In what follows $ \mathbb{R}^{n+m} = \{(x,y): x \in \mathbb{R}^n, \ y \in \mathbb{R}^m \} \ .$

Suppose $G: U \to V $ is a $C^1$-diffeomorphism from an open subset of a manifold to an open subset of $\mathbb{R}^{n+m}$. We write $$ (\xi,\eta)= G^{-1}(x,y) \ . $$

This is a local parameterization of $U$. We will think of $$ U_\eta: = G^{-1}(\mathbb{R}^n \times \{y\} \cap V) $$ as the "horizontal" fibers, which are $n$-dimensional $C^1$-manifolds. Similarly, we refer to $$ U_\xi = G^{-1}( \{x\} \times \mathbb{R}^m \cap V) $$ as the ``vertical" fibers, which are $m$-dimensional $C^1$-manifolds.

Question: Is there a version of the Fubini's theorem that equates the integral over $U$ to a double integral over fibers?

I am wondering if such a result exists in any textbook. In any other publication? It must be, because it is so natural to desire an integration over the original foliations.

This is a paraphrase of the question asked previously here.

I came up with the following:

Let$ DG_{|U_\xi} (\xi,\eta)$ be the derivative, at point $(\xi,\eta)$, of the map $G_{|U_\xi}$, i.e. the restriction of $G$ to $U_\xi$. This restriction is from an $m$-dimensional $C^1$-manifold to a subset of $\mathbb{R}^m$.} So its derivative and its Jacobian are defined.

Suppose for some fixed $\eta_0$, the union of vertical fibers along $U_{\eta_0}$ covers the set $U$.

Lemma Under the assumptions above, for any integrable function $f: U \to \mathbb{R} $, $$ \int_U f = \int_{U_{\eta_0}}\left(\int_{U_\xi} f(\xi,\eta) \frac{|\det DG_{U_\xi} (\xi,\eta)| \cdot |\det DG_{U_{\eta_0}} (\xi,\eta_0)|}{|\det DG(\xi,\eta)|} \ d\mathcal{H}^m(\eta)\right) d\mathcal{H}^n(\xi) \cdot $$

That is saying that to integrate over $U$ simply integrate along the fibers and then sum over the "base", which is what Fubini's theorem says in the standard orthogonal coordinates of the Euclidean space. The correcting factor is can be memorized as $$ \frac{\text{ Jacobian along fibers} \times \text {Jacobian along base}}{\text{full Jacobian}} \ .$$

Corollary: Integration in spherical coordinates. There, the fibers are orthogonal to the base, and so, the full Jacobian also factors, cancelling the numerator. Therefore, the correcting factor disappears (=1), and we get the familiar $$ \int_U f \ d \mathcal{L}^n = \int_0^\infty \left( \int_{\mathbb{S}^{n-1}(r)\cap U} f \ d \sigma \right) dr \ . $$

Example: Let $P$ be a parallelepiped in the plane, resulting from tilting a rectangle so that the acute angle between its edges $A$ and $B$ is $\theta$. Then, foliating $P$ by parallel copies of the edges $B$, indexed $B_x$, leads to $$ \int_P f \ d \mathcal{L}^2 = \int_A \left( \int_{B_x} f \sin(\theta) \ d \mathcal{H}^1 \right) dx \ = \sin(\theta) \int_A \left( \int_{B_x} f \ d \mathcal{H}^1 \right) dx \ . $$

Again my questions are:

1) Is there some alternative formula known out there?

2) Are there references giving specifically this? (I am not interested in "it can follow"s and "it can be done"s!)

  • 3
    $\begingroup$ Is this a question? $\endgroup$
    – LSpice
    Jan 22, 2020 at 17:14
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because it is not a question. $\endgroup$
    – Ben McKay
    Jan 22, 2020 at 17:39
  • 1
    $\begingroup$ Are there references in a textbook to this? That is the question. $\endgroup$ Jan 23, 2020 at 1:26
  • 1
    $\begingroup$ By the way, a recent edit fixed a number of small typos, but one of them wasn't actually: parallelopiped is an acceptable alternate spelling of parallelepiped. $\endgroup$
    – LSpice
    Jan 23, 2020 at 14:44

1 Answer 1


A nice version of the manifold version of Fubini's theorem is in Differentialgeometrie und Fasserbündel by Rolf Sulanke and Peter Wintgen:

Let $\phi \in C^1(M,N)$, where $M,N$ are smooth manifolds of dimensions $m,n$, respectively, with $m \ge n$. Let $\omega \in \Omega^{m-n}(M)$ and $\eta \in \Omega^n(N)$, and let $f : M \to {\bf R}$ be measurable (meaning, its superposition with any map is Lebesgue measurable). Assume the set of critical values of $\phi$ has measure zero in $N$ (again, this means the image under any map has Lebesgue measure zero).

If the $m$-form $f\omega\wedge\phi^*\eta$ is integrable on $M$, then for almost all $x \in N$ the integral:

$$ \int\limits_{\phi^{-1}(x)} f\omega $$

is well-defined and, moreover, when treated as a function of $x$ and multiplied by $\eta$, it is integrable on $N$ and:

$$ \int\limits_M f\omega\wedge\phi^*\eta = \int\limits_N \bigg( \int\limits_{\phi^{-1}(x)} f\omega \bigg)\, \eta $$


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