Prove: $A=\{x:f_n(x)\to 0\}$ is Lebesgue measurable Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of continuous function on $[0,1]$ 
Prove: $A=\{x:f_n(x)\to 0\}$ is Lebesgue measurable
What I know: from what I understand it is harder to show that a set is lebesgue measurable, it sufficient to show that it is Borel measurable and then $B([0,1])\subset L([0,1])$
especially when we have continuous function which we know that the pre-image of an open set it open too, and Borel is the smallest $\sigma$ algebra of open sets for a given set.
So we first need to look at the pre image of the functions, the functions is continuous so by definition  $$|f_n(x)-0|< \varepsilon\Rightarrow |f_n(x)|<\varepsilon$$
So $$A=\{x:|f_n(x)|< \varepsilon\}$$
let $f_n(x)\in E\subset [0,1]$ be an open group, then pre image of $E$ is open and $A$ is an open set and therefore $A\in B([0,1])$
Now there is a hint to use intersection and unions, which are called $G_\delta, G_\sigma$, how can it be used to prove the claim? and is my first reasoning is correct?
 A: $f_n(x) \to 0$ if and only if $\forall m \in \mathbb N , \exists N \in \mathbb N$ such that $\forall n > N , f_n(x) \in (-\frac 1m, \frac 1m)$.
Thus , we have :
$$
\{x : f_n(x) \to 0\} = \displaystyle\bigcap_{m \in \mathbb N}\bigcup_{N \in \mathbb N} \bigcap_{n > N} \left\{x : f_n(x) \in \left(-\frac 1m , \frac 1m\right)\right\}
$$
To prove this, read $\cap$ as "for every / for all" and $\cup$ as "for some / there exists" and you see the definition of the LHS on the RHS.
The innermost sets are open sets for each $m,n$ since $f_n$ are continuous. In particular, they are Borel. Now we are just taking intersections and unions, all of which are countable, therefore the resulting set is also Borel, in particular Lebesgue measurable.
A: If $(f_n)$ is a sequence of measurable functions, then $\liminf f_n$ and $\limsup f_n$ are also measurables and so is $H(x) = (\liminf f_n(x), \limsup f_n(x))$. Then,
$$
A = \big \{ x : \liminf f_n(x) = \limsup f_n(x) = 0 \big \} = H^{-1}(O)
$$
where $O= (0,0)$ which is measurable since closed and so $A$ is Borel.
This result generalizes to the set $A := \{ x : \lim f_n \ \text{exists}\}$ since $A = H^{-1}(U)$ where $U$ is a closed set of $[0,1]^2$.
