Quote by loom91
Firstly, they are equating volume integrals, not surface ones.

Yes, and the volume is bounded by a closed surface.
Secondly, I said why the argument was flawed in my previous post. Equality of ndimensional integrals, even over all possible nsurfaces, does not rule out pointwise inequality over a set of measure zero (n1surfaces, for example).
Molu

I don't understand what you're trying to say. Maybe someone with more expertise in math can take a look?