Proof of the monotonicity of probability

Proof

1. Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A, B \in \mathcal{F}$ be events with $A$ a subset of $B$.
2. The set $B \setminus A = B \cap A^c$ is in $\mathcal{F}$, because a σ-algebra is closed under complements and finite intersections.
3. By the complement rule applied to the event $\Omega$, $P(\varnothing) = 1 - P(\Omega) = 0$.
4. Let $E_1 = A$, $E_2 = B \setminus A$ and $E_n = \varnothing$ for $n \ge 3$. These sets are pairwise disjoint, because no point of $B \setminus A$ lies in $A$.
5. Their union is $A \cup (B \setminus A)$. This equals $B$, because $A \subseteq B$.
6. By the countable additivity of the probability measure $P$ and step 3, $P(B) = P(A) + P(B \setminus A)$.
7. All three numbers are real. So $P(B \setminus A) = P(B) - P(A)$.
8. The measure $P$ takes values in $[0, 1]$, so $P(B \setminus A) \ge 0$. Hence $P(A) \le P(B)$.