Probability of Limit of Sequence of Events/Increasing

Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. Let $\sequence {A_n}_{n \mathop \in \N}$ be an increasing sequence of events.

Let $\ds A = \bigcup_{i \mathop \in \N} A_i$ be the limit of $\sequence {A_n}_{n \mathop \in \N}$.

Then:
 * $\ds \map \Pr A = \lim_{n \mathop \to \infty} \map \Pr {A_n}$

Proof
Let $\ds B_i = A_i \setminus A_{i - 1}$ for $i \in \N: i > 0$.

Then:
 * $A = A_0 \cup B_1 \cup B_2 \cup \cdots$

is the union of disjoint events in $\Sigma$.

By definition of probability measure:

But we have:
 * $\map \Pr {B_i} = \map \Pr {A_i} - \map \Pr {A_{i - 1} }$ for $i \in \N: i > 0$.

So:
 * $\ds \map \Pr A = \map \Pr {A_0} + \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \paren {\map \Pr {A_i} - \map \Pr {A_{i - 1} } }$

The last sum telescopes.

Hence the result:
 * $\ds \map \Pr A = \lim_{n \mathop \to \infty} \map \Pr {A_n}$