I'm trying to prove the next statement.
If $\{A_n\}_{n\geq0}$ is a sequences of sets and $1_{A_n}$ it's he's indicator function respectively, then, for all $x:$
$lim_{n\rightarrow\infty}1_{A_n}(x)=1_{lim_{n\rightarrow\infty}A_n}(x)$
where
$1_{A_n}(x)=1$ if $x\in A_n$ or $1_{A_n}(x)=0$ if $x\notin A_n$
In my attempt, I consider cases:
For $x$ note that
Case 1. $\forall\hspace{2mm} m\geq n: \hspace{2mm}x\in A_m$
$\implies 1_{A_m}(x)=1\hspace{5mm}\forall \hspace{2mm}m\geq n$
$\implies \sup_{m\geq n}1_{A_n}(x)=1=\inf_{m\geq n}1_{A_n}(x)$
$\implies 1=\liminf_{n\rightarrow\infty}1_{A_n}=\lim_{n\rightarrow\infty}\inf_{m\geq n}1_{A_n}\leq\lim_{n\rightarrow\infty}1_{A_n}\leq \limsup_{n\rightarrow\infty}1_{A_n}=\lim_{n\rightarrow\infty}\sup_{m\geq n}1_{A_n}=1$
so $\lim_{n\rightarrow\infty}1_{A_n}=1$
but since $x\in A_m \hspace{2mm} \forall m\geq n$
$\implies x\in \cap_{m\geq n}A_m$
$\implies x\in \cup_{n\geq1}\cap_{m\geq n}A_m$
$\implies x\in \liminf A_n\subseteq \lim_{n\rightarrow\infty}A_n$
$\implies 1_{\lim A_n}(x)=1$
and then $lim_{n\rightarrow\infty}1_{A_n}(x)=1_{lim_{n\rightarrow\infty}A_n}(x)$
The other case, where exist $m\geq n$ such that $x\notin A_m$ is where I'm stuck. I got that
$\liminf_{n\rightarrow\infty}1_{A_n}(x)=0$ but I don't reach any relation with the limit or the superior limit, either of the sets or of the sequence of indicator's. It's there another way to prove it?
Any help it's appreciated. Thanks.