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A first stab at a definition of surface area might go like this:

Let S be a surface. Select finitely many points from S and make a bunch of triangles having these points as vertexes. Add up the areas of all of the triangles. This is a finite approximation to the surface area. Now to get the actual surface area, we increase the number of points so that they "densely cover" the surface (Maybe a good working definition would be that any open set of S should eventually contain some vertexes of triangles).

I seem to remember reading about a counterexample to this naive definition, but I can't find a reference. I believe there is even a natural looking polygonal approximation to the cylinder whose surface area diverges to infinity. Can anyone help me out?

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Are you really just randomly making triangles from a finite collection of points on the surface? One dimension lower, if you choose a finite collection of points on a curve and randomly make intervals out of them (ignoring the linear ordering along the curve), adding up the length of the interval, you almost always get a number very far from the length of the curve. Don't you want to triangulate the surface $S$? –  Ryan Budney Mar 1 '12 at 21:14
I have a related question for anybody to answer: Is there any deep reason why this attempt at extending "rectifiability" fails when the Hausdorff dimension is greater than 1? –  Christopher A. Wong Mar 1 '12 at 21:54
@Ryan: When I think about a "triangulation" of a surface, I usually think about the triangles living on the surface, but here the triangles are passing through the ambient space. So I didn't have a good word for it. You are right that my language is imprecise, but I hoped it would conjure the same mental image for my readers as it did for myself. –  Steven Gubkin Mar 1 '12 at 22:17
@Christopher: Although requiring a sequence of polygonal approximations to converge pointwise to the surface isn't by itself enough to get the definition of surface area to work correctly, imposing a mild additional requirement, such as convergence of normals, does suffice. So IMO, dimension 1 is special only because the limited amount of wiggle room means that pointwise convergence automatically implies some stronger convergence properties, so that you don't have to specify those stronger conditions explicitly. –  Timothy Chow Mar 1 '12 at 22:29
Your strategy, to succeed, should rely on an additional hypothesis: each triangle should ultimately be almost parallel of the tangent plan of any close point on the surface. –  Benoît Kloeckner Mar 2 '12 at 10:34

1 Answer 1

Perhaps you are thinking of the Schwartz Lantern? It converges to the cylinder in the Hausdorff metric but its area can be arranged to head toward $\infty$. It was mentioned in the earlier MO question, "Convergence of finite element method: counterexamples." There is nice applet here showing the lantern rotating. Here is an image from Conan Wu's blog:

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page 129, Calculus on Manifolds, by Michael Spivak. –  Will Jagy Mar 1 '12 at 21:51
@Joseph Excellent! Thanks a lot. @Will Yes that is where I must have seen it. –  Steven Gubkin Mar 1 '12 at 22:22

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