Applications of Fatou's lemma and Borel-Cantelli lemma 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.
 A: Can you prove the following inequality?
For a sequence of measurable sets $\{E_n\}$, show that
$$ P(\liminf_n E_n) \leq \liminf_n P(E_n) \leq \limsup_n P(E_n) \leq P(\limsup_n E_n), $$
where $\liminf_n E_n := \{E_n \text{ ev}.\} := \cup_{n=1}^{\infty} \cap_{k=n}^{\infty} E_k$ and $\limsup_n E_n := \{ E_n \text{ i.o.}\} := \cap_{n=1}^\infty \cup_{k=n}^\infty E_k$.
A: $0\leq \mathbb{E}\liminf_{n\rightarrow \infty}1_{\bar{E}_n}\leq \liminf_{n\rightarrow \infty}\mathbb{E}1_{\bar{E}_n}=\liminf_{n\rightarrow \infty}\mathbb{P}(\bar{E}_n)=\liminf_{n\rightarrow \infty}(1-\mathbb{P}(E_n))\leq 1-1=0$ 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.
A: I am just expanding on Kavi Rama Murthy's answer.
You have $P(\sum_{n}\mathbf{1}_{E_{n}}=\infty)>0$ iff $P(E_{n}\,\text{occurs infinitely many times})=P(\lim\sup E_{n})>0$.
But $P(\lim\sup E_{n})=P\bigg(\bigg(\lim\inf E^{c}_{n}\bigg)^{c}\bigg)=1-P(\lim\inf E_{n}^{c})$.
So $P(\lim\sup E_{n})>0$ iff $P(\lim\inf E_{n}^{c})<1$
Now You have $E(\lim\inf\mathbf{1}_{E_{n}^{c}})\leq \lim\inf P(E_{n}^{c})=\lim\inf(1-P(E_{n}))=1-\lim\sup P(E_{n})<1$ as $\lim\sup P(E_{n})\geq \lim\inf P(E_{n})>0$.
Hence You have $E(\lim\inf\mathbf{1}_{E_{n}^{c}})<1$ .
But $\lim\inf\mathbf{1}_{A_{n}}=\mathbf{1}_{\lim\inf A_{n}}$ . See wiki
Hence we have $E(\mathbf{1}_{\lim\inf E_{n}^{c}})<1$
$\implies P(\lim\inf E_{n}^{c})<1$ .
So $P(\lim\sup E_{n})>0$ and hence $E_{n}$ occur infinitely many times .
So $P(\sum_{n}\mathbf{1}_{E_{n}}=\infty)>0$
