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.

$\blacksquare$

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 2.1$: Exercise $3$