Probability Measure on Finite Sample Space

Theorem
Let $\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$ be a finite set.

Let $\Sigma$ be a $\sigma$-algebra on $\Omega$.

Let $p_1, p_2, \ldots, p_n$ be non-negative real numbers such that:
 * $p_1 + p_2 + \cdots + p_n = 1$

Let $Q: \Sigma \to \R$ be the mapping defined as:


 * $\forall A \in \Sigma: \map Q A = \ds \sum_{i: \omega_i \in A} p_i$

Then $\struct {\Omega, \Sigma, Q}$ constitutes a probability space.

That is, $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.

Proof
Recall the Kolmogorov axioms:

First we determine that $\Pr$ as defined is actually a probability measure.

By definition, we have that $\map \Pr A$ is the sum of some subset of $\set {p_1, p_2, \ldots, p_n}$.

Thus $0 \le \map \Pr A \le 1$ and Axiom $(1)$ is fulfilled trivially by definition.

Let $A \in \Sigma$ be such that:
 * $A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$

We have that:
 * $A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$

From Simple Events are Mutually Exclusive, $\set {\set {\omega_{r_1} }, \set {\omega_{r_2} }, \ldots, \set {\omega_{r_k} } }$ constitutes a set of pairwise disjoint events.

Hence:
 * $\map \Pr {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } } = \ds \sum_{i \mathop = 1}^k \map \Pr {\omega_{r_1} }$

and it is seen that axiom $(3)$ is fulfilled.

Then we have that:

Hence axiom $(2)$ is satisfied.