If the integrals of a sequence of integrable functions is bounded, why do the functions converge almost surely? Suppose $f_n$ is a sequence of Lebesgue integrable functions over $\mathbb R$ with $\int_{\mathbb R}|f_n|<\frac{1}{n^2}$. Then why is it true that $f_n\rightarrow 0$ $a.e.$?
 A: For any $\epsilon >0$, let $E_n = \{x: |f_n(x)| \geqslant \epsilon\}$.
Since, 
$$\frac{1}{n^2} > \int_{\mathbb{R}}|f_n| \geqslant  \int_{E_n}|f_n| \geqslant \epsilon \mu(E_n)$$
we have $\mu(E_n) < 1/(n^2 \epsilon)$ and
$$\sum_{n=1}^\infty \mu(E_n) < \infty.$$
Convergence of $f_n \to 0$ a.e. then follows from the first Borel-Cantelli lemma,
$$\mu(E_n \,\, \text{infinitely often})= \mu \left(\bigcap_{n =1}^\infty \bigcup_{k\geqslant n}E_k\right) = 0.$$
Except on a set of measure $0$ we have for any $\epsilon > 0$, $|f_n(x)| \geqslant \epsilon$ for at most finitely many $n$.
A: You can also argue as follows. Let $g_N$ be defined such that
$$
g_N =\sum_{n=1}^N |f_n|.
$$
Clearly, $0 \leq g_N \nearrow \sum_{n=1}^{\infty}|f_n|$. Therefore,
$$
\int \sum_{n=1}^{\infty}|f_n| = \lim_{N\to\infty}\sum_{n=1}^N \int |f_n| \leq \lim_{N\to\infty}\sum_{n=1}^N \frac{1}{n^2}  = \frac{\pi^2}{6} < \infty
$$
where the first equality follows from monotone convergence theorem. Hence, $\sum_{n=1}^{\infty}|f_n|$ is integrable, and therefore, for almost every $x$,
$$
\sum_{n=1}^{\infty}|f_n(x)| < \infty
$$
which is precisely saying that for almost every $x$, $\lim_{n\to\infty}|f_n(x)| = 0$, concluding the proof.
