 Quote by loom91
Firstly, they are equating volume integrals, not surface ones.
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Yes, and the volume is bounded by a closed surface.
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Secondly, I said why the argument was flawed in my previous post. Equality of n-dimensional integrals, even over all possible n-surfaces, does not rule out pointwise inequality over a set of measure zero (n-1-surfaces, for example).
Molu
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I don't understand what you're trying to say. Maybe someone with more expertise in math can take a look?