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 '13 at 17:46

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.

  • $\begingroup$ Thanks! I wasn't aware that such curves could be constructed. $\endgroup$ – Jim Belk May 11 '13 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 '13 at 20:36

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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