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.
Let $A=\{x \in X| \exists \lim\_{n \rightarrow \infty}f\_n(x) \in [- \infty,+ \infty] \}$

Then $A=A\_1 \cup A\_2 \cup A\_3$ where $A\_1=\{x \in X|f\_n(x)$ is Cauchy in $\mathbb{R}$ $\}$ 

$$A\_2=\{x \in X| \lim\_n f\_n(x) =+ \infty\}$$ $$A\_3=\{x \in X| \lim\_n f\_n(x) =- \infty\}$$

Now if you remember the definition of a Cauchy sequence you can prove that 

$A\_1= \bigcap\_{k=1}^{\infty} \bigcup\_{n\_0 \in \mathbb{N}} \bigcap\_{m,n \geqslant n\_0}^{ \infty} \{x \in X:|f\_n(x)-f\_m(x)|< 1/ k\}$ which is a measurable set because $f\_n$ are measurable for every $n \in \mathbb{N}$

Also from the definition $A\_2= \bigcap\_{k=1}^{\infty} \bigcup\_{N \in \mathbb{N}} \bigcap\_{n \geqslant N} \{x \in X:f\_n(x) \geqslant k\}$ which is also a measurable set because of the propertys of a sigma algebra and the measurabililty of $f\_n$.

The case of $A\_3$ i leave it to you. You just have to modify the idea for $A\_2$

Finally $A$ is measurable as a union of measurable sets $A\_1,A\_2,A\_3$

I hope it helps.

If you need further guidance,let me know.