is the <bounded-variation step outside W11> example.
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