A function of Locally Bounded Variation has a definition given on wikipedia here. (This question doesn't require knowing what it is)
I'm trying to follow the argument they give for showing that local variation is lower semicontinuous. The last equality confuses me:
Choose a Cauchy sequence of BV-functions $\{u_n\}_{n\in\mathbb{N}}$ converging to $u\in L^1_\text{loc}(\Omega)$. Then, since all the functions of the sequence and their limit function are integrable and by the definition of lower limit:
$$\liminf_{n\rightarrow\infty}V(u_n,\Omega) \geq \liminf_{n\rightarrow\infty} \int_\Omega u_n(x)\operatorname{div}\, \boldsymbol{\phi}\, \mathrm{d}x \geq \int_\Omega \lim_{n\rightarrow\infty} u_n(x)\operatorname{div}\, \boldsymbol{\phi}\, \mathrm{d}x = \int_\Omega u(x)\operatorname{div}\boldsymbol{\phi}\, \mathrm{d}x $$ $$\qquad\forall\boldsymbol{\phi}\in C_c^1(\Omega,\mathbb{R}^n),\quad\Vert\boldsymbol{\phi}\Vert_{L^\infty(\Omega)}\leq 1 $$
I'm not clear about the equality on the right. We only know that $\|u_n - u\|_{L^1_{loc}} \rightarrow 0$, not that $\lim u_n(x) = u(x)$ pointwise. Can someone clarify the equality on the right? Thanks!!