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Be $A_{1},A_{2},A_{3},...$ a sequence of events. When does the following equality is true?

$\mathbb{P}[ \lim_{n \to \infty} A_{n}] = \lim_{n \to \infty} \mathbb{P}[ A_{n}]$

The limit is defined in the following way:
1) Be $\{A_{n}\}_{n \in \mathbb{N}}$ an increasing sequence of events ( $A_{1} \subseteq A_{2} \subseteq A_{3} ... \,\,\,\, A_{n} \subseteq A_{n+1} \, \forall \, n$):
$ \,\,\,\,\,\,\, \lim_{n \to \infty} A_{n} := \bigcup_{n=1}^{ \infty} A_{n} $

2) Be $\{A_{n}\}_{n \in \mathbb{N}}$ a decreasing sequence of events ( $A_{1} \supseteq A_{2} \supseteq A_{3} ... \,\,\,\, A_{n} \supseteq A_{n+1} \, \forall \, n$):
$ \,\,\,\,\,\,\, \lim_{n \to \infty} A_{n} := \bigcap_{n=1}^{ \infty} A_{n} $

Thank you.

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  • $\begingroup$ How is defined $\lim A_n$ ? $\endgroup$ Feb 14, 2015 at 10:21
  • $\begingroup$ main post edited $\endgroup$
    – John M
    Feb 14, 2015 at 12:22
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    $\begingroup$ Ok. This is true and this is called the monotone continuity property, or something like that. $\endgroup$ Feb 14, 2015 at 12:47
  • $\begingroup$ Yes. You can also write down sensible definitions for $\limsup A_n$ and $\liminf A_n$ (it's what you'd guess), and say the limit is the common value, when they agree. $\endgroup$ Feb 14, 2015 at 16:44

2 Answers 2

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A sufficient condition is that the events are nested $A_1 \subset A_2 \subset \ldots$ or $A_1 \supset A_2 \supset \ldots$.

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    $\begingroup$ $\lim A_n$ is not defined in general for non-monotonic sequences, hence that makes no sense to wonder whether monotonicity is necessary (see @gui11aume's answer) $\endgroup$ Feb 15, 2015 at 9:10
  • $\begingroup$ @Stephane: Indeed, I commented as such above, before gui11aume edited his answer $\endgroup$ Feb 15, 2015 at 9:46
  • $\begingroup$ Nevertheless, I have now removed that part of my answer for clarity. $\endgroup$ Feb 15, 2015 at 9:49
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This is a basic property of probability measures.

One item of the definition for a probability measure says that if $B_n$ are disjoint events, then

$$ P \left(\bigcup_{n \geq 1} B_n \right) = \sum_{n \geq 1}P(B_n).$$

In the first case, you can define $B_n = A_n-A_{n-1}$, which gives the result immediately. Because $P(\Omega - A) = 1 - P(A)$, the converse is also true, as can be seen by taking the limit of the complement sets.

A fairly standard generalization is to say that $\lim A_n$ exists if $\lim \inf A_n = \lim \sup A_n$. In other words, if

$$ \bigcup_{n \geq 1} \bigcap_{k \geq n} A_k = \bigcap_{n \geq 1} \bigcup_{k \geq n} A_k = A.$$

Because $\lim \inf A_n$ is a non decreasing union of events, we have in general $P(\lim\inf A_n) = \lim_{n\rightarrow\infty} P(\cap_{k \geq n}A_k) \leq \lim_{n\rightarrow\infty}\inf_{k \geq n} P(A_k)$. Similarly, we also have $P(\lim\sup A_n) = \lim_{n\rightarrow\infty} P(\cup_{k \geq n}A_k) \geq \lim_{n\rightarrow\infty}\sup_{k \geq n} P(A_k)$.

If $\lim \inf A_n = \lim \sup A_n = A$ we thus have

$$P(A) = P(\lim\inf A_n) \leq \lim\inf P(A_n) \leq \lim\sup P(A_n) \leq P(\lim\sup A_n) = P(A).$$

This shows that $\lim P(A_n)$ exists and that $P(\lim_{n\rightarrow\infty}A_n) = \lim_{n\rightarrow\infty}P(A_n)$.

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