limit of measurable functions is measurable using preimage The task is to show that if $f_n$ is a sequence of measurable functions with respect to a sigma-algebra, $\mathcal{A}$, where $f_n: \Omega \rightarrow \mathbb{R}$, and $f_n$ converges pointwise to $f$ on $\Omega$, then $f$ is measurable with respect to $\mathcal{A}$. 
I'm not sure if this proof would be accepted but my strategy was to use the fact that given that $\mathcal{A}$ is closed under countable unions and countable intersections we have: 
$$
\forall c \in \mathbb{R}, \bigcap_{m=1}^\infty \bigcup_{n=m}^\infty f_n^{-1}((c,\infty))= \limsup_{n\to\infty} f_n^{-1}((c,\infty)) \in A\tag{1}
$$
$$
\forall c \in \mathbb{R}, \bigcup_{m=1}^\infty \bigcap_{n=m}^\infty f_n^{-1}((c,\infty))=\liminf_{n\to\infty} f_n^{-1}((c,\infty)) \in \mathcal{A}\tag{2}
$$
From this it follows that both:
$$ \limsup_{n\to\infty} f_n\tag{3}
$$
$$ \liminf_{n\to\infty} f_n\tag{4}
$$
are measurable. 
Further, given that $ \lim_{n\to\infty} f_n = f $:
$$ f=\limsup_{n\to\infty} f_n = \liminf_{n\to\infty} f \tag{5}$$
This proof is much shorter than what I've found on the internet so I wonder whether there might be an important gap I have failed to consider. 
 A: $\bigcap\limits^{\infty}_{m=1}\bigcup\limits^{\infty}_{n=m}f_n^{-1}((c,\infty)):=\limsup\limits_{n\rightarrow\infty}f_n^{-1}((c,\infty))=\inf\limits_{m\in\mathbb{N}}\sup\limits_{n\geq m}f_n^{-1}((c,\infty))$, and the $\liminf f_n$ is dually defined. The  $\lim\limits_{n\rightarrow\infty} f_n$ exists iff $\limsup\limits_{n\rightarrow\infty}f_n=\liminf\limits_{n\rightarrow\infty}f_n$ and in which case $\limsup\limits_{n\rightarrow\infty}f_n=\liminf\limits_{n\rightarrow\infty}f_n=\lim\limits_{n\rightarrow\infty}f_n$.  Can you show this?  So what you have done is to have shown $\limsup\limits_{n\rightarrow\infty}f_n$ to be measurable, you can similarly show the $\liminf\limits_{n\rightarrow\infty}f_n$ is measurable. Therefore the set on which $\{\omega\in\Omega:\lim\limits_{n\rightarrow\infty}f_n \space\space\text{exists}\}=\{\omega\in\Omega:\limsup\limits_{n\rightarrow\infty}f_n=\liminf\limits_{n\rightarrow\infty}f_n\}\in\mathcal{A}$, as the $\limsup\limits_{n\rightarrow\infty}f_n$ and $\liminf\limits_{n\rightarrow\infty}f_n$ are measurable, and in your case the set on which the limit exists is given to be the whole of $\Omega$.
So setting $E= \{\omega\in\Omega:\lim\limits_{n\rightarrow\infty}f_n \space\space\text{exists}\}$  $(=\Omega$ in this case), and as $\limsup\limits_{n\rightarrow\infty}f_n=\liminf\limits_{n\rightarrow\infty}f_n=\lim\limits_{n\rightarrow\infty}f_n$ we see that $\{\omega\in E: \lim\limits_{n\rightarrow\infty}f_n>c\}=\{\omega\in E:\liminf\limits_{n\rightarrow\infty}f_n>c\}=E\cap \{\liminf\limits_{n\rightarrow\infty}f_n>c\}\in\mathcal{A}$
