Let $\Omega \subset \mathbb{R}^n$ be a bounded domain and let $u \in L^2(0,T;H^1(\Omega))$ with $u_t \in L^2(0,T;H^{-1}(\Omega))$. Define the truncation function$$T_\epsilon(x) = \begin{cases} -\epsilon &: x \in (-\infty, -\epsilon]\\ x & x \in (-\epsilon, \epsilon)\\ \epsilon &: x \in [\epsilon, \infty) \end{cases}.$$ According to the second to last page in this paper,
clearly $$\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle_{H^{-1}(\Omega), H^1(\Omega)} = \int_\Omega |u(T)| - \int_\Omega |u(0)|$$
but I don't see this. We can write the LHS as $$\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle_{H^{-1}(\Omega), H^1(\Omega)} = \lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \left(\frac{d}{dt}\int uT_\epsilon(u) - \langle (T_\epsilon u)_t, u \rangle_{H^{-1}(\Omega), H^1(\Omega)}\right)$$ and the first term on the RHS is good (after using the FTOC) and we want to show the second term on the RHS vanishes. But I am not sure that the second term even makes sense since $u_t$ is a distribution not a function.
I tried to do this via a density argument (so let $u_n \to u$ where $u_n$ are smooth) but I was unable to pass to the limit on the left hand side of the desired equation.
Let $$ F_\varepsilon(x)=\int_0^x T_\varepsilon(s)ds. $$ Then \begin{eqnarray*} F_\varepsilon(x)=\left\{\begin{array}{ll} \frac{1}{2}\varepsilon^2+\varepsilon(x-\varepsilon), \text{ if } x>\varepsilon\\ \frac{1}{2}(x-\varepsilon)^2, \text{ if } 0<x\le\varepsilon\\ -\frac{1}{2}(x-\varepsilon)^2, \text{ if }-\varepsilon\le x\le 0\\ -\frac{1}{2}\varepsilon^2-\varepsilon(x+\varepsilon), \text{ if } x<-\varepsilon\\ \end{array}\right. \end{eqnarray*} and hence \begin{eqnarray*} \lim_{\varepsilon \to 0}\frac{1}{\varepsilon} \int_0^T \langle u_t(t), T_\varepsilon(u(t)) \rangle&=&\lim_{\varepsilon \to 0}\frac{1}{\varepsilon} \int_0^T\int_{\Omega} u_t(t) T_\epsilon(u(t) dxdt\\ &=&\lim_{\varepsilon \to 0}\frac{1}{\varepsilon} \int_0^T\int_{\Omega}\frac{d}{dt} F_\varepsilon(u(t) dxdt\\ &=&\lim_{\varepsilon \to 0}\frac{1}{\varepsilon} \int_0^T\frac{d}{dt}\int_{\Omega} F_\varepsilon(u(t) dxdt\\ &=&\lim_{\varepsilon \to 0}\frac{1}{\varepsilon} \left(\int_{\Omega} F_\varepsilon(u(T)) dx-\int_{\Omega} F_\varepsilon(u(0)) dx\right). \end{eqnarray*} We only need to show that $$ \lim_{\varepsilon \to 0}\frac{1}{\varepsilon}\int_{\Omega} F_\varepsilon(u)=\int_{\Omega}|u|dx. $$ In fact, \begin{eqnarray*} \lim_{\varepsilon \to 0}\frac{1}{\varepsilon}\int_{\Omega} F_\varepsilon(u)&=&\lim_{\varepsilon \to 0}\left(\int_{u>\varepsilon}F_\varepsilon(u)dx+\int_{u<-\varepsilon}F_\varepsilon(u)dx+\int_{0\le u\le \varepsilon}F_\varepsilon(u)dx+\int_{-\varepsilon\le u<0}F_\varepsilon(u)dx\right)\\ &=&\lim_{\varepsilon \to 0}\left(\int_{u>\varepsilon}(\frac{1}{2}\varepsilon^2+\varepsilon(u-\varepsilon))dx+\int_{u<-\varepsilon}(-\frac{1}{2}\varepsilon^2-\varepsilon(u+\varepsilon))dx\right.\\ &&\left.+\int_{0\le u\le \varepsilon}\frac{1}{2}u^2dx+\int_{-\varepsilon\le u<0}(-\frac{1}{2}u^2)dx\right)\\ &=&\int_{u>0}udx+\int_{u<0}(-u)dx\\ &=&\int_{\Omega}|u|dx. \end{eqnarray*}