If $\sum_{n=1}^\infty \mu\{x \in X : |f_n(x)| \geq \epsilon\}$ converges for each $\epsilon > 0$, then $f_n \to 0$ a.e. 
Let $\{f_n\}$ be a sequence of measurable functions on a measure space $(X, \mathcal{M}, \mu)$.  Suppose that the infinite series
  $\displaystyle \sum_{n=1}^\infty \mu\{x \in X : |f_n(x)| \geq \epsilon\}$ converges for each $\epsilon > 0.$  Prove that $f_n(x) \rightarrow 0$ a.e.

I am not really sure how to approach this problem.  Some help would be awesome.  Thanks.  It is a past qual problem.
 A: We have $$\displaystyle \sum_{n=1}^\infty \mu\{x \in X : |f_n(x)| \geq \epsilon\} =  \mu(\displaystyle \sum_{n=1}^\infty1_{ |f_n(x)| \geq \epsilon}) < \infty$$ which means $\displaystyle \sum_{n=1}^\infty1_{ |f_n(x)| \geq \epsilon}$ is finite $\mu$-a.e.
That is to say for any $\epsilon$, there are only finitely many $n$ such that $|f_n(x)| \geq \epsilon$
A: Denote $E =\{x : f_n(x) \to 0\}$. We prove that
$$E = \bigcap_{k=1}^\infty \bigcup_{m =1}^\infty \bigcap_{l=m}^\infty \left\{x: |f_l(x)| \leq \frac 1k\right\}.$$
Denote the set in right hand side by $F$. If $x\in E$ then $f_n(x) \to 0$, hence for any $k\geq 1$, there exists $m$ dependings on $k$ such that $|f_l(x)| \leq 1/k$ for any $l\geq m$. This shows that $x\in F$. Conversely, if $x\in F$, for any $\epsilon >0$, choosing $k_0$ such that $1/k_0 < \epsilon$ since $x\in F$ then 
$$x\in \bigcup_{m =1}^\infty \bigcap_{l=m}^\infty \left\{x: |f_l(x)| \leq \frac 1{k_0}\right\}.$$
This means that there exists $m$ such that $|f_l(x) |\leq 1/k_0 < \epsilon$ for any $l\geq m$. This means that $f_n(x) \to 0$, or $x\in E$. Hence $E =F$.
We have
$$E^c = \bigcup_{k=1}^\infty \bigcap_{m=1}^\infty \bigcup_{l=m}^\infty \left\{x : |f_l(x)| >\frac1k\right\}. $$
Hence
$$\mu(E^c) \leq \sum_{k=1}^\infty \mu\left(\bigcap_{m=1}^\infty \bigcup_{l=m}^\infty \left\{x : |f_l(x)| >\frac1k\right\}\right)\leq \sum_{k=1}^\infty \lim_{m\to\infty} \mu\left(\cup_{l=m}^\infty \left\{x : |f_l(x)| >\frac1k\right\}\right).$$
From our assumptions, we have
$$\lim_{m\to\infty} \mu\left(\cup_{l=m}^\infty \left\{x : |f_l(x)| >\frac1k\right\}\right) \leq \lim_{m\to \infty} \sum_{l=m}^\infty \mu\left(\left\{x : |f_l(x)| >\frac1k\right\}\right) = 0.$$
Therefor $\mu(E^c) = 0$.
