How to show that $\frac{f}{g}$ is measurable 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.

As you said, it suffices to show that $\dfrac{1}{g}$ is measurable.
Set $h(x)=\frac{1}{g(x)}$ if $g(x) \neq 0$ and $h(x)=0$ if $g(x)=0$
Let $a=0$ and $B=\{g=0\}$
Then $\{h>0\}=  \{x:\frac{1}{g(x)}>0\} \cap B^c$ 
If $a>0$ then $\{h>a\}=B^c \cap \{x:\frac{1}{g(x)}>a\}$ 
If $a<0$ then $\{h>a\}=(B^c \cap \{x:\frac{1}{g(x)}>a\} )\cup B$
In either case the sets are measurable because $g$ is measurable.
