$(f_n)$ Borel, $f(x)=\lim_{n\to\infty} f_n(x) \implies f$ Borel 
Prove that if $(f_n)$ is a sequence of Borel measurable functions and if $f(x)=\lim_{n\to \infty}f_n(x)$ exists in $\mathbb{R}$, then $f$ is Borel measurable. In fact $f$ is Borel measurable even if we only have $f(x)=\lim_{n\to\infty}$ almost everywhere on $D$, some measurable domain.

Attempt/Thoughts:
Suppose $(f_n)$ is Borel measurable. That is, the preimage of $f_n$ for every interval of the form, (for $\alpha\in\overline{\mathbb{R}})$,  $((\alpha,\infty])$ is a Borel set. We are given pointwise convergence, so by definition, $\forall\epsilon>0\exists N\in\mathbb{N}: \forall n\geq N: |f_n(x)-f(x)|<\epsilon$
I'm not sure how to proceed from here. 
I'm not sure how to start the second part at all. I know that it means that $f$ is Borel measurable even if $f(x)=\lim_{n\to\infty}f_n(x)$ on $D\setminus E$, where $m(E)=0$.
Any suggestions would be welcome.  Thanks.
 A: Part (1):
Equivalently, $f_n$ is Borel measurable if $\{x : f_n(x) < \alpha\} \in\mathcal{B}$ for every $\alpha$.
If $f(x) < \alpha$ (for some $x$), then there exists $m \in \mathbb{N}$ such that $f(x) < \alpha - 1/m$. Since $f_n(x) \to f(x)$ it follows that there exists $N \in \mathbb{N}$ such that $f_n(x) < \alpha - 1/m$ for $n \geqslant N$. The converse is true as well. 
Since $f_n$ is Borel measurable $\{x:f_n(x) < \alpha - 1/m\} \in \mathcal{B}$.  Since $\mathcal{B}$ is a $\sigma$-algebra we have
$$\{x: f(x) < \alpha\} = \bigcup_{m=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n= N}^\infty\{x:f_n(x) < \alpha - 1/m\} \in \mathcal{B},$$
and $f$ is Borel measurable.
A: I strongly suspect that the second part is not true. This because $(\mathbb R,\mathcal B,\lambda)$ where $\mathcal B$ denotes the Borel sigma algebra and $\lambda$ the Lebesgue measure, is not a complete measure space.
That makes it possible to find sets $D\subset E\subset\mathbb{R}$ such that
$D$ is a Borel set, $E$ is not a Borel set and $\lambda\left(\mathbb{R}-D\right)=0$. 
Define $f=1_{E}$ and
for $n=1,2,\dots$ define $f_{n}=1_{D}$. 
Then $\lim_{n\rightarrow\infty}f_{n}\left(x\right)=f\left(x\right)$
on $D$, hence almost everywhere. However $f$ is not Borel-measurable.
