I'm trying to prove this below result
Theorem: Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Let $f_n, f \in \mathcal L_1 (X, \mu, E)$ such that $f_n \to f$ $\mu$-a.e. and $\|f_n\|_1 \to \|f\|$ as $n \to \infty$. Then $\|f_n - f\| \to 0$.
Could you have a check if I correctly apply Fatou's lemma.
My attempt: We have $$ \begin{align} \limsup_n \int |f_n-f| &= - \liminf_n \int -|f_n-f|\\ &= - \liminf_n \int [|f_n| + |f|-|f_n-f|-(|f_n|+|f|)] \\ &\le - \liminf_n \int (|f_n|+|f|-|f_n-f|) - \liminf_n \int-(|f_n| +|f|) \\ &= - \liminf_n \int (|f_n|+|f|-|f_n-f|) + \limsup_n \int(|f_n|+|f|). \end{align} $$
- Because $f_n \to f$ $\mu$-a.e., we get $|f_n|+|f|-|f_n-f| \to 2|f|$ $\mu$-a.e. By Fatou's lemma, $$ \liminf_n \int (|f_n|+|f|-|f_n-f|) \ge 2\int |f| = 2 \|f\|_1. $$
- Because $\|f_n\|_1 \to \|f\|_1$, we get $$ \limsup_n \int (|f_n| + |f|) = \|f\|_1 + \limsup_n \|f_n\|_1= 2 \|f\|_1. $$
It follows that $$ \lim_n \|f_n-f\|_1 \le \limsup_n \|f_n-f\|_1 = \limsup_n \int |f_n-f| \le 0. $$
This completes the proof.