Measurability of $1/f$

Question

Let , $f:E\to \mathbb R$ be measurable and $f(x)\not=0$ in $E$. Then show that $1/f$ is measurable on $E$.

Let , $\alpha$ be any real number. Then ,

$$E\left(\frac{1}{f}>\alpha\right)=\begin{cases}E(f>0) &\text{ if } \alpha=0\\E(f>0)\cap E\left(f<\frac{1}{\alpha}\right) &\text{ if } \alpha>0\\E\left(f<\frac{1}{\alpha}\right) &\text{ if }\alpha<0\end{cases}$$

As each set of the R.H.S. is measurable so $1/f$ is measurable.

Please verify the calculation of $E\left(\frac{1}{f}>\alpha\right)$.

Answer 1

The first two lines are correct. But if $\alpha < 0$ and $1 / f > \alpha$, you either have

$$ f > 0 $$

or

$$ f < 1 / \alpha $$

So you are missing some things in the third line.