One direction is indeed true. In general convergence almost everywhere of $f_n \to f$ implies that every subsequence of $f_n$ has a further subsequence converging a.e. to $f$.
It is the converse that is false. Indeed, the problem lies in the fact that there are uncountably many subsequences of $f_n$, and an uncountable union of measure zero sets need not have measure zero! Or equivalently, an uncountable intersection of sets of full measure need not have full measure.
Your logic is presumably that if each subsequence of $f_n$ has a further subsequence converging to $f$ almost everywhere, then there exists a set $A$ whose complement has measure zero and such that for each $x \in A$, every subsequence of $f_n(x)$ has a further subsequence converging to $f(x)$. However, this is entirely false.
Indeed, we can try to build a "false proof" of the above fact as follows. Let $\Sigma$ denote the set of all increasing functions $\sigma: \Bbb N \to \Bbb N$, so that $\Sigma$ indexes all of the possible subsequences of $f_n$. Given a subsequence $(f_{\sigma(k)})$ there exists a set $A_{\sigma}$ whose complement has measure zero and such that $f_{\sigma(k)} \to f$ on $A_{\sigma}$, after passing to a further subsequence. Let $A = \bigcap_{\sigma \in \Sigma} A_{\sigma}$. If $\mu(A^c) = 0$ then the above fact is true. However, since the set $\Sigma$ is uncountable, we do not know whether $\mu(A^c)=0$ and therefore we cannot deduce anything.
You wrote:
$$\lim_{n\to \infty}\int_{X}f_n(x)\chi_{X\setminus N} (x)d\mu=\int_{X}f(x)\chi_{X\setminus N} (x)d\mu.$$
Since each $f_n(x)$ is nonnegative on $X$ then using linearity and $\mu(N)=0$ we see that $\int_{X}f_n(x)\chi_{X\setminus N} (x)d\mu=\int_{X}f_n(x)d\mu$.
But we cannot use the same reasoning for the $\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$ because $f(x)$ may not be non-negative on $X$. The only we know that $f(x)\geq 0$ a.e. on $X$.
So, using linearity, you can conclude that
$$\lim_{n\to \infty}\int_{X}f_n(x)(x)d\mu=\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$
Your question is how to deal with the right-hand size. Here is a simple solution:
Change the values of $f$ on a set of measure zero, to make it non-negative.
In detail:
Since $f(x)\geq 0$ a.e. on $X$, let us define the function $g$ by $g(x) = f(x)$ if $f(x) \geq 0$ and $g(x) = 0$ if $f(x) <0$.
Note that $g$ is non-negative, so $\int_X g(x) d\mu$ exists. Since $g$ is obtained from $f$ by just changing the values of $f$ on a set of measure zero, we have that $\int_X f(x) d\mu$ exists and $\int_X f(x) d\mu= \int_X g(x) d\mu$.
Note also that $f\chi_{X\setminus N}= g\chi_{X\setminus N}$ a.e.. So you can take care of the right-hand side as follows, using the linearity for $g$.
\begin{align*}
\lim_{n\to \infty}\int_{X}f_n(x)(x)d\mu &=\int_{X}f(x)\chi_{X\setminus N} (x)d\mu \\
&=\int_{X}g(x)\chi_{X\setminus N} (x)d\mu \\
&=\int_{X}g(x) d\mu \\
&=\int_{X}f(x) d\mu
\end{align*}
Best Answer
That depends on the exact context.
In general, $f$ will not be measurable. To see this, simply take $(X, \{\emptyset, X\}, \mu)$ as your measure space, with $\mu(\emptyset) = 0 = \mu(X)$.
Then $f_n \to f$ almost everywhere holds for every sequence $(f_n)_n$ and every map $f$, but only constant maps are measurable.
Now assume that your measure space $(X, \Sigma, \mu)$ is complete. This means that if $A \in \Sigma$ with $B \subset A$ and $\mu(A) = 0$, then also $B \in \Sigma$.
Then $f$ is measurable. To see this, first note that if $f = g$ almost everywhere (i.e. on $N^c$ with $\mu(N) = 0$) and $g$ is measurable, then so is $f$, because for every interval $I \subset \Bbb{R}$, we have
\begin{eqnarray*} f^{-1}(I) &=& [f^{-1}(I) \cap N^c] \cup [f^{-1}(I) \cap N] \\ &=& [g^{-1}(I) \cap N^c] \cup [f^{-1}(I) \cap N]. \end{eqnarray*}
But $N,N^c$ are measurable and $g$ is measurable. Hence, $g^{-1}(I) \cap N^c$ is measurable.
Finally, since your measure space is complete, $f^{-1}(I) \cap N$ is measurable (because it is a subset of the null-set $N$).
Hence, $f^{-1}(I)$ is measurable.
Now let $g := \liminf_n f_n$. Then $g$ is measurable and $f = g$ almost everywhere because of $f_n \to f$ almost everywhere. By the above, this implies that $f$ is measurable.
Finally note that the Lebesgue measure equipped with the $\sigma$-algebra of Lebesgue measurable sets is complete, but equipped with the $\sigma$-algebra of Borel measurable sets, it is not complete.