Probability Measure is Monotone/Proof 1

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Theorem

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


Then:

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


Proof

From Set Difference Union Second Set is Union:

$A \cup B = \paren {B \setminus A} \cup A$

From Set Difference Intersection with Second Set is Empty Set:

$\paren {B \setminus A} \cap A = \O$

From the Addition Law of Probability:

$\map \Pr {A \cup B} = \map \Pr {B \setminus A} + \map \Pr A$

From Union with Superset is Superset:

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

Thus:

$\map \Pr B = \map \Pr {B \setminus A} + \map \Pr A$

By definition of probability measure:

$\map \Pr {B \setminus A} \ge 0$

from which it follows that:

$\map \Pr B \ge \map \Pr A$

$\blacksquare$


Sources