If $f_n \rightarrow f$ in $L^\infty$ then $\| f_h \|_\infty \rightarrow \| f \|_\infty$

Question

Theorem: Given a sequence $f_n$ in $L^\infty$, $f_n \rightarrow f$ a.e. and $f\in L^\infty$. If $f_n \rightarrow f$ in $L^\infty$ then $\| f_h \|_\infty \rightarrow \| f \|_\infty$.

My attempt: $|f_n-f| \le 0 $ for almost every $ x\in \Omega, \forall n \ge 1$ and I want to prove that $|f_n|- |f| \le 0 $ for almost every $ x\in \Omega, \forall n \ge 1 $. By hypotesis $f_n \rightarrow f$ a.e., so I have that also $|f_n |\rightarrow |f|$ a.e. and I don't know what to do next. I think I should find some inequality, but I'm stucked.

Answer 1

$f_n \to f$ in $L^\infty$ means that $\|f_n - f\|_\infty \to 0$. The triangle inequality gives you $$ \bigg| \|f_n\|_\infty - \|f\|_\infty \bigg| \le \|f_n - f\|_\infty$$ so that $\|f_n - f\|_\infty \to 0$ implies $\|f_n\|_\infty \to \|f\|_\infty$.