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I will think of $ \mathbb{R}^{n+m}$ as $\mathbb{R}^n \times \mathbb{R}^m$.

Let $ V \subset \mathbb{R}^{n+m}$ be open and $g:V \to U \subset \mathbb{R}^{n+m} $ be a $C^1$ diffeomorphism. For a fixed ${y} \in \mathbb{R}^m$, the image $g(\mathbb{R}^n \times \{y\})$ is an $n$-dimensional $C^1$ manifold, and, similarly, for a fixed ${x}$, the image $g(\{x\} \times \mathbb{R}^m)$ is an $m$-dimensional $C^1$ manifold. Let $\mathcal{H}^n$ and $\mathcal{H}^m$ be, respectively, the Hausdorff measures on these with respect to the intrinsic metric on them induced from $\mathbb{R}^{n+m}$. As mentioned in the comments below, they will be different from fiber to fiber, and, for example, it is not true that all these measures are identifiable.

Original Question: I wonder if a "Fubini's theorem" can be formulated and proven using integrals on these manifolds directly.** I do NOT wish to pullback to $V$ via $g$,

Edit: Initially I stated "I do not want to contaminate my integral with the Jacobian!" In light of comments below, it will be impossible to bring in some type of Jacobian(S) into picture. Now, it looks obvious: We must take into account how fibers close in or expand away from one another at different neighborhoods. So, now I reiterate my question allowing this:

Edited Question: Is there "a Fubini's theorem" that equates an integral over $U$ to the iterated integrals (of the function probably multiplied with some Jaobian of the map $g$) over these fibers -- against their intrinsic Hausdorff measures.**

A cartoon of the sought-for identity will look like: for a continuous real-valued function $ \phi: U \to \mathbb{R}$, $$ \int_U \phi \ d\mathcal{L}^{n+m}= \int_{?} \left(\int_{?} \phi(x,y) \cdot Jacobian \ quantities \ from \ g \ d\mathcal{H}^n(x)\right) \ d\mathcal{H}^m(y) \ .$$

Note: I seem to have figured out one such formula but will wait longer for possible alternatives or references to known ones, if any exists.

I have the answer here: Fubini's Theorem on Arbitrary Foliations

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    $\begingroup$ There is a natural diffeomorphism between $g(\mathbb{R}^n \times \{y_1\})$ and $g(\mathbb{R}^n \times\{y_2\})$ by your mapping. For different values of $y$, the Hausdorff measure $\mathcal{H}^{n}(x)$ is not invariant under this diffeo (in the generic setting). Similarly $\mathcal{H}^m(y)$. In the inner integral you can explicitly write in the $y$ dependence. But you cannot just get rid of the $x$ dependence of the outer integral; hence the Jacobian factor... $\endgroup$
    – Willie Wong
    Commented Nov 19, 2019 at 19:55
  • $\begingroup$ What are the question marks? $\endgroup$
    – Piotr Hajlasz
    Commented Nov 20, 2019 at 3:20
  • $\begingroup$ @Piotr Hajlasz: To be determined. See Willie Wong's comment. Ideally the vertical ($m$-dim) fibers for the first and the horizontal fibers for the second integra. $\endgroup$
    – Behnam Esmayli
    Commented Nov 20, 2019 at 13:59
  • $\begingroup$ I have the answer here: mathoverflow.net/questions/350952/… $\endgroup$
    – Behnam Esmayli
    Commented Jan 22, 2020 at 17:12

3 Answers 3

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Coarea formula will do this for you. It is a "Fubini formula" relating the integral of a function $u$ on a Riemann manifold $(M_0,g_0)$ to the integrals along the fibers of a smooth map $F:(M_0,g_0)\to (M_1,g_1)$. The map $F$ need not be a submersion. See this note for details.

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    $\begingroup$ Though of course the coarea formula requires (the requisite) Jacobian factor $\endgroup$
    – Willie Wong
    Commented Nov 19, 2019 at 19:47
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    $\begingroup$ One cannot avoid Jacobians if we use the Hausdorff measure along the fivers. $\endgroup$
    – Liviu Nicolaescu
    Commented Nov 20, 2019 at 15:48
  • $\begingroup$ @LiviuNicolaescu Thanks for the nice notes, had them already! Do you have an explicit formula for how Jacobian transforms under a ($C^1$-)diffeomorphism that straightens the fibers? Is there a component that catches the distortion along the fibers in addition to the full Jacobian of the diffeomorphism. More precisly: Say you have a $C^1$ map $f:R^m \supset U \to R^n$, m>n, with rank df = n everywhere. You apply a diffeo $ G: V \to U$ (locally) such that $f \circ G:V \to R^n$ becomes a projection. How are the Jacobians (as used in the coarea formula) of $f$ and $f \circ G$ connected? $\endgroup$
    – Behnam Esmayli
    Commented Dec 11, 2019 at 0:58
  • $\begingroup$ @BehnamEsmayli The Jacobian is a metric quantity. To get something that is not metric you need to work with denisties such as in Prop. 9.1.8 of these notes . www3.nd.edu/~lnicolae/Lectures.pdf $\endgroup$
    – Liviu Nicolaescu
    Commented Dec 11, 2019 at 18:58
  • $\begingroup$ I don't see how this is a response to my last comment! Is it? $\endgroup$
    – Behnam Esmayli
    Commented Dec 12, 2019 at 0:09
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You can do this nicely with differential forms: see the chapter on Fubini's theorem in my lecture notes on Stokes's theorem.

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I have the answer here: Fubini's Theorem on Arbitrary Foliations

$$\int_U f = \int_{U_{\eta_0}} \left(\int_{U_\xi} f(\xi,\eta) \frac{|\det DG_{U_\xi} (\xi,\eta)| \cdot |\det DG_{U_{\eta_0}} (\xi,\eta_0)|}{|\det DG(\xi,\eta)|} \ d\mathcal{H}^m(\eta)\right) \ d\mathcal{H}^n(\xi) \ .$$

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