This is some crafty argument which I think I have seen somewhere on here (or something similar) Edit: AD. gave the link where I've got the argument from: Convergence of integrals in $L^p$.
First pick $D > 0$ and let choose the set $C_n := \{x \in \mathbf{R} : |f_n(x)g(x)| \leq D |g(x)|^2\}$. Then by Dominated Convergence we have that
$\int_{C_n} f_n g \, \text{d}\lambda \to 0.$
On the complement $\complement C_n$ we have that $|g(x)|^2 \leq D^{-1} |f_n(x) g(x)|$. So
$\int_{\complement C_n} |f_n g| \, \text{d}\lambda \leq \sqrt{\int_{\complement C_n} |f_n|^2 \, \text{d}\lambda}\sqrt{\int_{\complement C_n} |g|^2 \, \text{d}\lambda} \leq \frac{C}{\sqrt{D}} \sqrt{\int_{\complement C_n} |f_n g| \, \text{d}\lambda}.$
Hence,
$\int_{\complement C_n} |f_n| |g| \, \text{d}\lambda \leq \frac{C^2}{D}.$
So,
$\limsup_n \int_{\complement C_n} |f_n| |g| \, \text{d}\lambda \leq \frac{C^2}{D}.$
But $D$ was arbitrary, hence
$\lim_n \int |f_n g| \to 0.$