Probability Measure is Monotone/Proof 1

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Theorem

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


Proof

From Set Difference Union Second Set is Union:

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

From Set Difference Intersection with 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:

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

$\blacksquare$


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