gradient, then if it isn’t continuous at $x \in \Omega$, then a certain set of the form $\partial \{ u_c > t \}$ passes through $x$; otherwise $u_c$ would be constant in the neighbourhood of $x$. But in that case $u_c$ has a jump along the whole connected component of $\partial \{ u_c > t \}$ containing $x$, which is impossible as $(Du_c)_j = 0$.

6\. What is left is to prove uniqueness of such a decomposition. Let $u = u_c^1 + u_j^1 = u_c^2 + u_j^2$. Changing the order of summands we obtain

$$u_c^1 - u_c^2 = u_j^2 - u_j^1,$$

but the distributional derivative of the left hand side is a continuous measure, and the distributional derivative of the right hand side is supported on the set of zero measure with respect to $\mathcal{H}^{n-1}$, so both of them are zero measures. But the condition $Dv = 0$ implies $v = \text{const}$, so the functions $u_c^1$, $u_c^2$ differ by an additive constant.

In this decomposition $u_c$ isn’t necessarily continuous up to the boundary. Let us use the complex numbers notation for the plane. We take $\Omega = B(1,1)$. Let the boundary values be given by the formula $f(z) = \arg(z)$. 