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. 

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 $A\subset B \Rightarrow P(A) \le P(B)$
