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$
Best Answer
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.