Prove that $P(\bigcup_{i=1}^{\infty} A_i) = 1$ $A_i$ $(i=1,2,...)$ are independent events
$\sum_{i=1}^{\infty}P(A_i) = \infty.$ 
Prove that: 
$P(\bigcup_{i=1}^{\infty} A_i) = 1 $
Can someone please help me out with this question?

If $P\left( \bigcup_{n=1}^\infty A_n \right) < 1,$ then $P\left( \bigcap_{n=1}^\infty (\text{not } A_n) \right)>0.$ By independence, we have
$$
\prod_{n=1}^\infty P(\text{not } A_n) > 0, \text{ so } \log\prod_{n=1}^\infty (1-P(A_n))>-\infty.
$$
$$
\sum_{n=1}^\infty -\log (1-P(A_n)) <\infty.
$$
If you can show that when $0\le p<1$ then $p\le -\log(1-p),$ then you get
$$
\sum_{n=1}^\infty P(A_n) < \infty.
$$
The inequality $p\le -\log(1-p)$ depends on the base of the logarithm being $\le e$ but $>1;$ however, this is not essential, because for any base $>1$ one can find a suitable positive constant and say $p\le -(\text{constant}\cdot\log(1-p)).$
Observe that if $\sum_{n=1}^\infty p_n = \infty$ then $\sum_{n=\mathbf N}^\infty p_n=\infty$ no matter how be $\mathbf N$ gets. Thus after any such index $\mathbf N$, one can find some $n$ for which $A_n$ occurs. Therefore, infinitely many of them occur. Consequently this whole argument amounts to a proof of one of the Borel–Cantelli lemmas.
