I feel that there might be a problem with this question or there needs to be more conditions imposed on the functions $f_n$ i.e. integrability etc.
My initial thought was to use Monotone convergence on $f_n-g$. Hence, $\int f_n-g d\mu=\int f-g d\mu$ . Then I add $\int g d\mu$ on both sides. But to show that $\int f_n-gd\mu+ \int gd\mu=\int f_n-g+g d\mu$ requires that $f_n-g$ is integrable. Same goes for $f-g$.
Any suggestions or was my method wrong in the first place.
Integral is defined for all non-negative measurable functions, integrable or not. $\int (f-g) d\mu+\int g d\mu=\int f d\mu$ is valid whenever $f\geq g$ and $g$ is integrable. Note that $\int fd\mu$ makes sense because $\int f^{-} d\mu <\infty$. Indeed, $0\leq f^{-} \leq g^{-} \leq |g|$.