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.
 A: Using the fact that $fg$ is a measurable function and in view of the identity $$\frac{f}{g}=f\frac{1}{g}$$ it suffices to show that $1/g$ (with $g\not=0$) is measurable on $E$. Indeed, $$E\left(\frac{1}{g}<\alpha\right)=\left\{\begin{array}{lll} E\left(g<0\right) & \quad\text{if }\alpha=0\\ E\left(g>1/\alpha\right)\cup E\left(g<0\right) & \quad\text{if }\alpha>0\\ E\left(g>1/\alpha\right)\cap E\left(g<0\right) & \quad\text{if }\alpha<0 \end{array}\right.$$ Hence, $f/g$ ($g$ vanishing nowhere on $E$) is a measurable function on $E$.
For a better understanding of what is going on, I suggest to plot the function.
A: The function $h(x)=\frac{1}{x}$ is continuous so is Borel measurable, and $\frac{f}{g}$ is just $f(h\circ g)$.
A: 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.
