$|\{x \in O \ | \ \text{dist}(x, \mathbb{R} \backslash O) < \epsilon\}| = O(\epsilon) \ \ as \ \epsilon \downarrow 0$ Let $O \subset \mathbb{R}$ be a nonempty open set.
Is true that $|\{x \in O \ | \ \text{dist}(x, \mathbb{R}  \backslash O) < \epsilon\}| = O(\epsilon) \ \ as \ \epsilon \downarrow 0$?
 A: We can prove this result for an open interval $O$ in $\mathbb{R}$. 
For any point, $x\in\mathbb{R}$ and any $A\subseteq\mathbb{R}$, we define
$$\text{dist}(x,A)=\inf_{y\in A}|x-y|$$
Let $O=(a,b)$ for some $a,b\in\mathbb{R}$ with $a<b$. We can verify that for any given $x\in O$, $\text{dist}(x,\mathbb{R}\setminus O) = \min\{|x-a|,|x-b|\}$. Fix any $\epsilon >0$.
We then need only verify that $\mathcal{O}:=\{x\in O: \text{dist}(x,\mathbb{R}\setminus O)<\epsilon\} = (a,a+\epsilon)\cup (b-\epsilon,b)$
And then it is clear that $|\mathcal{O}|= O(\epsilon)$.
Edit: As seen with Teddy38's counter-example, this is not true in general. The most we can hope for is an open set consisting of a finite union of open intervals. For an open set consisting of a countable union of open intervals, we need a stronger condition: 
If we had an open set $U$ which consisted of a countable union of open intervals $U_i$. So, $\mathcal{O}=\bigcup_{i=1}^\infty \mathcal{O}_i,$ and if  $\mathcal{O}_i=\{x\in O: \text{dist}(x,\mathbb{R}\setminus O)<\epsilon\cdot 2^{-i}\}$, then by countable additivity we would have $|\mathcal{O}|=O(\epsilon)$.   
A: I don't think it is true: Take $O=\bigcup_{n\in\Bbb{Z}}(2n,2n+1)$, then $$\mu\big(\{x\in O\ |\ \text{dist}(x,\Bbb{R}\backslash O)<\epsilon\}\big)=\infty\ $$
for any $\epsilon>0.$
