Suppose we want to build/find a $BV$-function whose trace on the boundary of a set coincides with a (simple) $L^1$ function. In particular:
Let $D=B(0,1)$ the unit disk in $\mathbb{R}^2$ and let $u \in L^1(\partial D, \mathscr{H}^1)$ be defined by $u(x,y) := \text{sign}(xy)$. Provide a function $f \in BV(D)$ such that its trace on $\partial D$ is $u$.
Maybe we can split the disk into four parts (each part is a quadrant of the cartesian plane) called $A,B,C,D$ respectively. Then we define: \begin{equation} f:= \chi_A - \chi_B+\chi_C-\chi_D \end{equation} Can this work?