strengthen the condition of convergence in measure of sequence of functions 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 $\sum_{n=1}^\infty \mu\{x\in E: f_n(x)>\epsilon\}<\infty?$
 A: Everything that you wrote is correct. To summarize, your query is about different modes of convergence. Let's introduce the following (with the underlying assumption that here, the functions are nonnegative):


*

*$(f_n)$ converges almost completely to $0$ if $\sum_{n=1}^\infty \mu\{x\in E: f_n(x)>\epsilon\}<\infty$;

*$(f_n)$ converges a.e. to $0$ if for a.e. $x\in E$, $f_n(x)\xrightarrow[n\rightarrow+\infty]{}0$;

*$(f_n)$ converges in measure to $0$ if $\forall\epsilon>0,\ \mu\{x\in E: f_n(x)>\epsilon\}\xrightarrow[n\rightarrow+\infty]{}0$.


By Borell-Cantelli's lemma, 1. implies 2., and it is well-known that 2. implies 3. . This answers your first question. Your next two questions deal with the converse implications, which turn out to be false in general.
There are many examples illustrating that 3. need not imply 2. even if $\mu(E)<\infty$, see here for example.
I had not thought about the implication 2.$\Longrightarrow$1. previously, but have found a nice counterexample given by Did here. I'll sumarize what's relevant. Choose $E=[0,1]$ with the uniform measure $\mu([a,b])=b-a$ and define, for every $n\geqslant1$, the Bernoulli random variable $f_n$ by
$$
f_n(x)=1_{[0,1/n]}(x),\quad x\in[0,1].
$$
Then, $f_n$ tends to $0$ a.e., but for $\epsilon=1/2$, $\sum_{n=1}^\infty \mu\{x\in E: f_n(x)>\epsilon\}=\sum_{n=1}^\infty1/n=\infty$.
To summarize, (1) holds by the Borel-Cantelli lemma, and the answer to (2) and (3) is no.
