Characteristic Function on Event is Discrete Random Variable

Theorem
Let $$\left({\Omega, \Sigma, \Pr}\right)$$ be a probability space.

Let $$E \in \Sigma$$ be any event of $$\left({\Omega, \Sigma, \Pr}\right)$$.

Let $$\chi_E: \Omega \to \left\{{0, 1}\right\}$$ be the characteristic function of $$E$$.

Then $$\chi_E$$ is a discrete random variable on $$\left({\Omega, \Sigma, \Pr}\right)$$.

Proof
By definition of characteristic function, we have:
 * $$\forall \omega \in \Omega: \chi_E = \begin{cases}

1 & : \omega \in E \\ 0 & : \omega \notin E \\ \end{cases}$$

Then clearly:
 * $$\forall x \in \R: \chi_E^{-1} \left({x}\right) = \begin{cases}

E & : x = 1 \\ \Omega \setminus E & : x = 0 \\ \varnothing & : x \notin \left\{{0, 1}\right\} \end{cases}$$

So whatever the value of $$x \in \R$$, its preimage is in $$\Sigma$$.

Hence the result.