Let $\{f_n\}_{n=1}^\infty$ be a sequence of non-negative measurable functions on a measurable set $E$.
(1). Suppose for any $\epsilon>0$, $$\sum_{n=1}^\infty \mu\{x\in E: f_n(x)>\epsilon\}<\infty.$$ Prove $\lim_{n\to\infty}f_n(x)=0$ a.e. on $E$.
(2). Suppose for any $\epsilon>0$, $$\lim_{n\to\infty} \mu\{x\in E: f_n(x)>\epsilon\}=0.$$ Can we still obtain that $\lim_{n\to\infty}f_n(x)=0$ a.e. on $E$?
(3). Suppose $\mu(E)<\infty$ and $\lim_{n\to\infty}f_n(x)=0$ a.e. on $E$. Can we obtain $$\sum_{n=1}^\infty \mu\{x\in E: f_n(x)>\epsilon\}<\infty$$ for any $\epsilon>0$?
My attempt:
I want to prove (2) and consequently obtain (1).
Let $f=\limsup_{n\to\infty} f_n$. I want to prove $\mu\{x\in E:F(x)\neq 0\}=0$. Since $$\{x\in E:F(x)\neq 0\}=\bigcup_{j=1}^\infty\{x\in E:F(x)>\frac{1}{j}\},$$ I want to show for each $j\in\mathbb{N}$, $$\mu\{x\in E:F(x)>\frac{1}{j}\}=0.$$ Let $x\in\{x\in E:F(x)>\frac{1}{j}\}$. Then by definition, we can choose a subsequence $\{f_{n_k}\}_{k=1}^\infty$ such that $f_{n_k}(x)>\frac{1}{j}$ for all $k$. Hence $$ \{x\in E:F(x)>\frac{1}{j}\}\subseteq \bigcup_{1\leq n_1<n_2<n_3<\cdots}(\bigcap_{k=1}^\infty\{x\in E: f_{n_k}(x)>\frac{1}{j}\}). $$ By (2), we obtain $$ \mu(\bigcap_{k=1}^\infty\{x\in E: f_{n_k}(x)>\frac{1}{j}\})=0. $$ Now I confront with difficulties. Since the union is taken over all subsequences, The union is not countable! Thus the measure of the union may not be zero!
(3). By Egoroff theorem, for any $\eta>0$, there exists measurable subset $G\subseteq E$ such that $\mu(G)<\eta$ and $\{f_n\}$ converges uniformly to $0$ on $E\setminus G$. Thus there exists $N$ such that for any $n>N$, $f_n(x)<\epsilon$ for all $x\in E\setminus G$. Thus for all $n>N$, $\mu\{x\in E: f_n(x)>\epsilon\}<\eta$. Hence $\lim_{n\to\infty} \mu\{x\in E: f_n(x)>\epsilon\}=0$.
Can we obtain the stronger result $\sum_{n=1}^\infty \mu\{x\in E: f_n(x)>\epsilon\}<\infty?$