Brezis' Exercise 4.13

Question

I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space.

1. Check that $$ || a+b|-| a|-| b|| \leq 2|b| \quad \forall a, b \in \mathbb{R} . $$
2. Let $\left(f_n\right)$ be a sequence in $L^1(\Omega)$ such that (i) $f_n(x) \rightarrow f(x)$ a.e., (ii) $\left(f_n\right)$ is bounded in $L^1(\Omega)$ i.e., $\left\|f_n\right\|_1 \leq M \quad \forall n$. Prove that $f \in L^1(\Omega)$ and that $$ \lim _{n \rightarrow \infty} \int\left\{\left|f_n\right|-\left|f_n-f\right|\right\}=\int|f|. $$
3. Let $\left(f_n\right)$ be a sequence in $L^1(\Omega)$ and let $f$ be a function in $L^1(\Omega)$ such that (i) $f_n(x) \rightarrow f(x)$ a.e., (ii) $\left\|f_n\right\|_1 \rightarrow\|f\|$. Prove that $\left\|f_n-f\right\|_1=0$.

Could you confirm if my below attempt is correct?

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Proof

1. This is straightforward.
2. By Fatou lemma, $$ \int |f| =\int \liminf_n |f_n| \le \liminf_n \int |f_n| \le M<\infty. $$ So $f \in L^1 (\Omega)$. Let $a_n := f_n-f$ and $b:=f$ and $$ \varphi_n := \big | | a_n+b|-| a_n|-| b| \big| =\big | |f_n| - |f|-|f_n-f| \big|. $$ By (1.), $\varphi_n \le 2|b| = 2|f|$. Clearly, $\varphi_n \to 0$ a.e. By dominated convergence theorem, $\varphi_n \to 0$ in $L^1$ and thus $|f_n|-|f_n-f| \to |f|$ in $L^1$.
3. We have $$ \int |f_n-f| = \int (|f_n-f| +|f|-|f_n|) \le \int \varphi_n \to 0 \quad \text{by (2.)} $$ This completes the proof.