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 $L^2(d\mu)$.
Egorov's Theorem requires that $f$ is defined or supported on a set $E$ of finite measure, so that is implicitly given by the question. For any $\epsilon$, by Egorov's Theorem, there exists a $A_\epsilon \subset E$ such that $f_n$ converges uniformly to $f$ in $A_\epsilon$ and that $m(E-A_\epsilon) < \epsilon$. For large enough $n$, $|f_n - f| < \epsilon$ for all $x \in A_\epsilon$.
\begin{align*} \lim\limits_{n \to \infty} \left( \lVert f_n - f \rVert_2 \right)^2 &= \lim\limits_{n \to \infty} \int |f_n - f|^2 \, d\mu \\ &= \lim\limits_{n \to \infty} \int_{A_\epsilon} |f_n - f|^2 \, d\mu + \lim\limits_{n \to \infty} \int_{E - A_\epsilon} |f_n - f|^2 \, d\mu \\ &\le \epsilon^2 \cdot m(E) + \epsilon \cdot \lim\limits_{n \to \infty} \max(|f_n - f|^2) \\ \end{align*}
The left term is arbitrarily small. I am stuck on trying to show that the term on the right is arbitrarily small.
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.
I asked this question several days ago, but I didn't get an answer, so I'm rewriting a cleaned up version here.
Hint: $|f_n-f|^2 = |f_n|^2 + |f|^2 - 2\cdot\mathrm{Re}(f_n\overline{f})$. Since $f_n\to f$ almost everywhere, $\mathrm{Re}(f_n\overline{f})\to |f|^2$ almost everywhere. Use Egorov's theorem to show that for all large $n$, $$ \bigg|\int \mathrm{Re}(f_n\overline{f})\,d\mu - \int|f|^2\,d\mu\bigg| < \epsilon. $$ Then use that $\int |f_n|^2\,d\mu \to \int |f|^2\,d\mu$ to conclude that $f_n\to f$ in $L^2$.