Question: Let $(X, \mathcal{M})$ be a measurable space and let $f\_1, f\_2, ... : X \longrightarrow [-\infty, \infty]$ be measurable functions. Then $\{x \in X: \lim\limits\_{n \to \infty} f\_n(x) \text{ exists in } [-\infty, \infty]\} \in \mathcal{M}$. 

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My professor gave this as an extra problem during the spring term, and in reviewing my notes I realized I never finished this one.

I know that if for all $x \in X$, $f\_n(x) \to f(x)$, then $f(x)$ is measurable, but I'm having difficulty dealing with this situation, where the $f\_n(x)$ might not converge for all $x$. 

My professor suggested that we start by showing that $\{x \in X: f\_n(x) \to 12\} \in \mathcal{M}$, which I've done:

We know that $\limsup\limits\_{n \to \infty} f\_n(x)$ and $\liminf\limits\_{n \to \infty} f\_n(x)$ are measurable functions, and that for all $x$, $f\_n(x) \to 12$ if and only if $\limsup\limits\_{n \to \infty} f\_n(x) = \liminf\limits\_{n \to \infty} f\_n(x)=12$. Thus 

$$\{x \in X: f\_n(x) \to 12\} = \{x \in X: \limsup\limits\_{n \to \infty} f\_n(x)= 12\} \cap \{x \in X: \liminf\limits\_{n \to \infty} f\_n(x)= 12\}$$

is in $\mathcal{M}$ since it is the intersection of two sets in $\mathcal{M}$. 

I'm not sure how to adapt this to the situation at hand, though.

Answer: 
Here is an approach essentially from first principles: we can write
$$ \{x\in X:\lim\_{n\to\infty}f\_n(x)\;\mathrm{exists}\}=A\cup A\_{\infty}\cup A\_{-\infty} $$
where $A\_{\pm\infty}$ are the sets of $x$ such that the limit is equal to $\pm\infty$ respectively, and $A$ is the set of $x$ such that the limit exists and is real.

Now $\lim\_{n\to\infty}f\_n(x)=\infty$ if for all $M\geq 1$ there exists $N$ such that $f\_n(x)\geq M$ for all $n\geq N$. Therefore
$$ A\_{\infty}=\bigcap\_{M=1}^{\infty}\bigcup\_{N=1}^{\infty}\bigcap\_{n=N}^{\infty}\{x:f\_n(x)\geq M\}$$
hence is measurable. A similar argument works for $A\_{-\infty}$.

Finally, $\{f\_n(x)\}$ converges to a real number if and only if $f\_n(x)$ is real for all but finitely many $n$ and $\{f\_n(x)\}$ is a Cauchy sequence, hence
$$ A=\bigcap\_{k=1}^{\infty}\bigcup\_{N=1}^{\infty}\bigcap\_{m,n=N}^{\infty}\bigcup\_{r\in\mathbb{Q}}\{x:r-k^{-1}<f\_n(x),f\_n(x)<r\}$$
which is measurable.
For $\lim\_{n \to \infty} f\_n(x)$ to exist (and in $\mathbb{R}$), it is necessary and sufficient that $(f\_n(x))\_n$ is a Cauchy sequence, i.e.

$$\forall \varepsilon >0 : \exists N : \forall n,m \ge N: |f\_n(x) - f\_m(x)| < \varepsilon$$

Denote by $g\_{n,m}: X \to [-\infty,+\infty]$ the function $g\_{n,m}(t) = |f\_n(t) - f\_m(t)|$, and note that all these functions $g\_{n,m}$ are measurable, as the $f\_n$ all are.

To make everything countable, we use only $\varepsilon$ of the form $\frac{1}{n}$, which is sufficient. So we can rewrite the above Cauchy definition as that the limit of $f\_n(x)$ as $n \to \infty$ exists iff

$$x \in M:=\bigcap\_{k \in \mathbb{N}} \bigcup\_{N \in \mathbb{N}} \bigcap\_{n\ge N} \bigcap\_{m \ge N} g\_{n,m}^{-1}[[0, \frac{1}{k})]$$

and $M$ is a measurable set.