Discrete Random Variable is Random Variable

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

Let $X$ be a discrete random variable on $\left({\Omega, \Sigma, \Pr}\right)$.

Then $X$ fulfils the condition:
 * $\forall x \in \R: \left\{{\omega \in \Omega: X \left({\omega}\right) \le x}\right\} \in \Sigma$

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

Proof
Let $X$ be a discrete random variable.

Then by definition:
 * $\forall x \in \R: \left\{{\omega \in \Omega: X \left({\omega}\right) = x}\right\} \in \Sigma$

But see that:
 * $\displaystyle \left\{{\omega \in \Omega: X \left({\omega}\right) \le x}\right\} = \bigcup_{{y \in \Omega_X} \atop{y \le x}} \left\{{\omega \in \Omega: X \left({\omega}\right) = y}\right\}$

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

Hence, as $\Sigma$ is a sigma-algebra, $\left\{{\omega \in \Omega: X \left({\omega}\right) \le x}\right\} \in \Sigma$ as required.