Discrete Random Variable is Random Variable
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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.
$\blacksquare$