If $\{f_n\}$ is sequence of measurable functions on $X$, then $\{x: \lim f_n(x) \text{ exists}\}$ is a measurable set. 
If $\{f_n\}$ is sequence of measurable functions on $X$, then $\{x: \lim f_n(x) \text{ exists}\}$ is a measurable set.

My idea is to prove that
$$\{x: \lim f_n(x) \text{exists}\}=\{x: \liminf f_n(x)=a<\infty\}\cap \{x: \limsup f_n(x)=a<\infty\}=\bigg(\cup_{j=1}^{\infty}\cap_{n\geq j}\{x: f_n(x)=a\}\bigg)\cap\bigg(\cap_{j=1}^{\infty}\cup_{n\geq j}\{x: f_n(x)=a\}\bigg)$$
which two sets are measurable and their insection is also measurable. Is it correct?
Moreover,
Notice that function $g=\limsup f_n-\liminf f_n$ is measurable because $f_n$ is measurable and by proposition 2.7. So $$\{x: \lim f_n\text{ exists} \}=\{x: \limsup f_n=\liminf f_n\}=\{x: g(x)=0\}.$$ is measurable which because $g^{-1}(\{0\})$ is measurable.
Another question:
Do I need to consider that $ \limsup f_n=\pm \infty \text{ or } \liminf f_n=\pm \infty$
 A: Here's another way of doing it. Define
$$
g:=\liminf_{n\to\infty} f_n
$$
$$
h:=\limsup_{n\to\infty} f_n
$$
The functions $g$ and $h$ are both measurable.
Then note that
$$
E:=\{x\in X: \lim_{n\to\infty} f_n(x) \text{ exists}\}=\{x\in X: g(x)=h(x)\}
$$
As $g$ and $h$ are measurable, so is $E$.
Edit: If by "the limit exists" you mean also that it is finite, consider the set
$$
A=\{ x\in X: \limsup_{n\to\infty} f_n(x) <\infty\}
$$
Note that $A$ is measurable, and thus so is $E\cap A$, which is what you want.
A: Suppose $f_n: X\to \mathbb{\bar{R}}$ for all $n\in \mathbb{N}$. If we can write
$$ \{x\in X:\lim_{n\to\infty}f_n(x)\;\mathrm{exists}\}=E\cup E_{\infty}\cup E_{-\infty} $$
where $E_{\pm\infty}$ are the sets of $x$ such that the limit is equal to $\pm\infty$ and $E$ is the set of $x$ such that the limit exists. It is suffices to show that each term is measurable.
Notice that $\limsup f_n\text{ and} \liminf f_n$ are measurable because $f_n$ is measurable and by proposition 2.7. 
Define [g(x):= \limsup f_n(x)-\liminf f_n(x)  ] where $\limsup f_n(x)=\liminf f_n(x) \notin\{\pm\infty\}$. So $g(x)$ is measurable. 
Thus, [E={x: \limsup f_n=\liminf f_n\notin{\pm\infty}}]  implies the set $E$ can be written as $g^{-1}(\{0\})$ which is measurable.
Now $\lim_{n\to\infty}f_n(x)=\infty$ if for all $M\geq 1$ there exists $N$ such that $f_n(x)\geq M$ for all $n\geq N$, that is,
$$ E_{\infty}=\bigcap_{M=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n\geq N}\{x:f_n(x)\geq M\}$$
hence is measurable. A similar argument works for $E_{-\infty}$, that is,
$$ E_{\infty}=\bigcap_{M=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n\geq N}\{x:f_n(x)\leq -M\}$$
Suppose $f_n: X\to \mathbb{C}$ for all $n\in \mathbb{N}$. Since $f_n(x)$ converges if and only if $Re(f_n(x))$ and $Im(f_n(x))$ are both convergence, that is,
$$\{x: \lim f_n\text{ exists}\}=\{x: \lim Re(f_n)\text{ exists}\}\cap \{x: \lim Im(f_n)\text{ exists}\}$$
which two sets are both measurable by the previous argument and their intersection is also true. This gives the desired result.
