cycles. It also does not depend on the choice of $U_0$ up to an additive constant (if we chose some $U_1$ instead, the function would change by a summand $\sum_{\text{path connecting } U_0 \text{ with } U_1} a_k)$. We see that $u_j \in L^1(\Omega)$ and that it is piecewise constant.\
3. We notice that $Du_j = (Du)_j$, as $u_j$ is constant on each $U_i$, $J_{u_j} = J_u$ and the jumps along connected components of $J_u$ have the same magnitude. Thus we define $u_c = u - u_j$. We see that $(Du_c)_j = 0$.\
4. The $u_c$, $u_j$ defined above are functions of least gradient.

Suppose that $u_j$ is not a a function of least gradient, i.e. there exists $v \in BV(\Omega)$ such that $\int_\Omega |Dv| < \int_\Omega |Du_j|$ and $Tu_j = Tv$. Then we would get

$$\int_\Omega |Du| \leq \int_\Omega |D(u_c + v)| \leq \int_\Omega |Du_c| + \int_\Omega |Dv| < \int_\Omega |Du_c| + \int_\Omega |Du_j| = \int_\Omega |Du|,$$

where the first inequality follows from $u$ being a function of least gradient, and the last equality from measures $Du_c$ and $Du_j$ being mutually singular. The proof for $u_c$ is analogous.

5\. The function $u_c$ is continuous. As $u_c$ is of least 