In the common divergence theorem, shall the boundary (surface) not be smooth everywhere? Is there a version of this theorem where the boundary is nowhere differentiable?
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I do not think you can in the ordinary sense, because the divergence theorem states $$\int dFdv=\int F\cdot nds$$
Now if $S$ is not differentiable, then the integration of the differential form $F\cdot nds$ is not really well defined. You can still integrate $F\cdot nds$ as a measurable function on $S$, where $ds$ become a certain unit surface element. For example there are "exotic spheres" constructed from plumbing which does not admit a differentiable structure, and I assume a suitable modification of standard "area form" in polar coordinate might work. Then the theorem would carry through. But to write down such an integral explicitly would be very difficult.