pointwise convergence in $\sigma$-algebra 
Let $(T,\mathcal{A})$ be a $\sigma$-algebra, and $(f_{n})$ a sequence of $\mathcal{A}$-measurable functions from $T$ to $\mathbb{R}$. Show that the set
  $$\{ t \in T: \lim_{n\rightarrow \infty}f_{n}(t)\text{ exists and is finite}\}$$
  is in $\mathcal{A}$.

I try with the definition of pointwise convergence, that is, $\forall \epsilon >0$,  $\exists n_{0}$ such that $|f_{n}(t)-f(t)|<\epsilon$, now $f_{n}(t) \in (f(t)-\epsilon,f(t)+\epsilon)$ and like $f_{n}$ is measurable the set is in $\mathcal{A}$. Can I use this to resolve the problem, or does there exist another way?
Any help is appreciated.
 A: Hint: Given a $t\in T$, $\lim_{n\to\infty}f_n(t)$ exists if and only if $$\limsup_{n\to\infty}f_n(t)=\liminf_{n\to\infty}f_n(t),$$ and this quantity is finite. Show that the functions 
$$g(t)=\limsup_{n\to\infty}f_n(t),\quad\quad h(t)=\liminf_{n\to\infty}f_n(t)$$
are $\mathcal{A}$-measurable, hence the set where they are equal is measurable, hence the set where they are equal and finite is measurable.
A: Show the function $x \mapsto \sup_n f_n(x)$ is measurable. It follows from this that $x \mapsto \inf_n f_n(x)$ is measurable. From these, it follows that $\overline{f}(x) = \limsup_n f_n(x)$ and $\underline{f}(x) = \liminf_n f_n(x)$ are measurable. Then define the measurable function $\phi(x) = \overline{f}(x) - \underline{f}(x)$. Then consider the set $\phi^{-1} \{0\} \cap \smash{\overline{f}}^{-1} \mathbb{R} \cap \underline{f}^{-1} \mathbb{R}$, which is measurable.
A: The other answers using $\inf$, $\sup$, $\limsup$ and $\liminf$ are nice and efficient, but your idea can also be made into an argument:
Observe that a sequence of real numbers converges to a real number if and only if it is a Cauchy sequence. Thus
$$
\begin{align*}
\{t \in T\,&:\,\lim_{n\to\infty} f_n(t) \text{ exists and is in }\mathbb{R}\} \\
& =
\{t \in T\,:\,(\forall k \in \mathbb{N})\,(\exists N \in \mathbb{N})\,(\forall m,n \geq N) \; |f_n(t) - f_m(t)| \lt 1/k\} \\
& = \bigcap_{k = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{n,m \geq N} \{t \in T\,:\,\lvert f_n(t)-f_m(t)\rvert \lt 1/k\} \in \mathcal{A}
\end{align*}
$$
because $\mathcal{A}$ is a $\sigma$-algebra and $\lvert f_n(t) - f_m(t)\rvert$ is a measurable function, so that $\{t \in T\,:\,\lvert f_n(t)-f_m(t)\rvert \lt 1/k\} \in \mathcal{A}$.
