Suppose $(f_n)$ is a sequence of measurable functions on measure space $(X, \mathcal{A}, \mu)$. Suppose that $\mu(X) < \infty$, $f_n$ converges to $f$ almost everywhere and $\int_X |f_n|^2 d \mu \leq 1$ for all $n \in \mathbb{N}$. Prove that $f_n$ converges to $f$ in $L^1$.
Attempt:
By Egoroff's theorem, $f_n$ converges to $f$ almost uniformly (i.e. for all $\epsilon >0$, there exists $E \subseteq X$ such that $\mu(E) < \epsilon$ and $f_n$ converges to $f$ uniformly on $E^c$.)
\begin{align*} \lim_{n \rightarrow \infty} \|f_n-f\|_1 &= \lim_{n \rightarrow \infty} \int_X |f_n-f| d \mu \\ &= \lim_{n \rightarrow \infty} \left( \int_E |f_n-f| d \mu + \int_{E^c} |f_n-f| d \mu \right) \\ &= \lim_{n \rightarrow \infty} \int_E |f_n-f| d \mu + \lim_{n \rightarrow \infty} \int_{E^c} |f_n-f| d \mu \\ \end{align*}
Notice \begin{align*} \lim_{n \rightarrow \infty} \int_{E^c} |f_n-f| d \mu &\leq \lim_{n \rightarrow \infty} \int_{E^c} \sup_{E^c} |f_n-f| d \mu \\ &= \lim_{n \rightarrow \infty} \sup_{E^c} |f_n-f| \int_{E^c} d \mu \\ &= \left( \lim_{n \rightarrow \infty} \sup_{E^c} |f_n-f| \right) \mu(E^c) \\ &= 0 \end{align*} by uniform convergence of $f_n$ to $f$ on $E^c$.
My professor said to use Egoroff's theorem to do this proof but I am not sure how to proceed.