In any Riemann sum we use to prove any theorem, it does not matter how you divide your partitions. Why is that the case? Why, for example, the divergence theorem does not care how you divided the volume; in other words, it does not matter if you used tiny spheres or cubes.

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    $\begingroup$ It's very unclear what this question is asking. What is the connection between Riemann sums and the divergence theorem that you have in mind? $\endgroup$
    – Chris
    Dec 5, 2017 at 22:14
  • $\begingroup$ Doesn't the divergence theorem say that a surface integral of any volume is equal to the integral of the divergence multiplied by an infinitesimal volume. Isn't this integral or sum a riemann sum? $\endgroup$
    – Omar Ali
    Dec 5, 2017 at 22:16
  • $\begingroup$ The definition of Riemann integral implies that if a function is Riemann integrable, the upper and lower Riemann sums are converging to the same limit for any sequence of partitions with mesh converging to zero. You can prove it is enough to consider uniform partitions and you have that $\int$ does not depend on our personal taste in partitioning, which is reasonable for an object with the ambition of representing an area/measure. $\endgroup$ Dec 5, 2017 at 22:21
  • $\begingroup$ So it is not the divergence theorem (or any other theorem mentioning some integral) disregarding such aspect, it is the Riemann integral already. $\endgroup$ Dec 5, 2017 at 22:23
  • $\begingroup$ Ok, why it does not matter if you divide the volume of any shape into infinitesimal cubes or any shape? The diverngence theorem always work? $\endgroup$
    – Omar Ali
    Dec 5, 2017 at 22:27


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