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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.

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That is essentially correct, but you may want to first "throw out" the points where $g(x)=0$ explicitly; also, your equalities are not quite correct as stated, it would be better to write something like $$\left\{x\left| \frac{1}{g}\lt\alpha\right\}\right. = \left\{x\left| g\gt\frac{1}{\alpha}\text{ and }g\gt 0\right\}\right. \cup \left\{x\left|g\lt \frac{1}{\alpha}\text{ and }g\lt 0\right\}\right.$$instead. – Arturo Magidin Oct 13 '11 at 16:16
Thanks Arturo... – Jack Oct 13 '11 at 16:25

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.

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The function $h(x)=\frac{1}{x}$ is continuous so is Borel measurable, and $\frac{f}{g}$ is just $f(h\circ g)$.

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