The following set will be $G_\delta$ set. 
Can anyone please help me out? I know what is $G_\delta$ .
 A: $$\big\{x\in \Bbb R :\limsup \big|f_n(x)\big|=\infty\big\}$$$$=\bigcap_{n=1}^\infty\bigcup_{m=n}^\infty\big\{x\in \Bbb R:\big|f_m(x)\big|>n\big\}.$$

For a sequence of non-negative reals $\{x_n\}$ we say $\limsup
x_n=\infty\iff$ for each $G>0$ there is a natural number $n$ such that
$x_n>G\iff$ there is a subsequence $\{x_{n_k}\}_{k\in\Bbb N}$ with
$\lim_{k\to \infty}x_{n_k}=\infty$.

Note that for each $m\in\Bbb N$ the set $\big\{x\in \Bbb R:\big|f_m(x)\big|>n\big\}$ the sets are open for each $n$ as $f_m$ are continuous.
A: $\limsup_{m \to \infty} |f_m(x)| = +\infty$ precisely when there is a subsequence $n_k$ such that $|f_{n_k}(x)| \to +\infty$ as $k \to \infty$.
So for any boundary $n \in \Bbb N$ we can find $m > n$ such that $|f_{m}(x)| > m$.
This is exactly equivalent to the subsequence condition, but written in "countable terms".
So $x \in \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty O_m$ where $O_m= \{x\mid |f_m(x)| > m\}$ is open by continuity of $f_m$ and $|f_m|$. So the union (for each $n$) is open and we see $x$ is equal to an intersection of open sets, so is a $G_\delta$.
A: A $G_{\delta}$ set is a countable intersection of open sets. The set $S$ in question is just the set of all $x$ such that $|f_n(x)|$ is unbounded above. So
$\tag1 S=\{x:\forall m\in \mathbb N\ \exists n\ge m\in \mathbb N\ \ni|f_n(x)|>m|\}$
Thus,
$\tag2 S=\bigcap_{m=1}^\infty\bigcup_{n=m}^\infty\{x:|f_n(x)|>m\}$
which is a $G_{\delta}$
