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!
 A: By the Monotone Convergence Theorem, 
$$\sum_{n=1}^{\infty}\int f_{n}\ d\mu = \int \sum_{n=1}^{\infty} fn\ d\mu < \infty$$
The above implies that $\sum_{n=1}^{\infty}f_{n} < \infty$ a.e. $(\mu)$. Therefore $f_{n}(x) \rightarrow 0$ a.e. ($\mu$).
A: Just because a sequence of functions has integral tending to zero doesn't give you pointwise convergence. For example, one standard example is to take a sequence of indicator functions moving back and forth inside $[0,1]$ that have smaller and smaller support; for example, indicators on $[0, 1/2]$ and $[1/2, 1]$< then $[0, 1/3], [1/3, 2/3]$ and $[2/3, 1]$, and so on. The best result you can get is a subsequence which converges pointwise a.e.
For a start in a different direction that does lead to a proof, consider applying the Borel-Cantelli lemma.
A: 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 $x$,
$$
\lim_{n\to\infty}f_n(x)=0
$$
A: I'm going to assume


*

*$f_n$'s are $\mu$-integrable and in measure space $(S, \Sigma, \mu)$.

*By $\int X$ you mean $\int_S X d\mu$

Since the $f_n$'s are nonnegative because they are positive, we can switch summation and integral (this is because of MCT):
$$\sum_{n=1}^{\infty}\int_S f_n d\mu = \int_S \sum_{n=1}^{\infty} f_n d\mu < \infty$$
I recall a rule in probability theory that states $E[X] < \infty \to X < \infty \ \mu-\text{a.s.}$ (to see why, try to prove the contrapositive). I'm guessing this comes from measure theory: $\int_S X d \mu < \infty \to X < \infty \ \mu-\text{a.e.}$
Hence we have that $\sum_{n=1}^{\infty}f_{n} < \infty \ \mu-\text{a.e.}$.
We know from basic calculus that $\sum_n a_n < \infty \to \lim a_n = 0$
$a_n$ is a sequence of numbers, but it applies to a sequence of functions, something we know from real analysis.
$\therefore, \lim f_n = 0 \ \mu-\text{a.e.}$.
