I know the simple function f is defined as ai if x is in a sequence of measurable set Ei, 0 if it is not in the union of Ei.
I tried to prove that simple function is measurable.
It is measurable if B={x:f(x)<r} for all r in R is measurable.
My concern right now is if r is declared as 1 (1>0), then the set B must include the set of {x:f(x)=0}, which is not in the union of Ei, and which may be non-measurable. In this case, how do we prove that the function is measurable?
By definition a simple function is a function of the type $ \sum\limits_{i=1}^{n}c_iE_i$ where $E_i$'s are measurable. Note that the union of $E_1,E_2,...,E_n$ is measurable and so is the complement of this union. So the problem you are contemplating does not arise.
Let $E_0$ be the complement of the union of $E_i$'s. Then $f^{-1}(E)$ is a union of some of the sets $E_0,E_1,..,E_n$ for any Borel set $E$ in $\mathbb R$. Hence $f$ is measurable.
Complement of measurable set is measurable by definition.