Prove equality in liminf of set

Question

I need help in proving

$\chi_{\liminf A_n} = \liminf \chi_{A_n}$

Where $\liminf A_n=\bigcup \limits_{n=1}^{\infty}\bigcap \limits_{k=n}^{\infty}A_k$ and $\chi_{A}$ is the characteristic function of $A$.

Answer 1

If $x \in \liminf A_n$, then there exists an $n \in \Bbb N$ such that $x \in A_k$ for all $k \ge n$. Then $\chi_{A_k}(x) = 1$ for all $k \ge n$, and thus $\liminf \chi_{A_n}(x) = 1$. If $x \notin \liminf A_n$, then for each $n\in \Bbb N$, there exists $k \ge n$ such that $x \notin A_k$. So $x \notin A_k$ for infinitely many $k$, i.e., $\chi_{A_k}(x) = 0$ for infinitely many $k$. So $\liminf \chi_{A_k}(x) = 0$. Consequently, $\liminf \chi_{A_n} = \chi_{\liminf A_n}$.