how that if $P(\lim \sup A_n) = 1$ then, $P(\bigcup_{n=1}^\infty A_n)=1$ Question: 
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. 
 A: 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 $A\subset B \Rightarrow P(A) \le P(B)$
A: 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?
A: 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.
Generalising:
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. $\omega \in A_n$
$$\because A_n \subseteq \bigcup_{i=1}^{\infty} A_i, \omega \in \bigcup_{i=1}^{\infty} A_i$$
