Doubt on problem regarding limit infimum of indicator functions.

Question

For $n \in \mathbb{N}$, if $f_n = \chi_{E_n}$, then is:

$$\liminf f_n = \liminf \chi_{E_n} = \chi_{\liminf E_n} ?$$

It's the third inequality I am interested in.

Answer 1

$$\liminf_{n \to \infty} \chi_{E_n}(x) = \begin{cases} 1 & \text{$x \notin E_n$ for finitely many $n$}\\ 0 & \text{$x \notin E_n$ for infinitely many $n$} \end{cases} = \chi_{\bigcup_{n \ge 1} \bigcap_{k \ge n} E_k}(x) = \chi_{\liminf_{n \to \infty} E_n}(x).$$