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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.
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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 $(A^\varepsilon)^{-\varepsilon}$.

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  • $\begingroup$ 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 $A^\varepsilon$. $\endgroup$ – dohmatob Dec 25 '19 at 18:37

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