If $\{A_n\}_{n\geq 1}$ independent sets, then $\lim\sup \textbf{1}_{A_n}$ is constant

Question

I want to prove that if $\{A_n\}_{n\geq 1}$ are independent sets, then the following functions are constant almost everywere: $$\lim\sup \textbf{1}_{A_n} \quad \quad \lim\inf \textbf{1}_{A_n}$$ I don't know how to prove this. I don't know why does it matter the fact that $A_n$ are independent, since there are no probabilities here. I've tried using the fact that $w\in \lim \sup A_n \Longleftrightarrow w \in A_i$ for an infinite number if i's, and that $w\in \lim \inf A_n \Longleftrightarrow \exists n_0$ so that $w\in A_n \forall n\geq n_0$. That would mean that $\lim\sup \textbf{1}_{A_n}(x)=1 \Longleftrightarrow x \in A_i$ for an infinite number of i's, but it can be only for the even ones only (that is $x\in A_2, A_4, A_6,\ldots$) and that does't mean that the function is constant almost everywhere. Can someone help me?