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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.2011-10-13

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