Convergence in measure and related inequality

Question

Let $f_n$ be a sequence of bounded measurable functions on some probability space. Let $f_n$ does not converge to $0$ a.e. Does it mean that there exists a set of positive measure $E$ such that $\lim\sup\limits_{n\to\infty}|f_n(x)|>2\epsilon$ for all $x\in E$?

Answer 1

The correct statement is: There exists $\epsilon >0$ and a set $E$ of positive measure such that $\lim \sup |f_n(x)| >2\epsilon$ for all $x \in E$.

You can prove this by contradiction. If this is not true then $\mu \{x: \lim \sup |f_n(x)| > \frac 1 k\} =0$ for every $k \in \mathbb N$. Let $E =\cup_k \{x: \lim \sup |f_n(x)| > \frac 1 k\}$. Then $\mu (E)=0$ and $f_n \to 0$ for all $x \in E^{c}$ contradicting the hypothesis.