Error in my reasoning to prove a bounded convergence theorem for the Riemann integral I was trying to prove this fact:
If $f_n$ is a sequence of continuous functions from $[0;1]\to[0;1]$ such that $f_n(x)\to0\ \forall x\in[0;1]$ then $\lim_{n\to\infty}\int_0^1 f_n(x)dx=0$.
While trying to prove this, I reached the result that convergence is uniform, which sounds absurd. I'd like to see the error in my reasoning and see a proof of this fact (that doesn't depend on measure theory):
Fix $\varepsilon$
Let $E_n=\{x\in[0;1] / f_n(x)>\varepsilon\}$
Let $A_k=\cup_k^\infty E_n$.
Finally, let $A=\cap_1^\infty \overline{A_k}$.
There are two cases:
1) $A=\emptyset$. Then, since each $\overline{A_k}$ is compact and $\overline{A_{k+1}}\subset\overline{A_k}$ then $A_K$ is empty for some $A_K$. Thus, $E_n=\emptyset\ \forall n>K$. Thus, convergence is uniform.
2) $A\neq\emptyset$. Then, some $x$ belongs to $\overline{A_k}\forall k$. Then, $f_n(x)\geq\varepsilon\ \forall n_i$ and $n_i$ an infinite subsequence of the integers which means that $f_n(x)$ doesn't converge to $0$. 
I suspect that my error is in the last statement but I am not sure why and I have struggled to find a proof of the fact but I didn't reach anywhere. Thanks in advance.
 A: This is an example to make Giuseppe Negro's comment explicit.
Let $\{a_n\}\subset[0,1]$ be a strictly decreasing sequence such that $\lim_{n\to\infty}a_n=0$. Let $f_n\colon[0,1]\to[0,1]$ be a continuous function defined by $f_n(x)=0$ on $[0,a_{n+1}]\cup[a_n,1]$, $f(a_{n+1}+a_n)/2=1$, $f_n$ linear on $[a_{n+1},a_{n+1}+a_n)/2]$ and $[(a_{n+1}+a_n)/2,a_n]$. $E_n$ is an open interval centered at $(a_{n+1}+a_n)/2$ of length $(1-\epsilon)(a_{n+1}-a_n)$. $\overline A_k$ is the union of the closed intervals $\overline E_N$ and $\{0\}$, and $A=\{0\}$. However, $f_n(0)=0$ for all $n$.
You can find an elementary proof of the result in this paper by Nadish de Silva.
A: As GiuseppeNegro said, you can not choose the index "$n$" in your argument "$f_n(x) \geq \epsilon$". 
Because $x\in \overline{\cup_{n=k}^\infty E_n} = \overline{A_k}$, this does not imply $x\in \overline{E_N}$ for some $N \geq k.$ And you need $x\in \overline{E_N}$ in order to conclude $f_N(x) \geq \epsilon$. 
For counter example of the topological argument, $\overline{\cup_{q\in\mathbb{Q}} \{q\}}$ and $\cup_{q\in\mathbb{Q}} \overline {\{q\}}$.
