# Comparing $(((A^\varepsilon)')^\varepsilon)'$ and $int(A)$, where $A' := X\setminus A$ and $A^\varepsilon := \{x \in X \mid d(x,A) \le \varepsilon\}$

Disclaimer. This is follow up to the question https://math.stackexchange.com/q/3486130/168758.

Let $$X=(X,d)$$ be a Polish metric space equiped with the Borel $$\sigma$$-algebra and let $$\mu$$ be a nonnegative measure. For a nonempty measurable subset $$A$$ of $$X$$ and $$\varepsilon > 0$$, define the $$\varepsilon$$-enlargement of $$A$$ by $$A^\varepsilon := \{x \in X \mid \text{dist}(x,A) \le \varepsilon\}$$, where $$\text{dist}(x,A) := \inf_{a \in A} d(x,a)$$ is the distance of the point $$x$$ from the subset $$A$$. Finally, define $$A' := X\setminus A$$ and $$A^{-\varepsilon} := ((A')^\varepsilon)'$$.

Question 1. Under what general conditions on $$\mu$$ do we have $$\mu((A^\varepsilon)^{-\varepsilon}) \ge \mu(\overset{\circ}{A})$$ for every measurable $$A \subseteq X$$ ?

Of course, if $$((A^\varepsilon)')^\varepsilon \subseteq \overline{A'}$$ (the closure of $$A'$$), then the inequality holds. For example, take $$A = [0, 1]$$ in $$X=(\mathbb R,|\cdot-\cdot|)$$. Then for any $$\varepsilon > 0$$, we have $$(A^\varepsilon)^{-\varepsilon} = (0, 1) = \overset{\circ}{A}$$, the interior of $$A$$.

Question 2. Give an example of a metric space $$X$$ and measurable $$A \subseteq X$$ such that $$\overset{\circ}{A} \not\subseteq (A^\varepsilon)^{-\varepsilon}$$.

# Notes

• We may restrict the questions to closed sets $$A \subseteq X$$ if that helps.
• Any kind of insight which would help make progress on the above questions is more than welcome.

Here is an example for question 2. Given $$\varepsilon>0$$, consider $$X=[-1,0] \cup (\varepsilon,1+\varepsilon]$$ with the usual metric. Then $$A=[-1,0]$$ is closed and open in $$X$$, and $$A^\varepsilon =A$$. Note that $$0 \in (A')^\varepsilon$$, so $$(A^\varepsilon)^{-\varepsilon}=[-1,0)$$ does not contain $$\overset{\circ}{A} =A$$.
Insisting that $$X$$ is connected does not help, as there is a similar example where $$X$$ is an arc of a circle.
Perhaps a better way to avoid such examples is to redefine $$A^{\varepsilon}$$ to be $$\{x \in X \mid \text{dist}(x,A) < \varepsilon\}$$ (with a strict inequality). Then for any set $$A$$ in a metric space, $$A$$ is a subset of $$(A^\varepsilon)^{-\varepsilon}$$.
• Thanks. Looks like the end of the road. Concerning your last comment, indeed, the I'd showed in my old question math.stackexchange.com/q/3486130/168758 that the modified definition of $A^\varepsilon$ indeed avoids all issues. However, I really need to work with the "$\le$" definition of $A^\varepsilon$. – dohmatob Dec 25 '19 at 18:37