i found a theorem saying that if u belongs to $H^{1}(\Omega)$ where $\Omega$ is bounded regular domain of $R^{n}$ then $\nabla u^{+}=\chi_{u>0}\nabla u $ here $u^{+} =sup(0,u)$
the hint is to use this sequence: $\phi_{\epsilon}(u)=.\sqrt{\epsilon^{2}+ u^{2}}-\epsilon$ if $u\ge 0$ and equal $0$ if $u\le 0$ then $\phi_{\epsilon}(u)$ converge to $u^{+}$ in $L^{2}(\Omega)$
and $\nabla \phi_{\epsilon}(u)$ converge to $\chi_{u>0}\nabla u$ in $L^{2}(\Omega)$ in fact we have :
$\lvert \phi_{\epsilon}(u)\rvert \le \vert u\rvert $ because $\sqrt{\epsilon^{2}+ u^{2}}\le \vert u+ \epsilon \rvert$ so $\eta_{\epsilon}(u)$ converge to $u^{+}$ in $L^{2}(\Omega)$ And $\lvert \nabla \phi_{\epsilon}(u)\rvert \le \lvert \nabla u\rvert $ and i conclude by saying that convegence in $L^{2}$ imply convergence in distribution is my proof right? thanks