We said that a function $f:X \rightarrow \mathbb{R}$ is measurable iff we have that for all $I_a:=(a,\infty)$, $a \in \mathbb{R}$ $f^{-1}(a,\infty)$ is measurable.
Now I want to show that $\frac{1}{f}$ is also measurable, if $f$ was measurable.
Therefore I look at exactly those open intervals.
(i)If $a>0$ we have that $(\frac{1}{f})^{-1}(a,\infty) = f^{-1}(0,\frac{1}{a})$ which is measurable.
(ii)If $a = 0$ we have that $(\frac{1}{f})^{-1}(0,\infty) = f^{-1}(0,\infty)$ which is measurable.
(iii)But what happens for $a <0$ then we have that $(\frac{1}{f})^{-1}(a,\infty) =(\frac{1}{f})^{-1}(a,0] \cup (\frac{1}{f})^{-1}(0,\infty)$
The last one is known from the second case (ii), but what happens with the first one? I mean, you cannot have 0 as an actual result. $(\frac{1}{f})^{-1}(a,0)$ would be just $f^{-1}(-\infty,\frac{1}{a})$. But what does this 0 and are my calculations correct so far?