Q: Consider $f \in L^2(d\mu)$, $\{ f_n \} \in L^2(d\mu)$, $f_n(x) \to f(x)$ a.e. and $\int |f_n|^2 \, d\mu \mathop{\longrightarrow}\limits_{n \to \infty} \int |f|^2 \, d\mu$. Use Egorov's theorem to show that $f_n \to f$ in $L^2(d\mu)$.
For any $\epsilon$, by Egorov's Theorem, there exists a $A_\epsilon$ such that $f_n$ converges uniformly to $f$ in $A_\epsilon$ and that the measure of the complement $m(A_\epsilon^c) < \epsilon$. Then for large enough $n$, $|f_n - f| < \epsilon$ for all $x \in A_\epsilon$.
If $f_n$ and $f$ are supported on a set of finite measure, which is not given, then it would be simple to show that $\int_{A_\epsilon} (f - f_n)^2 \, d\mu$ is arbitrarily small.
If $f_n$ and $f$ are bounded, which is not given, then it would be simple to show that $\int_{A_\epsilon^c} (f - f_n)^2 \, d\mu$ is arbitrarily small.
However, we are not given that $f_n$ and $f$ are bounded or supported on a set of finite measure. So, I'm not sure how to approach this problem.
I'm not using the given that:
\begin{align*} \lim\limits_{n \to \infty} \int |f_n|^2 \, d\mu &= \int |f|^2 \, d\mu \\ \end{align*}
Which means that the $L^2$ norm of $f_n$ approaches that of $f$. I presume that I'm supposed to use that but I don't see how.