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

Here is a sketch. I'll let you fill in the details.

Both $\limsup f_n$ and $\liminf f_n$ exist for all $x \in X$ and are measurable functions. The set in question can also be written as:

$$ E = \left\{x \in X : \lim f_n(x) \text{ exists}\right\} = \left\{ x \in X : \limsup f_n(x) = \liminf f_n(x) \right\} $$

Define:

$$ h(x) = \begin{cases} 0 & \text{ if } \limsup f_n(x) = \liminf f_n(x) = \pm\infty\\ \limsup f_n(x) - \liminf f_n(x) & \text{ otherwise } \end{cases} $$ $h$ is a measurable function (why?). The set $E$ can also be written as $E = h^{-1}(\{0\})$. Thus, it's measurable.

By definition $\liminf f_n(x) = \sup_{n \geq 0}\inf_{m \geq n} f_n(x)$. Any $\sup$ or $\inf$ of a family of measurable functions is again measurable, therefore $g(x)=\liminf f_n(x)$ is measurable. The same is with $h(x) =\limsup f_n(x)$.

Your set is just $\{x : h=g\}$ and this is measurable as a consequence of the fact that the sets $\{x: g<h\}$ and $\{x : g>h\}$ are both measurable.

Another proof not using $\limsup$ and $\liminf$ (a bit lengthy, though).

We know that, given a fixed $x \in X$, the sequence $f_n(x)$ converges iff it is a Cauchy sequence. We will construct sets using this idea.

Let $E_{n,m,r} = \{ x: |f_n(x) - f_m(x)| \leq \frac{1}{r}\}$. This set is a preimage of $|f_n(x) - f_m(x)|$, which is a measurable function due to the fact that absolute value is a continuous function, and difference of measurable functions is measurable.

Now, we can define $T_{n_0, r} = \{ x: |f_n(x) - f_m(x)| \leq \frac{1}{r}, \forall n,m \geq n_0 \}$. This set is a countable intersection of $E_{n,m,r}$, which are measurable.

Then $S_{r} = \{ x: \exists n_0 \in \mathbb{N}: \forall n,m \geq n_0, |f_n(x) - f_m(x)| \leq \frac{1}{r} \}$. This set is a countable union of $T_{n_0, r}$.

Finally, the desired set is just $\bigcap_\limits{r=1}^{\infty} S_r$, which of course is measurable.