Elementary Properties of Probability Measure

Theorem
Let $\mathcal E$ be an experiment with probability space $\left({\Omega, \Sigma, \Pr}\right)$.

The probability measure $\Pr$ of $\mathcal E$ has the following properties:


 * $(1): \quad \Pr \left({\varnothing}\right) = 0$


 * $(2): \quad \forall A \in \Sigma: \Pr \left({\complement_\Omega \left({A}\right)}\right) = 1 - \Pr \left({A}\right)$

where $\complement_\Omega \left({A}\right)$ denotes the complement of $A$ relative to $\Omega$


 * $(3): \quad \forall A \in \Sigma: \Pr \left({A}\right) \le 1$.

Proof
From the conditions for $\Pr$ to be a probability measure, we have:


 * $(1): \quad \forall A \in \Sigma: 0 \le \Pr \left({A}\right)$


 * $(2): \quad \Pr \left({\Omega}\right) = 1$


 * $(3): \quad \displaystyle \Pr \left({\bigcup_{i \ge 1} A_i}\right) = \sum_{i \ge 1} \Pr \left({A_i}\right)$ where all $A_i$ are pairwise disjoint.

$(1)$: Probability of Empty Event
From the definition of event space, we have:
 * $\Omega \in \Sigma$
 * $A \in \Sigma \implies \complement_\Omega \left({A}\right) \in \Sigma$

From Intersection with Empty Set:


 * $\varnothing \cap \Omega = \varnothing$

Therefore $\varnothing$ and $\Omega$ are pairwise disjoint.

From Union with Empty Set:


 * $\varnothing \cup \Omega = \Omega$

Therefore we have:

As $\Pr \left({\Omega}\right) = 1$, it follows that $\Pr \left({\varnothing}\right) = 0$.

$(2)$: Probability of Non-Occurence of Event
Let $A \in \Sigma$ be an event.

Then $\complement_\Omega \left({A}\right) \in \Sigma$ from the definition of event space.

From Intersection with Relative Complement, we have that $A \cap \complement_\Omega \left({A}\right) = \varnothing$.

From Union with Relative Complement, we have that $A \cup \complement_\Omega \left({A}\right) = \Omega$.

So $\Pr \left({A}\right) + \Pr \left({\complement_\Omega \left({A}\right)}\right) = 1$ from above, and so $\Pr \left({\complement_\Omega \left({A}\right)}\right) = 1 - \Pr \left({A}\right)$.

$(3)$: Probabilty Not Greater than One
From the above: $\Pr \left({A}\right) + \Pr \left({\complement_\Omega \left({A}\right)}\right) = 1$.

We have that $0 \le \Pr \left({\complement_\Omega \left({A}\right)}\right)$, hence:
 * $\forall A \in \Sigma: \Pr \left({A}\right) \le 1$