Discrete Random Variable is Random Variable

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

Let $X$ be a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.

Then $X$ fulfils the condition:
 * $\forall x \in \R: \set {\omega \in \Omega: \map X \omega \le x} \in \Sigma$

That is, $X$ fulfils the condition for it to be a random variable.

Proof
Let $X$ be a discrete random variable.

Then by definition:
 * $\forall x \in \R: \set {\omega \in \Omega: \map X \omega = x} \in \Sigma$

But see that:
 * $\ds \set {\omega \in \Omega: \map X \omega \le x} = \bigcup_{\substack {y \mathop \in \Omega_X \\ y \mathop \le x} } \set {\omega \in \Omega: \map X \omega = y}$

This is the countable union of events in $\Sigma$.

Hence, as $\Sigma$ is a sigma-algebra, $\set {\omega \in \Omega: \map X \omega \le x} \in \Sigma$ as required.