$$\begin{gathered}
\label{sign_of_measure}
\int_{\mathbb{R}} \left((X(u_{\infty}^{+}\wedge a)-X(u_{\infty}^{-}\wedge a)\right)\cdot\omega_{K}\\=\operatorname{sign}(u^{-}_{\infty}-u^+_{\infty})\int_{\min\{u_{\infty}^{+},u_{\infty}^{-}\}}^{\max\{u_{\infty}^{+},u_{\infty}^{-}\}}\frac{f(u_{\infty}^{+})+f(u_{\infty}^{-})}{2}-f(a)\,da\,.\end{gathered}$$ By convexity of $f$ we get $$\frac{f(u_{\infty}^{+})+f(u_{\infty}^{-})}{2}>f(a)\quad\mbox{for}\quad a\in\left[\min\{u_{\infty}^{+},u_{\infty}^{-}\},\max\{u_{\infty}^{+},u_{\infty}^{-}\}\right]\,.$$ This and (\[sign\_of\_measure\]) imply, that sign of $\mu_{\infty}$ is completely determined by $\operatorname{sign}(u^{-}_{\infty}-u^{+}_{\infty})$. Henceforth $$\mu_{\infty}(B_{R_0}(0,0))\geq0$$ can only be fulfilled, if $$u^-_{\infty}\geq u^+_{\infty}\,.$$ But this implies that the measure is $\mu_{\infty}$ has a sign, i.e. $$\mu_{\infty}\geq0\,.$$ Let $\mu_{k}^{\pm}$ be the positive respective negative part of $\mu_{k}$, i.e. $\mu_{k}^{\pm}$ are non-negative measures such that $$\mu_{k}=\mu_k^+-\mu_k^-\,.$$ Then we can extract a further subsequence $k'$ such that $$\mu_{k'}^{+}\rightharpoondown\nu^{+}\quad\mbox{and}\quad\mu_{k'}^{-}\rightharpoondown\nu^{-}\quad\mbox{in}\quad\mathcal{M}_{loc}(\mathbb{R}^2)\,.$$ For $R>0$ and non-negative $\psi\in C^{\infty}_{c}(B_{R}(0,0))$ we get $$\int_{B_{R_0}(0,0)}\psi\,d\mu_{\infty}=\lim_{k'\rightarrow\infty}\int_{B_{R}(0,0)}\psi\,d\mu_{k'}=\int_{B_{R}(0,0)}\psi\,d\nu^+-\int_{B_{R}(0,0)}\psi\,d\nu^-\,.$$ Since $\mu_{\infty}$ is non-negative we get for all non-negative $\psi\in C^{\infty}_{c}(B_{R_0}(0,0))$ $$\int_{B_{R_0}(0,0)}\psi\,d\nu^-\leq\int_{B_{R}(0,0)}\psi\,d\nu^+\,.$$ Hence $$\label{inequality_for_negative_and_positive_part}
 \nu^{-}(B_{R}(0,0))\leq \nu^{+}(B_{R}(0,0))$$ By Theorem 1.2 in [@Le] (see also Theorem 1.1 in [@AKLR]) we have for a rectifiable set $J_u$ and an ${\mathcal H}^1$ measurable function $h:~J_u\to\mathbb{R}$ $$\label{rectifiable}
\int_{\mathbb{R}}|m(x,t,a)|\,da=h\cdot\mathcal{H}^1\res J_u+\delta_u\,,$$ where $\delta_u$ satisfies $$\forall~B\quad\mbox{Borel}\quad \mathcal{H}^1(B)<\infty\Longrightarrow\delta_u(B)=0\,.$$ Therefore we can choose $R_1$, such that for all $k'$ $$\mu^{-}_{k'}(\partial B_{R_1}(0,0))\leq\frac{1}{r_{k'}}\int_{D_{k'}^{-1}(\partial B_{R_{1}}(0,0))}h\,d\mathcal{H}^1\res J_u=0\,.$$ Hence $$\nu^{-}(\partial B_{R_1}(0,0))= \lim_{k'\rightarrow\infty}\mu^{-}_{k'}(\partial B_{R_1}(0,0))=0\,.$$ This and (\[inequality\_for\_negative\_and\_positive\_part\]) imply $$\begin{aligned}
 \limsup_{k'\rightarrow\infty}\mu_{k'}^-(B_{R_1}(0,0)&\leq\nu^-(\bar{B}_{R_1}(0,0))=\nu^-(B_{R_1}(0,0))\\
&\leq\nu^{+}(B_{R_1}(0,0))\leq\liminf_{k'\rightarrow\infty}\mu_{k'}^{+}(B_{R_1}(0,0))\,.\end{aligned}$$ $$\begin{aligned}
\limsup_{k'\rightarrow\infty}\mu_{k'}(B_{R_1}(0,0))\geq\liminf_{k'\rightarrow\infty}\mu_{k'}^+B_{R_1}(0,0))-\limsup_{k'\rightarrow\infty}\mu^-_{k'}(B_{R_1}(0,0))\geq0\,,\end{aligned}$$ which obviously contradicts (\[contradiction\]) and we get (\[negative\_of\_limit\]).

Inequality (\[negative\_of\_limit\]) implies, that the set $K$ in (\[structure\_of\_measure\]) is non-empty and $$\mu_{\infty}<0\,,$$ which gives, again from above considerations $$u_{\infty}^-<u_{\infty}^+\,.$$ Moreover the convexity of $f$ implies for every 