There's a computation in a proof that I'm having trouble following. I think this is a basic fact about limsup and liminfs that's escaping me. The proposition is as follows:
Assume $A$ is open and $B = \Omega - \overline{A}$ so that $B$ is open (here $\Omega \subset \mathbb{R}^n$). Suppose $\{f_j\}$ is a sequence of Bounded variation (definition here) s.t. $f_j \rightarrow f$ in $L_{loc}^1(\Omega)$ and $$\lim_{j \rightarrow \infty}\int_{\Omega}|Df_j| \rightarrow \int_{\Omega}|Df|$$ then $\forall A$ we have: $$\int_{\overline{A} \cap \Omega}|Df| \geq \limsup_{j \rightarrow \infty} \int_{\overline{A} \cap \Omega}|Df_j|$$
The proof proceeds as follows:
Previous results show that for $U$ open, $\int_{U}|Df| \leq \liminf_{j \rightarrow \infty} \int_{U}|Df_j|$. Then they show the following computation:
I don't understand how the second to last line is obtained.
There's nothing going on really. As happened in the first line of the computation, for each $j$ we have $\int_\Omega|Df_j|=\int_{\overline{A}\cap\Omega}|Df_j|+\int_B|Df_j|$, hence $$\lim_{j\to\infty}\int_\Omega|Df_j|=\lim_{j\to\infty}\big(\int_{\overline{A}\cap\Omega}|Df_j|+\int_B|Df_j|\big)$$ Now when a limit is existent it is equal to the limsup, so $$\lim_{j\to\infty}\big(\int_{\overline{A}\cap\Omega}|Df_j|+\int_B|Df_j|\big)=\limsup_{j\to\infty}\big(\int_{\overline{A}\cap\Omega}|Df_j|+\int_B|Df_j|\big)$$
Now if we show that given two (bounded) sequences of positive numbers $(a_n),(b_n)$ we have that $\limsup(a_n+b_n)\geq\limsup(a_n)+\liminf(b_n)$ we will be done (apply for $a_j=\int_{\overline{A}\cap\Omega}|Df_j|$ and $b_j=\int_B|Df_j|$).
We will use the classical inequality $\limsup(c_n+d_n)\leq\limsup(c_n)+\limsup(d_n)$ which can be found anywhere on MSE and the equality $\limsup(-c_n)=-\liminf(c_n)$ for positive sequences which is easy.
We have $$\limsup(a_n)=\limsup(a_n+b_n-b_n)\leq$$ $$\leq\limsup(a_n+b_n)+\limsup(-b_n)=\limsup(a_n+b_n)-\liminf(b_n).$$ You can rewrite this inequality as $\limsup(a_n+b_n)\geq\limsup(a_n)+\liminf(b_n)$, which is what we wanted.