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From Lee's introduction to smooth manifolds: $\DeclareMathOperator{\div}{div}$

The Divergence Theorem. Let $M$ be an oriented Riemannian manifold with boundary. For any compactly supported smooth vector field $X$ on $M$, $$\int_M\div X=\int_{\partial M}\langle X,n\rangle,$$ where $n$ is the outward-pointing unit normal vector field along $\partial M$.$^1$

Here $\partial M$ denotes the manifold boundary and thus this doesn't include the common scenerio where $M\subset N$ is an open subset of a manifold $N$ without boundary (e.g. $N=\mathbb{R}^n$) and $\partial M=\overline{M}\setminus M$ is the topological boundary. What about this case? Is the equation still correct? What are the precise requirements? Are there any references?

$^1$ Since $M$ and $\partial M$ are both Riemannian manifolds, we have a measure on both sets and hence we can integrate real valued functions.

Filippo
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    If you're taking $M$ to be open, its boundary as a manifold with boundary is empty. The point is that $X$ has compact support and so the boundary term vanishes. – Ted Shifrin Jan 24 '22 at 20:16
  • @TedShifrin Yes, Lee does explicitely say that a manifold with boundary can have an empty (manifold) boundary. By the way, I used $X$ to denote both the vector field and the manifold (now called $N$), I hope this didn't cause confusion. – Filippo Jan 24 '22 at 20:34
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    No, I didn't even notice that :P – Ted Shifrin Jan 24 '22 at 20:44
  • If $N$ is the real line, $K = \partial M$ a fat Cantor set (constructed analogously to the ternary set but by removing intervals whose total length is smaller than $1$, so having positive Lebesgue measure), and $M = [0,1] \setminus K$ the open complement, the "normal field" is rather nasty, the boundary integral is taken over infinitely many components, and "most of" the boundary integral does not come from endpoints of the removed open intervals (which have measure zero). At best, it appears there's work required even to make sense of the right-hand side. – Andrew D. Hwang Jan 25 '22 at 02:36
  • @TedShifrin As you already said, an open subset of a manifold without boundary is a manifold without boundary in its own right and the divergence theorem from Lee's book tells us that the integral of the divergence of each compactly supported vector field is zero. Thus, I would like to consider vector fields that are not compactly supported with respect to $M\subset N$, but compactly supported with respect to its closure. For me it would even be fine if we assume that the closure itself is compact such that we don't need to worry about the support. – Filippo Jan 25 '22 at 07:33
  • @AndrewD.Hwang Thank you for the example. As far as I understand, the set $M$ you constructed is even bounded. But it's not connected, is it? – Filippo Jan 25 '22 at 07:38
  • Another idea would be to find sufficient requirements such that the manifold boundary of the closure of $M$ equals the topological boundary of $M$ so that we can apply the divergence theorem from Lee's book. – Filippo Jan 25 '22 at 07:45
  • As written now, the question is ill-posed for several reasons. One reason is that there is no "normal vector field" to the topological boundary. The second is that you should be using $n-1$-dimensional measure and there might not be a finite one (just think of, say, Koch snowflake as $\partial M$ in the case of $N=R^2$) – Moishe Kohan Jan 25 '22 at 07:55
  • @MoisheKohan This is a good comment. I thought I had seen a proof of the divergence theorem for subsets of $\mathbb{R}^n$ which only involved the topological boundary. So I searched for it (it's chapter 15 in this german book) and it turned out that the theorem was formulated only for compact sets and it seems like the topological boundary and the manifold boundary are equal in this case. – Filippo Jan 25 '22 at 09:33
  • In addition, I found some lecture notes with a proof of the Stokes theorem for subsets of $\mathbb{R}^n$ which are not necessarely compact and there it is explicitly said that the boundary in that context is not the topological boundary. – Filippo Jan 25 '22 at 09:53
  • The linked answer by Chapman is sloppy and you should not use it in the context of your question. One more thing: When you are just learning the subject, it is best to use different words for the "manifold boundary" and the "topological boundary." The word "frontier" for the latter is a standard alternative. – Moishe Kohan Jan 25 '22 at 16:30
  • @MoisheKohan Thank you for the warning :) My take home message: Divergence theorem --> Manifold boundary (even if we consider subsets of $\mathbb{R}^n$). We may ask under what circumstances the manifold boundary and the topological one (with respect to the topology on $\mathbb{R}^n$) agree, but that's another question. – Filippo Jan 25 '22 at 18:01

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