Let $f_{n}, n \geq 1$ and $f$ be measurable functions on a measurable space $(\Omega, \mathcal{F})$. Show that the set $\{\omega : \lim_{n\to\infty} f_{n}(\omega)$ exists $\}$ is $\mathcal{F}$-measurable.
I know how to show the set $\{\omega : \lim_{n\to\infty} f_{n}(\omega) = f(\omega)\}$ is $\mathcal{F}$-measurable, and this problem appears to be very similar to that one. But something seems to not be clicking for me to be able to solve this problem. Here is the solution to the other set being $\mathcal{F}$-measurable, which may help with this problem: Proving a set is $\mathcal{F}$-measurable
I'm really not so sure how to deal with this notion of existence rather than convergence. I would really appreciate any help.
In the comments there is the answer in your problem in one-two lines.
If you want a solution with unions and intersections of sets,your way,then:
Let $x_0 \in X$
If $\lim_nf_n(x_0)$ exists then the sequence $f_n(x_0)$ is Cauchy and vice versa (if we assume that $f$ is real valued or complex valued).
Now you can easily see that.
$$\{x: \lim_nf_n(x) \text{ exists }\}=\bigcap_{k=1}^{\infty}\bigcup_{m=1}^{\infty}\bigcap_{n,p \geq m}\{x:|f_n(x)-f_p(x)|<\frac{1}{k}\}$$
which is measurable as countable set theoretic operations of measurable sets.