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Apologies if I am missing something obvious here. I am trying to understand counterexamples to the segment property. From reading Adams' book titled Sobolev Spaces, the definition is given as follows:

Definition: We say that a domain $\Omega$ satisfies the segment condition if for every $x \in \partial\Omega$ has a neighborhood $U_x$ and a nonzero vector $y_x$ such that if $z \in \overline{\Omega} \cap U_x$, then $z+ty_x\in \Omega$ for $0<t<1$.

After the definition, it states that "the domain [satisfying this condition] cannot lie on both sides of any part of its boundary." A similar geometric intuition is given in segment condition in sobolev space. But I don't understand why this intuition works at all. If I take a domain that is kidney shaped, or I take an annulus, this seems to satisfy the condition that Adams and the Stack question are proposing as types of domains that wouldn't have the segment property. However, it seems pretty self-evident that both of those domains have the segment property. The reason the counterexample domain in the stack question works is because the boundary is enclosing a point of $0$ Lebesgue measure. Any help understanding the general intuition of categorizing domains that don't have the segment property in the way Adams and the Stack question seem to intuit would be greatly appreciated.

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    Perhaps the sentence should be interpreted as a local characterization: Yes, the annulus lies on both sides of some boundary points, but this is only true in the large, not in an arbitrarily small neighborhood of the boundary point. I think the example of removing a segment from a disk illustrates this notion — it's not the points on the boundary of the disk that cause a problem, it's the points on the (removed) segment. – Ted Shifrin Apr 11 '21 at 21:41
  • Mm, but the neighborhood is arbitrary, we get to control what it looks like. I think the example of a line through a disk works, so maybe Lebesgue measure 0 is a poor characterization; a better one might be if the set it is surrounding is of dimension at least one less than the dimension of the set? Is that how I should be understanding it? I think maybe that is it... – mathishard.butweloveit Apr 11 '21 at 21:45
  • Oh yes, I think that is exactly what you are saying. – mathishard.butweloveit Apr 11 '21 at 21:47
  • Perhaps the right notion to think about is the Lebesgue density of these boundary points. If the density is $1$ (as in the case of the removed segment), then the segment condition will fail. If the density is less than $1$, it should hold fine. (That may be too glib, but seems like something worthwhile for you to investigate.) – Ted Shifrin Apr 11 '21 at 22:08
  • I don't recall having heard of the concept of Lebesgue density, but I'm not surprised it exists. I was trying to figure out why in general Poincar{'e}'s inequality is written with the average subtracted from the function, but this is unnecessary in settings where the domain is bounded in a direction isomorphic to a straight line/direction in $\mathbb{R}^n$. It seemed directly related to the segment property that was discussed in the section immediately preceding, hence my somewhat dumb question. I will investigate, thank you very much for all your input. – mathishard.butweloveit Apr 12 '21 at 00:25

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