Assume that we have a measure space $(X, \Omega, \mathbb{P})$, if $\mathbb{P}(\lim \sup_{n \rightarrow \infty}) = 0$, and $\mathbb{P}(A_n \cap A_m)= \mathbb{P}(A_n)\mathbb{P}(A_m),$ for $m \neq n,$ also, the measure of $X$ is 1..then how can I show that $\sum_{n=1}^{\infty} \mathbb{P} (A_n) < \infty. $
Basically, I just know that $$\mathbb{P}(\lim \sup_{n \rightarrow \infty} A_n) = \mathbb{P}(\bigcap_{m=1}^{\infty} \bigcup_{n=m}^{\infty} A_n)$$
and that if we let $B_m = (\bigcap_{m=1}^{\infty} \bigcup_{n=m}^{\infty} A_n),$ $\{B_m\}_{m=1}^{\infty}$ is a decreasing sequence so $\mathbb{P}(\lim \sup_{n \rightarrow \infty} A_n) = \lim_{m \rightarrow \infty} \mathbb{P} (\bigcup_{n=m}^{\infty} A_n)$.
Also, I know that $\mathbb{P}(\emptyset) = 0$ and $\mathbb{P}(\bigcup_{n=1}^{\infty} A_n) \leq \sum_{n=1}^{\infty} \mathbb{P}(A_n)$ (since $\mathbb{P}$ is a measure space hence $\mathbb{P}$ is countably subadditive.)
Then I don't know if this is useful and I don't know what to do next.. Please help.. thank you in advance!
If $0<c_i<1$ for all $i$ then $\sum c_i$ converges if and only if $\prod_i (1-c_i) >0$.
So we have to show that $\prod (1-P(A_i)) >0$. (or $\prod_k^{\infty} (1-P(A_i)) >0$ for some $k$). For this note that $$\prod_{i=n}^{N} (1-P(A_i))$$ $$=\prod_{i=n}^{N} (P(A_i^{c}))$$ $$=P(\bigcap _{i=n}^{N} A_i^{c})$$ by the independence assumption. Now let $N \to \infty$. You will see that $\prod_n^{\infty} (1-P(A_i)) = P(\bigcap _{i=n}^{N} A_i^{c}) \to 1$ since $P(\bigcup _{i=n}^{\infty} A_i) \to 0$.