$(\Omega,\mathscr{F},P)$ is a probability space,$A_n\in\mathscr{F}$
If $\sum_nP(A_n)=\infty$, and $\{A_n\}_{n\geq1}$ are independent, then $P(\limsup A_n)=1$.
How can I prove $P(\limsup A_n)=1$?
============================================================================== I have thought about it for a long time, and my intuition tells me that $P(A_n)$ converges to $0$ or $1$ , because they are independent.And with $\sum_nP(A_n)=\infty$, $P(\limsup A_n)$ must be $1$.$(*)$
In fact, given $A_1(\not= \emptyset,\not= \Omega)$, if $A_2$ and $A_1$ are independent,$A_2$ can be $\emptyset$ or $\Omega$(the two are what we want) or some set $A_2\not\subset A_1,A_1\not\subset A_2,A_1\cap A_2\not= \emptyset.$ Then, $(A_1\cup A_2)^c$ is smaller than $(A_1)^c$……So, if excluding $ \emptyset $ and $\Omega$, $\cup_{n=1}^N A_n$ strictly increase, so it converges, assuming to $A$. If $P(A)=1-\epsilon<1$, Let $x_1=P(\cup_{n=1}^N A_n),x_2=P(A_{n+1})$,then \begin{equation} x_1+x_2-x_1x_2\leq1-\epsilon \end{equation}
\begin{equation} x_2\leq1-\frac{\epsilon}{1-x_1} \end{equation} We can choose $x_1>1-k\epsilon(k>1)$,then \begin{equation} x_2\leq1-\frac{1}{k} \end{equation} when $k$ close to $1$ enough,$x_2$ is small. Which may cause contradiction with $\sum_nP(A_n)=\infty$.$(**)$
I don't know if my thought is right. If it is, how can I write it down clearly. $(*)$ and $(**)$ are where I am not clear mostly.
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After considering, with the assumptions $A_1(\not= \emptyset,\not= \Omega)$(prove the situation from $1$ is enough), let $B_n=\cup_{i=1}^nA_i$ , assume $P(\lim B_n)=1-\epsilon<1$,let $P(B_n)=1-k_n\epsilon$,where $k_n>1$ and $k_n$ monotonically decreasing to 1.
\begin{equation} P(B_n)=P(A_n)+P(B_{n-1})-P(A_n)*P(B_{n-1}),n\geq 2 \end{equation} then ,I got \begin{equation} P(A_n)=\frac{P(B_n)-P(B_{n-1})}{1-P(B_{n-1})}=\frac{k_n-k_{n-1}}{k_n},n\geq 2 \end{equation} then \begin{equation} \sum_n^\infty P(A_n)=P(A_1)+\sum_{n=2}^\infty\frac{k_n-k_{n-1}}{k_n}\leq P(A_1)+\sum_{n=2}^{\infty}k_n-k_{n-1}=P(A_1)+k_2-1<\infty \end{equation} which causes a contradiction.
At first, I try to use $P(A_n)$ and $P(\cup_{i=1}^{n-1}A_{i})$ to figure out $P(\cup_{i=1}^nA_i)$, then to estimate $P(A_{n-1})$. However, it shows, it's much easier to use $\{P(\cup_{i=1}^nA_i)\}_{n=1}^\infty$ to estimate $\{P(A_{n})\}_{n=1}^\infty$.