I don't understand the "idea" of the following proof, as well as some of the steps. As I'm not sure about its "ways", I'm not editing it much and as such it might be in the wrong order. My sincere apologies. If you got any idea where to find that proof in the literature, I'd be very thankful!

The proof works only on certain sets, equipped with the following properties:

Definition "Set of integration"

Let $G\subset \mathbb R^n$ be an open, bounded set and his boundary $\partial G:= \partial_r G \cup G_0$ so that the following holds true:

  1. $\partial G$ is a $(n-1)$-dimensional submanifold of $\mathbb R^n$
  2. dim$ G_0 < n-1$, (hausdoff-dimension or lower dimensional manifold)
  3. vol$(\partial_r G < \infty)$
  4. For every $x \in \partial_r G$ exists an open subset $U \subset \mathbb R^n$, so that $x \in U$ and a diffeomorphism $\phi:U\to B^n\subset \mathbb R^n$, so that $\phi(U\cap G)=\{x \in B^n|x_1 > 0\}$.

Stokes Theorem

Let $G\subset \mathbb R^n$ be a set of integration. Let $U \supset \overline{G}$ be an open subset and $w \in \Lambda^{n-1} (U)$ a differential form. Then the following holds true:

$\int_G d\omega = \int _{\partial_r G} \omega $



$G$ is a union of finitely many intervalls.

Question one:Do i get the finitely many intervalls because of the following argument?

As $\partial_r G$ is of dimension zero according to 2., it's charts carry the discrete topology and therefore the charts domains are only single points. And with 3. and the counting measure the finity follows.

Without the loss of substance we can limit ourself to one open intervall.

Question two:With the exactness of the cohomolgy group on the intervall and and the fundamental theorem of integration follows:

$\int_a^b f'(x) dx = f(b)-f(a)$

$\underline{1\to n}$:

Let $H:=\{x \in R^n | x_1 = 0\}$, $pr_H : R^N \to H$ the orthgonal projection and $p:= pr_H|_\overline{G}$.

Then $p(\overline{G})=p(\partial G)$:

Question three:About: $p(\overline{G})=p(\partial G)$

$p(x)$ is closed, $p$ is continuous and $G$ is bounded therefore $p^{-1} (x)$ is compact. With compactness we get the existence of a finite cover of $p^{-1}(x)$, so we get finitely many intervals.But how does that help me, if it's true?

So that now we can write:(Fubini)

$\int_G d\omega = \int _{p(G)} \int_{p^{-1}(x)} d\omega $

Here i will omit the proof and only name his next two claims. The author proves that:

$p(G) \backslash p(\partial_r G)$ and $p(\partial_r G) \backslash p(G)$ are both null sets.

And concludes with following lemma the proof of stokes theorem:

Lemma Let $X,Y$ be oriented manifolds, dimension of $X$ and $Y$ should be $n$. And $\varphi:Y\to X$ a $C^1$ function, $\Omega \in \Lambda^n(X)$ and $f: Y \to \mathbb R$ be $\varphi ^* \Omega $ integratable. Then the following holds true:

$\int_Y f\varphi^* \Omega = \int_X (\sum_y\in \varphi^{-1} f(y))\Omega$

where $\epsilon(y)$:= sign of the determinant of $d\varphi_y$

We can simplify the differential form $\omega$ to $\omega = f dx_2 \wedge ... \wedge dx_n$

and $d\omega = \frac{\partial f}{\partial x_1} dx_1^...dx_n:=\omega_1 \wedge \Omega, \omega_1 \in \Lambda^1(U) $. For $\Omega = dx_2 \wedge ... \wedge dx_n$ there is $p^* \Omega = dx_2 \wedge...\wedge dx_n$ and we can write:

$\int_{\partial_r G} \omega$ $= \int_{\partial_r G}p^*\Omega $ = $\int_{\partial_r G} f\Omega $= $\int_{p(\partial_r G)} (\sum_{y \in (p|_{\partial _r G})^{-1} (x) } \epsilon(y) f(y))\Omega$

And we have already shown that: $(p|_{\partial_r G})^{-1}= \{a_1,b1\} \cup ...\cup \{a_m,b_m\}$

And $\epsilon(a_j)=-1$,$\epsilon(b_j)=1$, so that: $\sum_{y \in p^{-1}(x)}\epsilon(y)f(y)=\sum_{j=1,..,n} f(b_j)-f(a_j)=\int_p^{-1}(x) \frac{\partial f}{\partial x_1}dx_1$.

Now we conclude that:

$ \int_G d\omega $= $ \int_{p(\partial_r G)}(\int_{p|_G)^{-1}(x)} \frac{\partial f}{\partial x_1}\omega $ =$ \int_{p(\partial_r G)} (\sum_{y \in (p|_{\partial _r G})^{-1} (x) } \epsilon(y) f(y))\Omega = \int_{\partial_r G} \omega$

Question four: We now know that $p(G) \backslash p(\partial_r G)$ and $p(\partial_r G) \backslash p(G)$ are both null sets. I suppose we're using the Lebesgue measure to quantify a nullset. When doing the actual integration in the stokes theorem, we use the integral of a differential form, not the Lebesgue measure. Or am i wrong, in that we use the Lebesgue measure to define the integral of a differential form? And we do need to know that $G_0$ is a nullset for both sides $\omega$ and $d\omega$ because we use a Lebesgue measure of degree $n$ and one of degree $n-1$ ? I hope $p(G) \backslash p(\partial_r G)$ is a nullset for the measure of degree $n$ and $p(\partial_r G) \backslash p(G)$ for $n-1$.

Question five: One more thing about the sort of interval $p^{-1}(x)$. He's basically using the fact that $p^{-1}(x) \in \mathbb R x \mathbb R^{n-1} \cong \mathbb R$, that means for orthogonal projections the intervall in $\mathbb R^n$ is canonically isomorph to $\mathbb R$. So if we integrate now, we measure $p^{-1}(x)$ with the Lebesgue measure in $\mathbb R$? I guess that's some sort of fubini? Thanks again!

  • $\begingroup$ May I know where did you see this proof? $\endgroup$
    – Shuhao Cao
    Jun 8, 2013 at 18:54
  • $\begingroup$ i got it from a global analysis lecture (differential forms, poincare, homotopy invariance,cohomolgy and then this was the proof for stokes - 5th semester). $\endgroup$
    – Marcel
    Jun 9, 2013 at 10:16

1 Answer 1


Question 1: The set $G_0$ is empty, since it has dimension less than zero. Then the rest of your argument works, because the volume of the rest of the boundary is in fact counting measure.

Question 2: This one is good.

Question 3: Just look at the line of points that map to some $x$ in $p(\overline{G})$. The intersection of this line with $\overline{G}$ is compact, so has a largest point, which must be a boundary point, because all neighborhoods about it intersect the set and its complement.

  • $\begingroup$ Ah, ok that's where the orthonormal projection comes into place. We can make sense of the word " large" on the one dimensional line.So to say $ \mathbb R x \mathbb R^{n-1}$ $\endgroup$
    – Marcel
    Jun 9, 2013 at 13:35
  • $\begingroup$ Exactly! I feel my answer is somewhat incomplete, so if you want any additional information, let me know and me or someone else will answer it in more detail. $\endgroup$ Jun 10, 2013 at 14:05
  • $\begingroup$ The answer was great Brian, it got me the missing pieces and training in applying them. I might ask two more things though, but ill do so in my own post - makes your insight more useful for others. $\endgroup$
    – Marcel
    Jun 10, 2013 at 19:36

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.