t: x \in \partial^{*} \{ u > s \} \cap \partial^{*} \{ u > t \} \};$$

$$u^{-}(x) = \inf \{ t: x \in \partial^{*} \{ u > s \} \cap \partial^{*} \{ u > t \} \}.$$

More precisely, the trace is uniquely determined up to a measure zero set by the mean integral property from [@EG Theorem 5.3.2]. But it holds for all $x \in \partial^{*} \{ u > s \} \cap \partial^{*} \{ u > t \}$; from the weak maximum principle this set divides $\Omega$ into two disjoint parts, $\Omega^+$ and $\Omega^-$. Let $\Omega^+$ be the part with greater values of $u$ in the neighbourhood of the cut. If $u^+(x) < \sup \{ t: x \in \partial^{*} \{ u > s \} \cap \partial^{*} \{ u > t \} \} = s$, then for sufficiently small neighbourhoods of $x$ we would have $u \geq s$, so $\fint_{B(x,r) \cap \Omega^+} |u^+(x) - u(y)| \geq |u^+(x) - s| > 0$, contradiction. The other cases are analogous.

Suppose $u \in BV(\Omega)$ is a least gradient function. Then $J_u = \bigcup_{k=1}^{\infty} S_k$, where $S_k$ are pairwise disjoint minimal surfaces. In addition, the trace of $u$ from both sides is constant 