Probability Measure is Monotone/Proof 1

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

Let $A, B \in \Sigma$ such that $A \subseteq B$.

Then:


 * $\Pr \left({A}\right) \le \Pr \left({B}\right)$

where $\Pr \left({A}\right)$ denotes the probability of event $A$ occurring.

Proof
From Set Difference Union Second Set is Union:


 * $A \cup B = \left({B \setminus A}\right) \cup A$

From Set Difference Intersection Second Set is Empty Set:


 * $\left({B \setminus A}\right) \cap A = \varnothing$

From the Addition Law of Probability:
 * $\Pr \left({A \cup B}\right) = \Pr \left({B \setminus A}\right) + \Pr \left({A}\right)$

From Union with Superset is Superset:


 * $A \subseteq B \implies A \cup B = B$

Thus:
 * $\Pr \left({B}\right) = \Pr \left({B \setminus A}\right) + \Pr \left({A}\right)$

By definition of probability measure:
 * $\Pr \left({B \setminus A}\right) \ge 0$

from which it follows that:
 * $\Pr \left({B}\right) \ge \Pr \left({A}\right)$