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

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.