Green's theorem with no differentiability
  The result is established for a Jordan measurable region with rectifiable boundary. The integrand F for the new plane integral to be used is a function of axis-parallel rectangles, finitely additive on non-overlapping ones, hence unambiguously defined and additive on "figures" (i.e. finite unions of axis-parallel rectangles). Define its integral over Jordan measurable S as the limit of its value on the figures, which contain a subfigure of S and are contained in a figure containing S, as the former/complements of the latter expand directedly to fill out S/the complement of S. The integral over every Jordan measurable region exists when additive F is "absolutely continuous" in the sense of converging to zero as the area enclosed by its argument does, or with F the circumferential line integral $\oint P dx + Q dy$ for $P$, $Q$ continuous at the rectifiable boundary of S and integrable along axis - parallel line segments. Thus the equality of this area integral with the line integral around the boundary, to be proved, follows for the various integrals of divergence presented in: Pfeffer, W.F. The Riemann Approach to Integration, Cambridge Univ. Press, New York, 1993. 
I. Fleischer