Let $\{A_n\}$ be a sequence of independent events in a probability space $(\Omega, F, P)$

show that if $P(\lim \sup A_n) = 1$ then, $P(\bigcup_{n=1}^\infty A_n)=1$

I tried solving this question, i think that i need to use the following inequalities;

$$P( \lim\sup A_n) = P(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k) \le P(\bigcup_{k=n}^\infty A_k) \le \sum_{k=n}^\infty P(A_k)$$

my thought may be false or not.I'm not sure. please help me solving this question. thank you.

  • 1
    $\begingroup$ "[...] then $P (\bigcup_n)=1$" ... do you mean $P(\bigcup_n A_n)=1$? $\endgroup$ – saz Mar 28 '15 at 15:18
  • $\begingroup$ I editted it. @ThomasAndrews $\endgroup$ – user315 Mar 28 '15 at 15:19

Note that $$\bigcup_{n=1}^\infty A_n \supset \bigcup_{n=1}^\infty A_n \cap \bigcup_{n=2}^\infty A_n \cap \ldots \cap \bigcup_{n=k}^\infty A_n \cap \ldots = \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_n = \limsup_{n\to\infty} A_n$$ And use $A\subset B \Rightarrow P(A) \le P(B)$

  • $\begingroup$ ... that's exactly what @B11b has written/used in his question, isn't it? $\endgroup$ – saz Mar 28 '15 at 15:45
  • $\begingroup$ I wrote this $ \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k \subset \bigcup_{k=n}^\infty A_k$ so, $P(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k ) \le P(\bigcup_{k=n}^\infty A_k)$ I have already used $ A \subset B \to P(A) \le P(B) $. Will I use this for what ? In addition to this, how to get $P(\bigcup_{k=n}^\infty A_k ) =1$ $\endgroup$ – user315 Mar 28 '15 at 15:53
  • $\begingroup$ @saz not quite. Take a second look. $\endgroup$ – AlexR Mar 28 '15 at 15:56
  • $\begingroup$ @B11b Don't overcomplicate. We have shown that $\limsup_{n\to\infty} A_n \subset \bigcup_{n=1}^\infty A_n$, thus $1 = P(\limsup A_n) \le P(\bigcup_n A_n) \le 1$. $\endgroup$ – AlexR Mar 28 '15 at 15:57
  • $\begingroup$ @B11b Is there anything left unclear? If so, feel free to ask for further clarification. If not, please mark the question as answered to remove it from the unanswered list and idicate that you need no further assistance. $\endgroup$ – AlexR Mar 28 '15 at 23:43

I think you've almost got it:

$1 = P(\lim \sup A_n)=P(\cap_{n=1}^\infty \cup_{k=n}^\infty A_k) \le P(\cup_{k=n}^\infty A_k) \leq P(\cup_{n=1}^\infty A_n).$

So you have $P(\cup_{n=1}^\infty A_n) \ge 1,$ and hence the probability must be 1.

Now to be sure you understand the meaning of lim sup (other than as a bunch of cups and caps), can you construct a sequence where the lim sup and union obviously have different probabilities?

  • $\begingroup$ Dear B.Trumbo thank you for helping. I have understood what I need to do. But I have one more question. No body gives any answer. Please can you look at my question? thank you. math.stackexchange.com/questions/1210275/… $\endgroup$ – user315 Mar 28 '15 at 21:37
  • $\begingroup$ Since we're all missing your question, maybe you need to state it again--separately and more explicitly. I have looked several times and I honestly do not see any unanswered questions up there. [To me, $\sum P(A_k)$ has nothing to do with proving the main result.] $\endgroup$ – BruceET Mar 28 '15 at 22:02

This is obvious if you think about the definitions of limsup and union.

Suppose we flip a coin infinitely. What is the probability that we will get heads at least once? Well, it is obvious that we will get heads infinitely often. Therefore, it is obvious it will happen at least once.


If a sample point is an element of infinitely many of the events $A_1, A_2, ...$, then it is an element of at least one of the events $A_1, A_2, ...$

Proof involving quantifiers:

Suppose $\omega \in \limsup A_n$.

Then $\forall m \ge 1, \exists n \ge m$ s.t. $\omega \in A_n$

$$\because A_n \subseteq \bigcup_{i=1}^{\infty} A_i, \omega \in \bigcup_{i=1}^{\infty} A_i$$


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