Limits of Measurable Functions Let $(X, \mathcal{M})$ be a measurable space and let $f_1, f_2, ... : X \longrightarrow [-\infty, \infty]$ be measurable functions. Then $\{x \in X: \lim\limits_{n \to \infty} f_n(x) \text{ exists in } [-\infty, \infty]\} \in \mathcal{M}$. 

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. 
 A: 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.
A: 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.
A: 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.
