# Probability Measure is Monotone/Proof 1

## Theorem

$\map \Pr A \le \map \Pr B$

## Proof

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

From the Addition Law of Probability:

$\Pr \left({A \cup B}\right) = \Pr \left({B \setminus A}\right) + \Pr \left({A}\right)$
$A \subseteq B \implies A \cup B = B$

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:

$\Pr \left({B}\right) \ge \Pr \left({A}\right)$

$\blacksquare$