Simple function is measurable

Question

I want to show that a simple function is measurable. I know that a simple function is a function whose range set is a finite set. Let $f$ be a simple function defined on a measurable set $E$ such that it's range set is $\{a_1,a_2,\ldots , a_n\}$. Then help me show that $f$ is measurable.

Answer 1

Simple functions are precisely linear combinations of indicator functions of measurable sets.

If $E$ is a measurable space and $f\colon E\to\mathbb{C}$ is a simple function, there exists $a_1,...,a_n\in\mathbb{C}$ and measurable sets $A_1,...,A_n\subseteq E$ such that $A_i\cap A_j=\emptyset$ if $i\neq j$, $E=A_1\cup\cdots\cup A_n$ and $f=\sum_{k=1}^n a_k {\bf 1}_{A_k}$.

To see that $f$ is a measurable function, just notice that $$f^{-1}(V) = \bigcup\{A_j : a_j\in V\},$$ is measurable (being empty or finite union of measurable sets) for every $V\subseteq\mathbb{C}$.