Let $E_1, E_2, \cdots$ be a sequence of events such that $\inf_n \mathbb{P}(E_n)>0$. Prove that $\mathbb{P}\left(\sum_n 1_{E_n}=\infty\right)>0$. i.e. there is a positive probability that an infinite number of the $E_n$ hold.
Here is what I have tried so far: By Fatou's lemma, $$0\leq \mathbb{E}\liminf_{n\rightarrow \infty}1_{\bar{E}_n}\leq \liminf_{n\rightarrow \infty}\mathbb{E}1_{\bar{E}_n}=\liminf_{n\rightarrow \infty}\mathbb{P}(\bar{E}_n)=\liminf_{n\rightarrow \infty}(1-\mathbb{P}(E_n))\leq 1-1=0.$$
Hence $\mathbb{E}\liminf_{n\rightarrow \infty}1_{\bar{E}_n}=0$ and $\mathbb{E}\liminf_{n\rightarrow \infty}1_{E_n}=1$
Hence by Monotone Convergence Theorem, $$\mathbb{E}\sum_n 1_{E_n}=\sum_n \mathbb{E} 1_{E_n}\geq \sum_n \mathbb{E} \liminf_{n\rightarrow \infty}1_{E_n}=\sum_n\mathbb{E}1=\infty.$$
But then I was stuck and I have no idea how to proceed. Thank you so much for your help.
Best Answer
$0\leq \mathbb{E}\liminf_{n\rightarrow \infty}1_{\bar{E}_n}\leq \liminf_{n\rightarrow \infty}\mathbb{E}1_{\bar{E}_n}=\liminf_{n\rightarrow \infty}\mathbb{P}(\bar{E}_n)=\liminf_{n\rightarrow \infty}(1-\mathbb{P}(E_n))\leq 1-1=0$ is not correct. You can only get $0\leq \mathbb{E}\liminf_{n\rightarrow \infty}1_{\bar{E}_n}\leq \liminf_{n\rightarrow \infty}\mathbb{E}1_{\bar{E}_n}=\liminf_{n\rightarrow \infty}\mathbb{P}(\bar{E}_n)=\liminf_{n\rightarrow \infty}(1-\mathbb{P}(E_n)<1.$
[Because you only know that $\inf P(E_n) >0$ not that $\inf P(E_n)=1$]. It now follows that $$\mathbb{E}\limsup_{n\rightarrow \infty}1_{E_n} >0.$$ Hence, the set of all $\omega$ such that $$\limsup_{n\rightarrow \infty}1_{E_n}(\omega) >0$$ has positive probability. This means that with positive probability we have $1_{E_n} (\omega) =1$ for infinitely many values of $n$ which forces $\sum 1_{E_n}(\omega)$ to be $\infty$ with positive probability.