What is the weakest sufficient condition for $\chi_{A_n}\xrightarrow{a.e.} 0$?

Question

Let $A_1, A_2, \ldots$ be events in some probability space. What is the weakest sufficient condition for $\chi_{A_n}\xrightarrow{a.e.} 0$?

The condition $P(\limsup A_n)=0$ seems to be sufficient. I wonder if there is a weaker one.

Answer 1

Notice that $\lim_{n\to\infty}\chi_{A_n}(\omega)=0$ is equivalent to $\omega\notin A_n$ for $n$ large enough, i.e., $\omega\in\liminf_{n\to \infty}\Omega\setminus A_n$. Hence $$\left\{\omega\mid \lim_{n\to\infty}\chi_{A_n}(\omega)=0\right\}=\liminf_{n\to \infty}\Omega\setminus A_n,$$ and we deduce that the condition $\displaystyle\mathbb P\left(\limsup_{n\to \infty} A_n\right)=0$ is a necessary and sufficient condition for having $\chi_{A_n}\to 0$ almost everywhere.