$\limsup$ of indicator function of converging sets Let $\{A_n\}_{n=1}^{\infty}$ be a sequence of bounded sets of $\mathbb{R}^n$, such that $A_n \rightarrow A \subset \mathbb{R}^n$.
Let $\mathbb{1}: \mathbb{R}^n \rightarrow \{0,1\}$ be the indicator function.
1) I am wondering if the following statement holds.
Assume $A_n$'s compact. For all $x \in \mathbb{R}^n$, it holds that 
$$\limsup_{n \rightarrow \infty} \mathbb{1}_{A_n}(x) \leq \mathbb{1}_{A}(x). $$
2) What if the $A_n$'s are open?
Comments.
I tried to follow this post. I am not clear whether the compactness is actually needed.

The post to which you linked and the answer to it contain the answers to your questions: if $\langle A_n:n\in\Bbb N\rangle\to A$, then $\langle 1_{A_n}(x):n\in\Bbb N\rangle\to 1_A(x)$ for each $x\in\Bbb R^n$. It follows that
$$\limsup_{n\to\infty}1_{A_n}(x)=\lim_{n\to\infty}1_{A_n}(x)=1_A(x)$$
for all $x\in\Bbb R^n$. No special properties of the sets $A_n$ (like boundedness or compactness) are needed: all that’s necessary is that the sequence $\langle A_n:n\in\Bbb N\rangle$ converge to some $A\subseteq\Bbb R^n$.
