Here is my attempt to show that $\frac{f}{g}~,g\neq 0$ is a measurable function, if $f$ and $g$ are measurable function. I'd be happy if someone could look if it's okay.
Since $fg$ is measurable, it is enough to show that $\frac{1}{g}$ is measurable. $ \left\{x\;\left|\;\frac{1}{g}\lt \alpha \right\}\right.=\left\{x\;\left|\;g \gt \frac{1}{\alpha} \right\}\right., \qquad g\gt 0,\quad \alpha \in \mathbb{R},$ which is measurable, since the right hand side is measurable.
Also, $ \left\{x\;\left|\;\frac{1}{g}\lt \alpha \right\}\right.= \left\{x\;\left|\;g\lt\frac{1}{\alpha} \right\}\right.,\qquad g\lt 0,\quad \alpha \in \mathbb{R},$ which is measurable since the right hand side is measurable.
Therefore, $\frac{1}{g}$ is measurable, and so $\frac{f}{g}$ is measurable.