Existence of Probability Space and Discrete Random Variable

Theorem
Let $$I$$ be some indexing set.

Let $$S = \left\{{s_i: i \in I}\right\} \subset \R$$ be a countable set of real numbers.

Let $$\left\{{\pi_i: i \in I}\right\} \subset \R$$ be a countable set of real numbers which satisfies:
 * $$\forall i \in I: \pi_i \ge 0, \sum_{i \in I} \pi_i = 1$$

Then there exists a probability space $$\left({\Omega, \Sigma, \Pr}\right)$$ and a discrete random variable $$X$$ on $$\left({\Omega, \Sigma, \Pr}\right)$$ such that the probability mass function $$p_X$$ of $$X$$ is given by:

$$ $$

Proof
Take $$\Omega = S$$ and $$\Sigma = \mathcal P \left({S}\right)$$.

Then let:
 * $$\Pr \left({A}\right) = \sum_{i:s_i \in A} \pi_i$$

for all $$A \in \Sigma$$.

Then we can define $$X: \Omega \to \R$$ by:
 * $$\forall \omega \in \Omega: X \left({\omega}\right) = \omega$$

This suits the conditions of the assertion well enough.

Comment
What this theorem allows us to do is ignore all the detail of sample spaces, event spaces and probability measure, and merely say:


 * "For each $$i \in I$$, let $$X$$ be a random variable which takes value $$s_i$$ with probability $$\pi_i$$"

and we know that such a random variable exists without having construct it every time.