Must the topological boundary of an embedded manifold be a set of Lebesgue measure zero? Why is this question closed? Let $X\subset\mathbb{R}^n$
be a bounded connected $\mathcal{C}^1$
embedded k
dimensional manifold (k<n
); i.e., for each x∈X
, there exists an open (in the subspace topology) neighborhood $U_x$
of x
which is diffeomorphic to the unit open ball in $\mathbb{R}^k$
. Must the topological boundary ∂X
of X
be a set (possibly empty, of course) of Lebesgue measure zero in $\mathbb{R}^n$? Equivalently, must cl(X)
be a set of Lebesgue measure zero? If so, how does one prove it?
 A: To convert my comments to an answer:

*

*Start with an Osgood curve $J\subset {\mathbb C}$. (An Osgood curve is a Jordan curve in the plane that has positive 2d Lebesgue measure.) By the Jordan-Schoenflies theorem, $J$ bounds an open topological disk $U\subset {\mathbb C}$.


*Let $D$ be the open unit disk (centered at the origin) in ${\mathbb C}$. Let $f: D\to U$ be a Riemann mapping (a conformal diffeomorphism).


*By Caratheodory's theorem, $f$ extends to a homeomorphism between the closures of $D$ and $U$.


*Now, let $S\subset D$ be a $C^\infty$ spiral in $D$, diffeomorphic to ${\mathbb R}$, such that the closure of $S$ consists of the union of $S$ itself, the unit circle $\partial D$ and the center of $D$.
If you want a concrete equation, consider the polar coordinates $(r,\theta)$ in $D$, where $0\le r<1$, $\theta\in {\mathbb R})$. Take the function
$$
\sigma: {\mathbb R}\to D,
$$
$$
\sigma(\theta)= \left(\frac{2}{\pi}\arctan(\theta) +1, \theta\right).
$$
(Here I am using the principal branch of $\arctan$.)
Then $S$ is the image of $\sigma$.


*Lastly, take $M=f(S)\subset U$. Then $M$ is a connected 1-dimensional submanifold of the plane and the "topological boundary" (aka the "frontier") of $M$ contains the Osgood curve $J$. In particular, it has positive 2d Lebesgue measure.

