Let $(f_n)_{n \in \Bbb N}$ be a sequence of measurable functions on a measure space $(X, M, \mu)$. Prove that the set $\{x \in X \; | \; \lim_n f_n(x) \text{ exists} \text{in } [-\infty, \infty]\}$ is a measurable set, {i.e}., belongs to the $\sigma$-algebra $M$.
My attempt:
Let $k(x)=$ limsup $f_n (x)$ and $m(x)=$ liminf $f_n (x)$. Both $k$ and $m$ are measurable.
$lim_n\to \infty f(x)$ exists if $k(x)=m(x)=lim_n\to \infty f(x)$
$E=$ {$x \in X : k(x)=m(x)$}.
We know limsup and liminf exists for all $x$ in extended real numbers but the points at which limit exists is a subset of this. Now my question is how is $E$ measurable, because the points at which $k(x)=m(x)$ is a subset of X and we do not have any claim about its measurability.
Please suggest.
Here is a brute force approach with nothing elegant:
Let $E = \{x \in X : \lim f_n(x)\;$ exists and is finite$\}$.
Let $f=\limsup f_n$. Notice that $x \in E$ if and only if $f_n(x) \to f(x)$. Letting $h_n := f-f_n$, we see that $h_n$ is measurable and that $E = \{x \in X: h_n(x) \to 0 \}$.
By the definition of a convergent sequence, we know that $h_n(x) \to 0$ if and only if for all $k \in \mathbb{N}$, there exists some $N \in \mathbb{N}$ such that $|h_n(x)|<2^{-k}$ whenever $n \geq N$. Therefore we can write $$E = \{x \in X: h_n(x) \to 0 \} = \bigcap_{k\in \mathbb{N}} \;\bigcup_{N \in \mathbb{N}} \; \bigcap_{n \geq N} h_n^{-1}\big((-2^{-k},2^{-k})\big)$$ i.e, the set of $x$ such that $\forall k, \exists N; \forall n \geq N : |h_n(x)|<2^{-k}$. Since the $h_n$ are measurable, we know that $h_n^{-1} (-2^{-k},2^{-k}) \in \mathcal{M}$ for all $n,k$, so by the above equation $E \in \mathcal{M}$
Also, $E' = \{x \in X: \lim |f_n(x)|= +\infty\} = \big|f\big|^{-1}(\{+\infty\})$ is measurable, which covers the case of infinite limits (or you can use the arctan metric to cover all cases at once!)