Probability of Limit of Sequence of Events/Increasing
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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:
\(\ds \map \Pr A\) | \(=\) | \(\ds \map \Pr {A_0} + \map \Pr {B_1} + \map \Pr {B_2} + \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \Pr {A_0} + \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \map \Pr {B_k}\) |
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:
- $\ds \map \Pr A = \lim_{n \mathop \to \infty} \map \Pr {A_n}$
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 1.9$: Probability measures are continuous: Theorem $1 \ \text{C}$