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I found this question quite interesting, but its answers were disappointingly non-geometric. I'd be interested to know whether there exists a geometric example.

To be precise about what I mean by a geometric example, recall that an open set $U\subset \mathbb{R}^n$ is called regular if it is equal to the interior of its closure. Here is my question:

Question: Does there exist a regular open set $U\subset \mathbb{R}^n$ whose boundary has nonzero Lebesgue measure?

I would guess that the answer is no in $\mathbb{R}^1$, but it ought to be yes in $\mathbb{R}^2$. Part of why this is interesting is that it seems to me that the Mandelbrot set might fall into this class (although according to this MathOverflow post, it is an open question whether the boundary of the Mandelbrot set has positive Lebesgue measure).

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    $\begingroup$ Doesn't a Jordan curve of positive measure work? Unless I'm missing something, the Schoenflies theorem enssures that both its interior and its exterior are regular open and their boundaries are the curve. $\endgroup$
    – Martin
    May 11, 2013 at 17:46

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Osgood constructed an example of a Jordan curve of positive area in Trans. Amer. Math. Soc. 4 (1903), 107-112. Knopp's construction of such a curve is the content of a demonstration on Wolfram|Alpha.

By the Schoenflies theorem we can extend such a Jordan curve $\gamma$ to a homeomorphism $\mathbb{C} \to \mathbb{C}$ such that the interior and exterior of the unit circle are sent to the interior and exterior of $\gamma$, respectively. Since homeomorphisms preserve regular open sets, the interior and exterior of such a Jordan curve will be examples of regular open sets whose boundary has nonzero volume.


Added: Brian M. Scott shows in Cantor set is boundary of regular open set how to get a Cantor set as boundary of a regular open set. The argument appears to apply just as well for a fat Cantor set, so it is even possible in one dimension.

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  • $\begingroup$ Thanks! I wasn't aware that such curves could be constructed. $\endgroup$
    – Jim Belk
    May 11, 2013 at 20:11
  • $\begingroup$ Such Jordan curves are sometimes called Osgood curves. They come up on MathOverflow from time to time, e.g. here and here. This answer by Andrey Rekalo recommends the book Space Filling Curves by Hans Sagan for a good survey on them. $\endgroup$
    – Martin
    May 11, 2013 at 20:36
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I know I'm answering an old question, but for completeness of MSE and since the fact is not too easy to find in the literature let me give an example showing that there is an example in $\mathbb{R}$ of a regular open set whose boundary has positive measure. The following construction is mostly taken from Tomkowicz & Wagon, The Banach-Tarski Paradox (2d ed, 2016), prop. 10.33:

Let $(q_n)$ be an enumeration of the rationals in $(0,1)$. Fix $0<\varepsilon<1$. Inductively define an interval $I_n$ (possibly empty) as follows: if $q_n \in I_m$ for some $m<n$ then let $I_n = \varnothing$, otherwise, let $I_n$ be an interval centered at $q_n$, with length less than $2^{-n}\,\varepsilon$, having irrational endpoints, and disjoint from all $I_m$ with $m<n$ (which is possible since no $I_m$ has $q_n$ as an endpoint). Also, let $I'_n$ be the left open half of $I_n$ (i.e., has the same left endpoint and $q_n$ as right endpoint, with of course $I'_n=\varnothing$ when $I_n=\varnothing$) and $I''_n$ be the similarly defined right open half.

Let $A = \bigcup_{n\in\mathbb{N}} I_n$ and $A' = \bigcup_{n\in\mathbb{N}} I'_n$ and $A'' = \bigcup_{n\in\mathbb{N}} I''_n$. Note that $A'$ and $A''$ are open and disjoint — so the closure of $A'$ is disjoint from $A''$ (and conversely).

Claim: $A'$ is regular open. In other words, we must show that any point $x$ belonging to an open interval $J$ contained in the closure $\overline{A'}$ of $A'$ does, in fact, belong to $A'$. Note that $J$ is disjoint from $A''$ (since $J$ is contained in the closure of $A'$). Now let $0<r<x$ be a rational number in $J$: this $r$ belongs to exactly one of the $I_n$; and the center $q_n$ of this $I_n$ must be to the right of $J$ (i.e., $\sup J\leq q_n$), since $J$ is disjoint from $I''_n$. Thus, $r$ is in $I_n$ and less than $q_n$, so it belongs to $I'_n$; and so does $x$ since $r<x<q_n$. This shows that $x$ both belongs to $I'_n$, and in particular, to $A'$, as claimed.

Of course, similarly, $A''$ is regular open. On the other hand, the closure $\overline{A'\cup A''}$ of $A'\cup A''$ contains every rational number between $0$ and $1$, so it is $[0,1]$ (i.e., $A' \cup A''$ is dense).

Now by construction, the (Lebesgue) measure of $I_n$ is $<2^{-n}\,\varepsilon$, so that of $A$ is $<\varepsilon$, and in particular, so is that of $A'\cup A''$. Since $A'\cup A''$ is dense, the boundary $\partial(A'\cup A'')$ of $A'\cup A''$ has positive measure; but since $\partial(A'\cup A'') = (\overline{A'}\cup\overline{A''}) \setminus (A'\cup A'')$ equals $(\overline{A'}\setminus A') \cup (\overline{A''}\setminus A'') = (\partial A') \cup (\partial A'')$, at least one of $A'$ and $A''$ has a boundary with positive measure.

(Equivalently: recall that the Jordan-measurable sets in $[0,1]$ are exactly those whose boundary has (Lebesgue, or equivalently, Jordan) measure zero; so we are looking for an example of a regular open set in $[0,1]$ that is not Jordan-measurable. Now $A'\cup A''$ is not Jordan-measurable since it is dense with Lebesgue-measure $<\varepsilon$, and since Jordan measure is finitely additive, this means that at least one of $A'$ and $A''$ is not Jordan-measurable.)

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  • $\begingroup$ See also: Reinhard Börger, A non-Jordan-measurable regularly open subset of the unit interval, Archiv der Mathematik (Basel) 73 #4 (October 1999), pp. 262- 264 (The definition of "regularly open" in the abstract is not correct, but the correct definition is used in the body of the paper.) (continued) $\endgroup$ Mar 28, 2022 at 15:40
  • $\begingroup$ AND Ivan Netuka, Regularly open sets with boundary of positive volume, Seminarberichte des Fachbereichs Mathematik der Fernuniversität Hagen 69 (2000), pp. 95-97. If anyone is interested, I wrote an unpublished manuscript back in March or April 2001 simplifying their proofs and providing more exposition and references. $\endgroup$ Mar 28, 2022 at 15:47
  • $\begingroup$ Sorry for my late follow-up, but may I ask, is $A'$ supposed to be a disjoint union since the length of a left half with a greater index is less than half the length of a previous left one? $\endgroup$
    – PinkyWay
    Apr 13, 2022 at 19:02
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    $\begingroup$ @Spring: Yes, the $I'_n$ and the $I''_n$ are all pairwise disjoint. $\endgroup$
    – Gro-Tsen
    Apr 13, 2022 at 22:14

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