Let $\{X_n\}_{n=1}^\infty$ is a succession of indepedent random variables, such that for all $n\geq 1$, $\mathbb E(X_n) =0$ and $\mathbb E(|X_n|) = 1$,
Prove or disprove that $\mathbb P(\lim \inf_{n} X_n < 0) > 0.$
I tried to handle it like this
Let us consider the succession of events $A_n = \{X_n < 0\}$. Since $(X_n)_n$ are independent then $(A_n)_n$ and $A^c_n = \{X_n \geq 0\}$ are also indepedent events. We have \begin{align*} \mathbb P(\lim \inf_{n} X_n < 0) &= \mathbb P(\lim \inf_{n} A_n)\\ & = \mathbb P(A_n \, \text{ e.v.})\\ &= 1 - \mathbb P\big((A_n \, \text{ e.v.})^c\big)\\ &= 1 - \mathbb P(A_n^c \, \text{ i.o.}\big). \end{align*} On the other hand, we have \begin{align*} \mathbb P(A_n^c) &= \mathbb P(X_n \geq 0)\\ & = ... \end{align*}
Here some recall of the notations used and my intention is to use the second BorelCantelli lemma that also I recall it here
First let's recall some definitions. Let $(A_n)_n$ be a sequence of events, we define \begin{align*} A_{n} \text{ infinitely often (i.o.) } &\equiv\left\{\omega: \omega \text { is in infinitely many } A_{n}\right\}\equiv \limsup _{n} A_{n} \equiv \bigcap_{m}^{\infty} \bigcup_{n=m}^{\infty} A_{n} \end{align*}
Note that $$ \mathbb {I}_{A_{n} \,i.o. }=\lim_{n} \sup \mathbb{I}_{A_{n}} $$ Similarly, \begin{align*} A_{n}\text{ eventually (e.v.) } \equiv\left\{\omega: \omega \text { is in } A_{n} \text { for all large } n\right\} \equiv \liminf _{n} A_{n} \equiv \bigcup_{m} \bigcap_{n=m}^{\infty} A_{n}. \end{align*} Note that $$ \mathbb{I}_{A_{n} \,e.v.} =\liminf _{n} \mathbb{I}_{A_{n}} $$ Also we have $\left(A_{n} \text { e.v.}\right)^{c}=\left(A_{n}^{c} \text { i.o. }\right)$. Moreover recall the second Borel-Cantelli Lemma:
If the events $(A_n)_n$ are independent, then $\sum_{n} \mathbb{P}(A_{n})=\infty$ implies $\mathbb{P}(A_{n} \text{ i.o.})=1$
