Let $A\subset\mathbb{R}$ measurable and denote the set of density points $$\tilde{A}:=\{x\in\mathbb{R}\mid \lim_{\epsilon\to 0}\frac{m([x-\epsilon,x+\epsilon]\cap A)}{2\epsilon}=1\}$$Porve/Dsiprove this set is open.
I thought building a set $A$ which it's density points are a finite number of singletons (or countable) and then it'll contradict the claim above.
My question is how can I build a set based on number of density points? More specific: Does exist a set $A\subset \mathbb{R}$ with finitely number of density points or s.t $|\tilde{A}|<\infty$?
In general it is false. Here is a way to construct a counter-example, albeit light on details. Let $\{a_n\}$ be a sequence of positive numbers decreasing to zero, define $$B = \cdots \cup (a_6,a_5) \cup (a_4,a_3) \cup (a_2,a_1)$$ and $$ A = B \cup (-B).$$ If you choose the $\{a_n\}$ correctly you can force $0$ to be a point of density, although by construction $0$ is not interior to $\tilde A$.
Interestingly, the collection of measurable sets $A$ for which $A = \tilde A$ forms a topology, finer than the standard topology.