$f_n \to f$ a.e. and $\lvert f_n(x) \rvert \le \lvert f(x) \rvert $ where $f_n, f \in L^\infty$

Question

If we have $f_n \to f$ a.e. and $\lvert f_n(x) \rvert \le \lvert f(x) \rvert $ a.e. where $f_n, f \in L^\infty$

Does this give enough to ensure $\lvert \vert f_n - f \rvert \rvert_\infty \to 0 $?

This question provides a suitable counterexample if we omit the condition that $\lvert f_n(x) \rvert \le \lvert f(x) \rvert $ a.e.

In the above link $f_n \to 0$ a.e. but always a set of positive measure where the bounding fails. So I am trying to find a potential counterexample.

Answer 1

No. Define $f: (0,1] \to \mathbb R$ by $f(x) = \frac 1 x$ for $x \in (0,1]$, and let $f_n(x) = \min(n,f(x))$ for any $x \in (0,1]$. Then $\lvert f_n(x) \rvert \le \lvert f(x)\rvert$ for all $x$ and $f_n(x) \to f(x)$ for all $x \in(0,1]$, but $f_n \not \to f$ in $L^\infty$.