$E=\{x:f(x)=0\}$ is measurable How to prove that
$E_1=\{x:f(x)=0\}$
and $E_2=\{x:f(x)>0\}$ are measurable sets.Here $f:X\rightarrow \mathbb C$.
I know this may be a very basic question but as a new in Measure Theory I am not able to find the proof. Please help.
 A: There are some assumptions you need for your question to make some sense. What is $X$? What is the measure defined in $X$?
For the sake of giving you an answer I will assume $f$ is a continuous function and $X$ is a measure space whose $\sigma$-algebra contains the borel sets (that means, every borel set is measurable). Another thing, if $f$ is a complex-valued function then saying $f(x)>0$ is nonsense, so I will assume you wanted to say that the real part of $f(x)$ is positive.
Under these assumptions, you can have the following answer. I will expect you to modify your question though.
Assuming $f$ is continuous then $E_1=f^{-1}(\{ 0 \}$ is the inverse image of the closed set $\{ 0 \}$ which means it is closed, hence a Borel set, hence measurable.
In the same fashion, note that the region in $\mathbb{C}$ of all complex numbers with strictly positive real part is an open set. If you prove this then $E_2$  can be seen as the inverse image under f of an open set, hence it Is open,  hence borel, hence measurable.
