Consider $f_n \to f$ a.e., $f_n \ge 0$, $f_n \in L^2(d\mu), f \in L^2(d\mu)$ and $\int |f_n|^2 \, d\mu \mathop{\longrightarrow}\limits_{n \to \infty} \int |f|^2 \, d\mu$.
Find an elementary argument to show that $f_n \to f$ in $L^2(d\mu)$.
We need to show that:
\begin{align*} \lim\limits_{n \to \infty} \lVert f_n - f \rVert_2 &= 0 \\ \lim\limits_{n \to \infty} \left(\lVert f_n - f \rVert_2\right)^2 &= 0 \\ \end{align*}
Starting with:
\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 (f_n^2 + f^2 - 2 f f_n) \, d\mu \\ &= \lim\limits_{n \to \infty} \int f_n^2 + \int f^2 - 2 \lim\limits_{n \to \infty} \int f f_n \, d\mu \\ \end{align*}
Since $f_n \to f$ in $L^2(d\mu)$, we can simplify:
\begin{align*} &= 2 \int f^2 - 2 \lim\limits_{n \to \infty} \int f f_n \, d\mu \\ \end{align*}
If $f_n \nearrow f$ we could use the monotone convergence theorem to conclude this proof. But we don't have that. I'm not sure how to conclude this.
Indeed you can't use Beppo Levi, but you can use Fatou. Specifically, $$-2\lim_{n\to\infty} \int f_nf\,d\mu\le-2\int f^2\,d\mu$$ and thus $\lim_{n\to\infty} \lVert f_n-f\rVert^2\le 0$.
Added: Techincally, you chose for yourself the bad notation, and so now I have to rewrite everything.
\begin{align}\limsup_{n\to\infty} \lVert f-f_n\rVert^2&=\limsup_{n\to\infty}\int f_n^2+f^2-2ff_n\,d\mu=\\&=\lim_{n\to\infty} \int f_n^2+f^2\,d\mu+\limsup_{n\to\infty}\int -2ff_n\,d\mu=\\&=2\lVert f\rVert^2-2\liminf_{n\to\infty}\int f_nf\,d\mu\le\\ \text{(Fatou) }&\le 2\lVert f\rVert^2-2\int \liminf_{n\to\infty} f_nf\,d\mu=0\end{align}