Show that $\prod (1- P(A_n))=0$ iff $\sum P(A_n) = \infty$ Let $A_n$ be independent events with $P(A_n) \neq 1$. Show that $\prod_{n=1}^{\infty} (1- P(A_n))=0$ iff $\sum P(A_n) = \infty$
It kind of looks obvious but I really have no idea how to prove it. Can someone give me help?
 A: By the second Borel-Cantelli lemma, if $A_n$ are independent and $\sum P(A_n) = \infty$, then $P(A_n \text{ infinitely often})=1$. Then
$$
0=P \left( \bigcap_{n=1}^\infty A_n^c\right) = \prod_{i=1}^\infty (1- P \left( A_n\right))
$$
since this is the probability that none of $A_n$ would happen. 
Conversely, if $\sum P(A_n) < \infty$, by first Borel-Cantelli lemma, $P(A_n \text{ infinitely often})=0$, meaning a.s. there exists $N<\infty$ such that $A_n$ does not happen for $n >N$. Therefore
$$
\begin{split}
\prod_{i=1}^\infty (1- P \left( A_n\right)) &=P \left( \bigcap_{n=1}^\infty A_n^c\right) \\
&= P \left( \bigcap_{n=1}^N A_n^c\right) \\
& =E_N \left[ E\left[ \left.1_{\bigcap_{n=1}^N A_n^c}\right| N\right]\right] \\
& =E_N \left[ E\left[ \left. \prod_{n=1}^N 1_{A_n^c}\right| N\right]\right] \\
& =E_N \left[ \prod_{n=1}^N E\left[ \left. 1_{A_n^c}\right| N\right]\right] \\
& =E_N \left[ \prod_{n=1}^N P(A_n^c)\right] > 0\\
\end{split}
$$
since $P(A_n^c)>0$ for all $n$, and $N<\infty$. ($1_{(\cdot)}$ denotes indicator function in the above).
Perhaps cleaner and a bit more rigorous solutions are possible.
A: The proof of this result has nothing to do with probability theory in general or with Borel-Cantelli lemmas in particular, it is only plain-old-deterministic real analysis:

Let $(x_n)$ denote a real valued sequence such that $0\leqslant x_n\lt1$ for every $n$, then $\prod\limits_n(1-x_n)=0$ if and only if $\sum\limits_nx_n$ diverges.

Proof: If $\sum\limits_nx_n$ diverges, note that $\log(1-x)\leqslant-x$ for every $x\lt1$ to deduce that $$\prod\limits_n(1-x_n)=\exp\left(\sum_n\log(1-x_n)\right)\leqslant\exp\left(-\sum_nx_n\right)=0.$$
If $\sum\limits_nx_n$ converges, note that $x_n\leqslant\frac12$ for every $n$ large enough, say every $n\geqslant N$, and that $\log(1-x)\geqslant-2x$ for every $x\leqslant\frac12$ to deduce that $$\prod\limits_{n\geqslant N}(1-x_n)=\exp\left(\sum_{n\geqslant N}\log(1-x_n)\right)\geqslant\exp\left(-2\sum_{n\geqslant N}x_n\right),$$ hence $$\prod\limits_{n\geqslant N}(1-x_n)\ne0.$$ Since $x_n\lt1$ for every $n$, in particular for every $n\lt N$, this implies the desired result, namely, that $$\prod\limits_n(1-x_n)\ne0.$$
