Show that $\sum^\infty_{n=1} \mu(\{x : |f_n(x) - f(x)| > \epsilon\}) < \infty$ implies $f_n \to f$ a.e. Show that $\sum^\infty_{n=1} \mu(\{x : |f_n(x) - f(x)| > \epsilon\}) < \infty$ implies $f_n \to f$ a.e, where $f_n$ and $f$ are measurable functions.
My attempt: The Borel Cantelli lemma gives us that almost all $x$ are in at most finitely many $A_n = \{x : |f_n(x) - f(x)| > \epsilon\}$, so, for each $x$, there exists some $N$ such that for all $n \ge N$, $|f_n(x) - f(x)| \le \epsilon$. Since this is true for all $\epsilon$, we let $\epsilon$ go to zero, and we are done.
This is what I have, but someone told me that I have an issue with countability, and that in order to fix it, I need to consider $\epsilon = 1/k$. But, I don't see why my proof is incorrect, so I don't see how that would fix it.
 A: Your proof is essentially fine. The countability issue is on $\epsilon$. Your proof works for a fixed $\epsilon>0$, but how extend it to all $\epsilon>0$?  
Here is how to solve it. 
Let $A_{n,m} = \{x : |f_n(x) - f(x)| > \frac{1}{m}\}$ and let $[f_n \nrightarrow f]$ be the set of $x$ where $f_n(x)$ does not converge to $f$. It is easy to see that 
$$[f_n \nrightarrow f] = \bigcup_{m=1}^\infty\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_{k,m}$$
Since, for all $m$,  $\sum^\infty_{n=1} \mu(A_{n,m}) < \infty$, by Borel-Cantelli, we have that 
$$\mu \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_{k,m} \right)=0$$
So we have 
$$\mu([f_n \nrightarrow f]) = \mu \left (\bigcup_{m=1}^\infty\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_{k,m}\right)=0$$
So $f_n \to f$ a.e.
A: Let $A_n = \{x : |f_n(x) - f(x)| \geqslant\epsilon\}$. By Borel Cantelli lemma, we have
$$
\sum^\infty_{n=1} \mu(\{x : |f_n(x) - f(x)| \geqslant \epsilon\}) < \infty\implies\mu(\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k)=0
$$
Since
$$
x\in\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k\iff x\in A_n \hspace{3 mm}\text{infinitely often}
$$
Which means that there are infinitely many $n$ that $|f_n(x) - f(x)| \geqslant\epsilon\,$ or $\,f_n(x) \nrightarrow f(x)$. So we have $f_n \to f$ a.e.
