# Probability Measure is Monotone

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## Theorem

Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.

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

Then:

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

where $\map \Pr A$ denotes the probability of event $A$ occurring.

## Proof 1

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:

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

$\blacksquare$

## Proof 2

As by definition a probability measure is a measure, we can directly use the result Measure is Monotone.

$\blacksquare$