Probability of Limit of Sequence of Events/Increasing

Theorem
Let $\left({\Omega, \Sigma, \Pr}\right)$ be a probability space. Let $\left \langle{A_n}\right \rangle_{n \mathop \in \N}$ be an increasing sequence of events.

Let $\displaystyle A = \bigcup_{i \mathop \in \N} A_i$ be the limit of $\left \langle{A_n}\right \rangle_{n \mathop \in \N}$.

Then:
 * $\displaystyle \Pr \left({A}\right) = \lim_{n \to \infty} \Pr \left({A_n}\right)$

Proof
Let $\displaystyle B_i = A_i \setminus A_{i-1}$ 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:
 * $\Pr \left({B_i}\right) = \Pr \left({A_i}\right) - \Pr \left({A_{i-1}}\right)$ for $i \in \N: i > 0$.

So:
 * $\displaystyle \Pr \left({A}\right) = \Pr \left({A_0}\right) + \lim_{n \to \infty} \sum_{k \mathop = 1}^n \left({\Pr \left({A_i}\right) - \Pr \left({A_{i-1}}\right)}\right)$

The last sum telescopes.

Hence the result:
 * $\displaystyle \Pr \left({A}\right) = \lim_{n \to \infty} \Pr \left({A_n}\right)$