Given a sequence $u_n\in BV(\Omega)$ and $u\in BV(\Omega)$, where $\Omega\in R^n$ is open. We assume $u_n\to u$in $L^1_{loc}(\Omega)$ and we also assume that $$ \lim_{n\to\infty}|D u_n|(\Omega)=|Du|(\Omega).$$
Now, give another open set $V\subset \Omega$, we want to prove $$ \lim_{n\to\infty}|D u_n|(V)=|Du|(V)$$ given that $|Du|(\partial V \cap \Omega)=0$.
I don't know where should I get start. A hint would be very useful!
This follows more-or-less from definitions.
The variation of a function $u \in L^1_{\rm loc}(\Omega)$ is the quantity $$|Du|(\Omega) = \sup \left\{ \int_\Omega u\ {\rm div\, }v \, dx: v \in C_0^\infty(\Omega;\mathbb R^n),\ |v| \le 1 \right\}.$$ Thus if $u_k \to u$ in $L^1_{\rm loc}(\Omega)$ you have $$ \int_\Omega u\ {\rm div\, }v \, dx = \lim_{k \to \infty} \int_\Omega u_k\ {\rm div\, }v \, dx \le \liminf_k |Du_k|(\Omega)$$ whenever $v \in C_0^\infty(\Omega;\mathbb R^n)$ and $|v| \le 1$. Take the sup over all such $v$ to get $$|Du|(\Omega) \le \liminf_k |Du_k|(\Omega).$$ Similarly, if $V \subset \Omega$ is open you have $$|Du|(V) \le \liminf_k |Du_k|(V) \quad \text{and} \quad |Du|(\Omega \setminus \overline V) \le \liminf_k |Du_k|(\Omega \setminus \overline V).$$
Now suppose that $u_k,u \in BV(\Omega)$ and that $|Du_k|(\Omega) \to |Du|(\Omega)$. Then \begin{align*}\limsup_k|Du_k|(V) + |Du|(\Omega \setminus \overline V) &\le \limsup_k|Du_k|(V) + \liminf_k |Du_k|(\Omega \setminus \overline V) \\ &\le \limsup_k \left( |Du_k|(V) + |Du_k|(\Omega \setminus \overline V) \right) \\ &\le \limsup_k |Du_k|(\Omega) \\ &= |Du|(\Omega).\end{align*} But $$ |Du|(\Omega) = |Du|(V) + |Du|(\Omega \setminus \overline V) + |Du|(\Omega \cap \partial V)$$ so that $|Du|(\Omega \cap \partial V) = 0$ implies $$\limsup_k |Du_k|(V) \le |Du|(V).$$