$$\begin{aligned}
\int_{\Gamma_{t_{1}}^{t_{2}}}q(x,t,a)\psi(x,t)\,dx\,dt&=\int_{\Gamma_{t_{1}}^{t_{2}}}\left[\partial_{t}v(x,t)\wedge a+\partial_{x}f(v(x,t)\wedge a)\right]\psi(x,t)\,dy\,ds\\
&=\int_{\Gamma_{t_{1}}^{t_{2}}}\left[\partial_{t}v(x,t)+\partial_{x}f(v(x,t))\right]\psi(x,t)\,dx\,dt\\
&=-\int_{\Gamma_{t_{1}}^{t_{2}}}v(x,t)\partial_{t}\psi(x,t)+f(v(x,t))\partial_{x}\psi(x,t)\,dx\,dt\\
&=0\,.\end{aligned}$$ Consider now a convex function $\eta(a)$ in $C^1$, which satisfies $$\lim_{a\rightarrow-\infty}\eta(a)=0$$ and denote $$\xi(a)=\int \eta'(a)f'(a)\,da\,.$$ We claim that for all $(x,t)\in\Gamma_{t_1+C\varepsilon}^{t_2-C\varepsilon}$ the following holds $$\label{entropy_equality}
 \partial_t \eta_{\varepsilon}(x,t)+\partial_x \xi_{\varepsilon}(x,t)=-\int_{\Gamma_{t_1}^{t_2}}\eta''(a)q_{\varepsilon}(x,t,a)\,da\,,$$ where $$\eta_{\varepsilon}(x,t)=\int_{\Gamma_{t_{1}}^{t_{2}}}\eta(v(y,s))\varphi_{\varepsilon}\left(x-y,t-s\right)\,dy\,ds$$ and $$\xi_{\varepsilon}(x,t)=\int_{\Gamma_{t_{1}}^{t_{2}}}\xi(v(y,s))\varphi_{\varepsilon}\left(x-y,t-s\right)\,dy\,ds\,.$$ Later will make special choices of $\eta$ in order to get (\[Linfty\_bound\]) and (\[control\_of\_the\_defect\_measure\]).

**Proof of claim (\[entropy\_equality\])**. We multiply (\[regularized\_equation\]) by $\eta'(a)$ $$\eta'(a) \partial_{t}\chi_{\varepsilon}+\eta'(a)f'(a)\partial_{x}\chi_{\varepsilon}=\partial_{a}q_{\varepsilon}(x,t,a)\,.$$ Then integrating this equation with respect to $a$ gives $$\label{zwischen1}
 \int_{\mathbb{R}}\eta'(a)\partial_{t} \chi_{\varepsilon}+\eta'(a)f'(a)\partial_{x}\chi_{\varepsilon}\,da= \int_{\mathbb{R}}\eta'(a)\partial_{a}q_{\varepsilon}(x,t,a)\,da\,.$$ We compute for the left-hand side $$\begin{aligned}
\int_{\mathbb{R}}\eta'(a)\chi_{\varepsilon}\,da&=\int_{\Gamma_{t_{1}}^{t_{2}}}\int_{\mathbb{R}}\eta'(a)\chi(v(y,s);a)\varphi_{\varepsilon}\left(x-y,t-s\right)\,da\,dy\,ds\\
&=\int_{\Gamma_{t_{1}}^{t_{2}}}\eta(v(y,s))\varphi_{\varepsilon}\left(x-y,t-s\right)\,dy\,ds=\eta_{\varepsilon}(x,t)\end{aligned}$$ and similarly for the second term $$\int_{\mathbb{R}}\eta'(a)f'(a)\chi_{\varepsilon}\,da=\int_{\Gamma_{t_{1}}^{t_{2}}}\xi(v(y,s))\varphi_{\varepsilon}\left(x-y,t-s\right)\,dy\,ds=\xi_{\varepsilon}(x,t)\,.$$ Thus (\[zwischen1\]) reduces to $$\label{zwischen2}
\partial_{t}\eta_{\varepsilon}(x,t)+\partial_{x}\xi_{\varepsilon}(x,t)=\int_{\mathbb{R}}\eta'(a)\partial_{a}q_{\varepsilon}(x,t,a)\,da\,.$$ Integrating the right-hand side by parts gives $$\int_{\mathbb{R}}\eta'(a)\partial_{a}q_{\varepsilon}(x,t,a)\,da=-\int_{\mathbb{R}}\eta''(a)q(x,t,a)\,da\,,$$ where we have used the fact, that $q_{\varepsilon}$ is compactly supported in $a$. This gives the result (\[entropy\_equality\]) as claimed.

