I am looking for a generalized version of the Gauss-Green theorem also known as the divergence theorem:


A quick search on MathSciNet suggests that there are generalizations for bad domains and nonsmooth functions. However, they seem to rely on heavy machinery and not to be suited for the special case I am interested in.

For example, I found this formula on PlanetMath:$$ \int_E \mathrm{div} f(x)\, dx = \int_{\partial^* E} \langle \nu_E(x),f(x)\rangle \,d\mathcal H^{n-1}(x)$$

See http://planetmath.org/?method=l2h&from=objects&name=GaussGreenTheorem&op=getobj for the details.

Let $\Omega \subset \mathbb{R}^n$ be open and bounded and $f\in C^1(\Omega, \mathbb{R}^n) \cap C^0(\overline\Omega, \mathbb{R}^n)$.

Question: What conditions do we have to impose on $\Omega$ (or $f$) to ensure that the divergence theorem holds true?

To clarify my question: I know that requiring the boundary of $\Omega$ to be piecewise regular is sufficient for the Gauss-Green theorem to be true. I wondered if this condition is also necessary. If so: is the an other "version" of Gauss-Green (e.g. the one cited above) which holds true under weaker conditions and is especially suited for the case of an open and bounded domain

  • $\begingroup$ Rather than a generalization of Gauss-Green theorem, the divergence theorem is the $3$-dimensional version of Stokes theorem, of which the Gauss-Green theorem itself is the $2$-dimensional version. $\endgroup$
    – Qfwfq
    Jun 7, 2011 at 21:51
  • $\begingroup$ Wikipedia says that the divergence theorem is also known as Gauss' theorem, Ostrogradsky's theorem, or Gauss–Ostrogradsky theorem. My professor called it the Gauss-Green theorem. $\endgroup$
    – Peter
    Jun 7, 2011 at 22:09
  • $\begingroup$ You are yet to say precisely what conclusions you want to hold. Exactly what theorem do you want to be true? And do you have reason to doubt the first statement on the PlanetMath page? $\endgroup$
    – Spencer
    Jun 7, 2011 at 22:11

4 Answers 4


I sympathise on the "machinery". The general Stokes theorem is known to work with quite a lot of singularity on the boundary. I only know about this from (trying to read about it in) volume 9 of Dieudonné's massive treatise on analysis, XXIV.14. There are some criteria there for sets to be "differentially negligible" as he terms it, so they can be thrown out of the boundary. And the criteria he gives for that are quite broad: one is in terms of measure of small neighbourhoods, another says anything in codimension 2 doesn't matter. The approach is not very abstract, and assured me that reasonable results on "Stokes with singularities" can probably be proved.

A bit more abstractly, in terms of De Rham currents, you are trying to compute the derivative of the characteristic function of an open set. With the flavour of distribution theory, the derivative is something that will exist, and you will find it is supported on the boundary much as you'd expect unless you have done enough to construct a "counter-example" to Stokes; which will be something a bit more exotic but still describable. This for me calibrates the issue. (I don't know the general theory, but am pretty confident that much more than all this is in books by Whitney et al. on geometric measure theory.)


Rather general versions of these theorems have been established for "sets of finite perimeter." You can find a recent paper on this subject here: http://www.math.northwestern.edu/~gqchen/10-Papers/ChenTorresZiemer.pdf

On the other hand, you say you don't want "heavy machinery," so it is hard to guess what you are looking for.


In these lecture notes you will find the statement for set of finite perimeter which is AFAIK the most general form you can get.


I think you look at Serge lang analysis I. there you may get an answer gneeral enough and one which requires limited machinery. also look on net for macdonalds paper on Proof of satokes theorem. the conditions on f are very weeak if you employ gauge integral in fact differential forms should be defined using integral .The prototype is efinition of divergence given in many physics texts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.