The infinite sum of integral of positive function is bounded so function tends to 0 
Let $f_n(x)$ be positive measurable functions such that 
  $$\sum_{n=1}^\infty \int f_n \lt \infty.$$
  Show that $f_n \to 0$ almost everywhere.

Attempt:
Let $\displaystyle K = \sum_{n=1}^\infty\int f_n$ and $\displaystyle S_m = \sum_{n=1}^m \int f_n$. Then,  $\forall \epsilon \gt 0$, $\exists L$ such that $\forall m \gt L$, $|S_m - L| \le \epsilon$.
That is, $\displaystyle \sum_{n=m+1}^\infty \int  f_n \lt \epsilon$. Therefore, $\forall n \gt L$ we have  $\displaystyle \int f_n \lt \epsilon $, then the result should follow. 
I don't know why the grader of my class said this proof is wrong.
If I am truly wrong, where is my error?
Thanks!

Define
$$
E_k=\left\{x:\limsup_{n\to\infty}f_n(x)\ge\frac1k\right\}
$$
For each $x\in E_k$, $f_n(x)\ge\frac1{2k}$ infinitely often. Therefore, for each each $x\in E_k$
$$
\sum_{n=1}^\infty f_n(x)=\infty
$$
Thus, if the measure of $E_k$ is positive,
$$
\int_{E_k}\sum_{n=1}^\infty f_n(x)\,\mathrm{d}x=\infty
$$
Therefore, the measure of each $E_k$ must be $0$. Thus,
$$
\left|\bigcup_{k=1}^\infty E_k\right|\le\sum_{k=1}^\infty\left|E_k\right|=0
$$
However,
$$
\left\{x:\limsup_{n\to\infty}f_n(x)\ne0\right\}\subset\bigcup_{k=1}^\infty E_k
$$
Therefore, for almost every $x$,
$$
\lim_{n\to\infty}f_n(x)=0
$$
