If $\int_0^1 f_n dx \leq a_n $ where $\sum_{n=1}^{\infty}a_n < \infty$ show $f_n \to 0$ 
Suppose $\{f_n\}_{n=1}^{\infty}$ is a sequence of nonnegative measurable functions on $[0,1]$ with 
  $$\int_0^1 f_n \, dx \leq a_n \qquad \text{for all}   \quad  n\geq 1$$
  where $$\sum_{n=1}^{\infty}a_n < \infty.$$ Prove that $f_n \to 0$ a.e on $[0,1]$

What I am able to do so far
$f_n\geq 0$, measurable , $x\in [0,1]$
$f_n \geq 0 \Rightarrow 0 \leq \int_0^1 f_ndx \leq a_n$ , $\Rightarrow a_n \geq 0, \forall n$
Hence $\{a_n\}$ is  convergent.
thus $\sum_{n=1}^{\infty}|a_n| <\infty$
Therefore $\forall \epsilon >0$, $\exists N $such that  $\forall n\geq N$ $$\sum_{n=N}^{\infty}a_n<\epsilon$$
and so $$n\geq N, 0\leq  \int_0^1 f_n \leq \epsilon$$
But $\epsilon$ is arbitrary $\Rightarrow \int_0^1 f_n \to 0$
That is to say the integral converges to $0$, but how do I show that the sequence itself converges to $0$

By the monotone convergence theorem,
$$
\int_0^1\sum_{n=1}^\infty f_n = \sum_{n=1}^\infty\int_0^1 f_n \le \sum_{n=1}^\infty \int_0^1a_n =\sum a_n< \infty.
$$
Thus, for almost every $x$ we have that $\sum_{n=1}^\infty f_n(x) < \infty$. Since the series $\sum f_n(x)$ is convergent for almost every $x$, we can conclude that $f_n(x)\to 0$ for almost every $x$.
