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 $0$

Convergence of the integrals $\int_0^1 f_n \to 0$ does, in general, not imply $f_n \to 0$; therefore your approach doesn't work.
Hints:


*

*Recall the Borel-Cantelli lemma: Let $\mu$ be a finite measure. If $(A_n)_{n \in \mathbb{N}}$ is a sequence of measurable sets such that $$\sum_{n \geq 1} \mu(A_n)<\infty$$ then $$\mu \left( \limsup_{n \to \infty} A_n \right) = 0.$$

*Define $$A_n := \{x \in [0,1]; |f_n(x)| \geq 1/k\}$$ for fixed $k \geq 1$. Use Markov's inequality and the Borel-Cantelli lemma to show that $\limsup_{n \to \infty} A_n$ has Lebesgue measure $0$.

*By step 2 there exists for any $k \in \mathbb{N}$ a null set $N_k$ such that for any $x \in [0,1] \backslash A_k$ it holds that $$|f_n(x)| \leq \frac{1}{k}$$ for $n \gg 1$ sufficiently large. Conclude that $f_n(x) \to 0$ for any $x \in [0,1] \backslash \left( \bigcup_{k \geq 1} N_k \right)$.

