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:


 * $$\Pr \left({\varnothing}\right) = 0$$;

where $$\mathcal{C}_\Omega A$$ denotes the complement of $$A$$ relative to $$\Omega$$;
 * $$\forall A \in \Sigma: \Pr \left({\mathcal{C}_\Omega A}\right) = 1 - \Pr \left({A}\right)$$


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


 * If $$A_1, A_2, \ldots$$ are disjoint events in $$\Sigma$$, then $$\Pr \left({\bigcup_{i-1}^\infty}\right) = \sum_{i=1}^\infty \Pr \left({A_i}\right)$$.

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


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


 * $$\Pr \left({\Omega}\right) = 1$$;


 * $$\forall A, B \in \Sigma: A \cap B = \varnothing \implies \Pr \left({A \cup B}\right) = \Pr \left({A}\right) + \Pr \left({B}\right)$$.

Probability of Empty Event
From the definition of event space, we have:
 * $$\Omega \in \Sigma$$;
 * $$A \in \Sigma \implies \mathcal{C}_\Omega A \in \Sigma$$.

Hence as $$\varnothing \cap \Omega = \varnothing$$ and $$\varnothing \cup \Omega = \Omega$$, we have:

$$ $$

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

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

Then $$\mathcal{C}_\Omega A \in \Sigma$$ from the definition of event space.

From Intersection with Relative Complement, we have that $$A \cap \mathcal{C}_\Omega A = \varnothing$$.

From Union with Relative Complement, we have that $$A \cup \mathcal{C}_\Omega A = \Omega$$.

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

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

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

Probability Countable Union of Events
This can be proved by induction.