$(f_n(x))_n$ is cauchy means that for any positive integer $p$, there exists one integer $k$, such that for all $m > k$ and $n > k$ we have $|f_n(x) - f_m(x)| < \frac{1}{p}$.
Since it's for any positive integer p, we should have something like $\cap_{p=1}^{\infty}$.
And there exists a $k$ such that blabla means $\cup_{k=1}^{\infty}$ blabla..., i.e. for one k in $\{1,2, \cdots\}$, blabla is ok
For all $m,n$ greater than $k$ is $\cap_{m > k, n>k}$.
So finally $(f_n(x))_n$ is cauchy means that $x$ is in the set
$\cap_{p=1}^{\infty} \cup_{k=1}^{\infty}\cap_{m > k, n>k}\{x: |f_m(x) - f_n(x)| < \frac{1}{p}\}$
To resume, when there is "for any, for all", use intersection; when there is "exists", use union
Hint : Show that $\liminf f_n$ is measurable. The proof for $\limsup f_n$ is similar, and
$$
\{ x \in X \, | \, \lim_{n \to \infty} f_n \text{ exists} \} = \{ x \in X \, | \, \liminf_{n\to \infty} f_n = \limsup_{n\to \infty} f_n\}.
$$
Hope that helps,
Best Answer
I preassume that $E^{*}:=\{x\in E: (F_n(x))_n\text{ is a convergent sequence}\}$ and mention that $\mathbb N$ denotes the set of positive integers in this answer.
Then for $x\in E$:
$$x\in E^{*}\iff\forall k\in\mathbb{N}\exists n\in\mathbb{N}\forall r,s\in\mathbb{N}\left[r,s\geq n\implies\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right]$$ So if $\mathbb N_n$ denotes $\{n,n+1,\dots\}$ then:
$$E^{*}=\bigcap_{k\in\mathbb{N}}\bigcup_{n\in\mathbb{N}}\bigcap_{\left(r,s\right)\in\mathbb{N_n}^{2}}\left\{ x\in E:\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right\}$$
The sets $\left\{ x\in E:\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right\} $ are measurable and we are dealing with countable unions and intersections.
So $E^{*}$ is measurable.