Next we integrate inequality (\[entropy\_equality\]) over the set $\Gamma_{t_1+C\varepsilon}^{\bar{t}}$, where $\bar{t}\in(t_1+C\varepsilon,t_2-C\varepsilon)$. We will abbreviate $t_1+C\varepsilon$ by $\bar{t}_1$. We have $$\label{inequality_for_contraction}
   \int_{\Gamma_{\bar{t}_1}^{\bar{t}}}\partial_{t}\eta_{\varepsilon}(x,t)+\partial_{x}\xi_{\varepsilon}(x,t)\,dx\,dt=- \int_{\Gamma_{\bar{t}_1}^{\bar{t}}}\int_{\mathbb{R}}\eta''(a)q(x,t,a)\,da\,.$$ For the first term on the left-hand side of (\[inequality\_for\_contraction\]) we compute $$\begin{aligned}
\int_{\Gamma_{\bar{t}_1}^{\bar{t}}}\partial_{t}\eta_{\varepsilon}\,dx\,dt&=\int_{-\delta}^{\delta}\eta_{\varepsilon}(x,\bar{t})-\eta_{\varepsilon}(x,t_{1})\,dx\\
&\quad+\int_{\delta}^{\frac{(\bar{t}-\bar{t}_{1})}{\hat{\lambda}}+\delta}\int_{\hat{\lambda}(x-\delta)+\bar{t}_{1}}^{\bar{t}}\partial_{t}\eta_{\varepsilon}(x,t)\,dt\,dx\\
&\quad+\int^{-\delta}_{-\frac{\bar{t}-\bar{t}_{1}}{\hat{\lambda}}-\delta}\int_{-\hat{\lambda}(x+\delta)+\bar{t}_{1}}^{\bar{t}}\partial_{t}\eta_{\varepsilon}(x,t)\,dt\,dx\,.\end{aligned}$$ This gives $$\begin{aligned}
\int_{\Gamma_{t_1}^{\bar{t}}}\partial_{t}\eta_{\varepsilon}\,dx\,dt&=\int_{-\delta}^{\delta}\eta_{\varepsilon}(x,\bar{t})-\eta_{\varepsilon}(x,\bar{t}_{1})\,dx\\
&\quad+\int_{\delta}^{\frac{\bar{t}-\bar{t}_{1}}{\hat{\lambda}}+\delta}\eta_{\varepsilon}(x,\bar{t})-\eta_{\varepsilon}\left(x,\hat{\lambda}(x-\delta)+\bar{t}_{1}\right)\,dx\\
&\quad+\int^{-\delta}_{-\frac{\bar{t}-\bar{t}_{1}}{\hat{\lambda}}-\delta}\eta_{\varepsilon}(x,\bar{t})-\eta_{\varepsilon}\left(x,-\hat{\lambda}(x-\delta)+\bar{t}_{1}\right)\,dx\,.\end{aligned}$$ A regrouping of the terms together with a change of variable leads to $$\label{zwischen3}\begin{split}
\int_{\Gamma_{t_1}^{\bar{t}}}\partial_{t}\eta_{\varepsilon}\,dx\,dt&=\int_{\theta^{-}_{\varepsilon}(\bar{t})}^{\theta^{+}_{\varepsilon}(\bar{t})}\eta_{\varepsilon}(x,\bar{t})\,dx-\int_{-\delta}^{\delta}\eta_{\varepsilon}(x,\bar{t}_{1})\,dx\\
&\quad+\hat{\lambda}\int_{\bar{t}_{1}}^{\bar{t}}\eta_{\varepsilon}(\theta^{-}_{\varepsilon}(t),t)-\eta_{\varepsilon}(\theta^{+}_{\varepsilon}(t),t)\,dt\,,
\end{split}$$ where $$\label{theta}
\theta_{\varepsilon}^{\pm}(t)=\pm\frac{\bar{t}-\bar{t}_1}{\hat{\lambda}}\pm\delta\,.$$ Integrating now the second term on the left-hand side 