This is actually a part b), where part a) was to show $f\in L^2$. This was simple: as $f_n$ converges in measure to $f$, there exists a sub-sequence which converges $\mu$-a.e., thus Fatou's gives: $\int \liminf f_{n_k}^2d\mu\leq \liminf \int f_{n_k}^2d\mu$ and we see $\int \liminf f_{n_k}^2d\mu=\int f^2d\mu$ and $\liminf\int f_{n_k}^2d\mu\leq 1$ by supposition, thus $\int f^2d\mu\leq 1$ from Fatou's result, so $f\in L^2$. (We neglected absolute values as squaring would make this redundant.)
So, we now wish to show $\lim \int |f_n-f|d\mu=0$.
Showing our limit goes to $0$ is defined to be: $\forall \epsilon>0$, $\exists N$ such that $\forall n> N$, $\int |f_n-f|d\mu<\epsilon$. So, let $\epsilon/\mu(\Omega)>0$: as $f_n$ converges in measure to $f$, then $\lim \mu(|f_n-f|>\epsilon)=0$, that is $\forall \delta>0$, $\exists N$ such that $\forall n>N$, $\mu(|f_n-f|>\epsilon)<\delta$. Let $N_{\epsilon/\Omega}$ be such an $N$ for $\delta=\epsilon/\Omega$. Thus, we may consider $\lim\int|f_n-f|d\mu=\lim \int_{>\epsilon}|f_n-f|d\mu+\int_{<\epsilon}|f_n-f|d\mu$ (omitting $|f_n-f|$ in the set of integration for readability).
If $|f_n-f|$ was bounded we could make conclusions about the size of this integral, which together with the finite measure of the space would let us bound the integral as small as we could hope to be less than our $\epsilon$, yet I do not know how to immediately show that this is bounded. There are a few ideas floating around here, but I am not sure how to proceed.