Showing that a set is measurable I am trying to prove the following:

Let $(X, \mathcal{A})$ be a measurable space and let $\left\{f_n\right\}_{n\geq 1}$ be a sequence of $[-\infty, \infty]$-valued measurable functions on $X$ (i.e., $f_n:(X, \mathcal{A})\rightarrow ([-\infty,\infty],\mathcal{B}([-\infty, \infty])$). Prove that $\left\{x\in X: \lim_{n\rightarrow \infty} f_n(x) \text{ exists and is finite}\right\}\in \mathcal{A}$. 

There is a lemma that states the following:

Let $f_n\colon X\rightarrow [-\infty, \infty]$ ($n\geq 1$) be a sequence of measurable functions. Suppose that $f(x)=\lim_{n\rightarrow \infty} f_n(x)$ exists for every $x\in X$. Then $f$ is a measurable function. 

I am stuck. All I have done is
Let $C=\left\{x\in X: \lim_{n\rightarrow \infty} f_n(x) \text{ exists and is finite}\right\}$ and let $f(x)=\lim_{n\rightarrow \infty} f_n(x)$ for every $x\in C$. By the lemma above, we know that $f$ is measurable. 
I need to show that $C$ is the preimage of some measurable set in $\mathcal{B}([-\infty, \infty])$, but I have no idea how! Perhaps there is something I am missing? Any help is greatly appreciated!
 A: We don't need to show that $C$ is the preimage of a measurable set; we can show it belongs to the sigma algebra directly. 
It helps translating the conditions for the limit to exist:
$$C=\{x\in X: \lim f_n(x) \text{ exists and is finite}\}$$
$$=\{x\in X: (f_n(x))_{n=1}^\infty \text{ is a Cauchy sequence}\}$$
$$=\{x\in X: \text{for all }\epsilon>0\text{ there exists }N \text{ such that } |f_n(x)-f_m(x)| <\epsilon \text{ for all }n,m>N\}.$$
As we know, countable unions and intersections of measurable sets are measurable. We can try to express the last set as countable unions/intersections of sets that we know are measurable, like the sets $\{x\in X: |f_n(x)-f_m(x)|<\epsilon\}$ for fixed $n,m$ and $\epsilon$. 
The set of "all positive epsilon" is not countable, but we can overcome this difficulty: the condition need only be satisfied for  $\epsilon$ of the form $\frac 1 k$ ($k\in\mathbb N$) in order to hold for all $\epsilon>0$. Rewriting it in this way we get
 $$C=\{x\in X: \text{for all }k>0\text{ there exists }N \text{ such that } |f_n(x)-f_m(x)| <\frac 1k \text{ for all }n,m>N\}$$
 $$=\bigcap_{k\in\mathbb N}\{x\in X: \text{there exists }N \text{ such that } |f_n(x)-f_m(x)| <\frac 1k \text{ for all }n,m>N\}$$
$$=\bigcap_{k\in\mathbb N}\bigcup_{N\in\mathbb N}\{x\in X: |f_n(x)-f_m(x)| <\frac 1k \text{ for all }n,m>N\}$$
$$=\bigcap_{k\in\mathbb N}\bigcup_{N\in\mathbb N}\bigcap_{n\in\mathbb N}\bigcap_{m\in\mathbb N}\{x\in X: |f_n(x)-f_m(x)| <\frac 1k \}.$$
A: It is not too hard to show that $\overline{f_n}(x) = \sup_{k \ge n} f_k(x)$ and $\underline{f_n}(x) = \inf_{k \ge n} f_k(x)$ are measurable functions.
Note that these are non increasing, non decreasing sequences respectively, hence they have measurable limits
$\overline{f}(x) = \lim_n \overline{f_n}(x)$, $\underline{f}(x) = \lim_n \underline{f_n}(x)$.
Hence the sets $\overline{L} = \{ x | \overline{f}(x) < +\infty \}$, 
$\underline{L} = \{ x | \underline{f}(x) > -\infty \}$ and
$E= \{ x | \overline{f}(x) = \underline{f}(x) \}$ are all measurable.
We see that $C= \overline{L} \cap \underline{L} \cap E$.
A: The hypothesis implies that $f_{n}$ is a Cauchy sequence. Observe that $C=\bigcap_{k=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{m=N}^{\infty}\bigcap_{m=N}\{x \in X: |f_{m}- f_{n}| <  \frac{1}{k} \}$
The sets $A_{m,n,k} = \{x \in X: |f_{m} - f_{n}| < \frac{1}{k} \}$ $ \in \mathcal{A}$. It follows from the properties of $\mathcal{A}$ that $C \in \mathcal{A}$.
