Characteristic Function on Event is Discrete Random Variable

Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $E \in \Sigma$ be any event of $\struct {\Omega, \Sigma, \Pr}$.

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

Then $\chi_E$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.

Proof

By definition of characteristic function:

$\forall \omega \in \Omega: \chi_E = \begin{cases} 1 & : \omega \in E \\ 0 & : \omega \notin E \\ \end{cases}$

Then clearly:

$\forall x \in \R: \map { {\chi_E}^{-1} } x = \begin{cases} E & : x = 1 \\ \Omega \setminus E & : x = 0 \\ \O & : x \notin \set {0, 1} \end{cases}$

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

Hence the result.

$\blacksquare$

Sources

• 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.1$: Probability mass functions: Exercise $3$