Are smooth surfaces embedded in R3 , with finite area, always the boundary of a finite perimeter set?

Question

Are smooth compact surfaces embedded in R3 (with no boundary) , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections. There are some problems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

(finite perimeter set is on the distributional sense of GMT)

Answer 1

Your question is worded a bit confusingly, in particular I can't tell if you want compactness or not.

If you don't want compactness, it's not entirely clear what you mean by "bounding": take the unit sphere with a point removed, which is smooth, has finite area, and doesn't have a boundary (in the sense of the Gauss-Green theorem, anyways), but it's not not a closed subset of $\mathbb R^3$ so it's not a boundary in most senses of the word.

If you do want compactness, then the answer to your question is yes. In fact the much stronger below theorem holds:

Let $d \geq 1$, and let $S$ be a compact embedded $C^1$ $d - 1$-dimensional submanifold of $\mathbb R^d$ without boundary. Then $S$ is the boundary of an open subset of $\mathbb R^d$ of finite perimeter.

I think you can weaken $C^1$ to Lipschitz here, if you are careful enough about what the word "embedded" means in that setting.

The proof is standard, but it's easier for me to recite the proof than to dig through my bookshelf for a reference, because the argument isn't long.

Under the hypotheses on $S$, $S$ is orientable and has finitely many connected components. If we show that each of them bounds a set of finite perimeter, we are done, so we may assume that $S$ is connected. In particular, we may assume that $$H^{d - 1}(M, \mathbb Z) \cong \mathbb Z.$$ By taking a one-point compactification of $\mathbb R^d$, we can think of $S$ as an embedded submanifold of $\mathbb S^d = \mathbb R^d \setminus \{\infty\}$; since $S$ is compact, it follows by Alexander duality that $$H_0(\mathbb S^d \setminus S, \mathbb Z) \cong H^{d - 1}(M, \mathbb Z) \oplus \mathbb Z \cong \mathbb Z^2$$ where we picked up the extra direct summand $\mathbb Z$ because Alexander duality is concerned with reduced homology, but we are using unreduced homology. Anyways, the point is that $\mathbb S^d \setminus S$ is an open subset of $\mathbb S^d$ with two components and $\infty$ is in one of those components; let $U$ be the component that doesn't have $\infty$. Then $U$ is open and $\partial U = S$. Also, the euclidean geometry of $\mathbb R^d$ induces a $C^1$ Riemannian metric $g$ on $S$, and since $S$ is compact and $\sqrt{\det g}$ is continuous, $$\int_S \sqrt{\det g} < \infty.$$ So $S$ has finite area, and by the Gauss-Green theorem $U$ has finite perimeter.