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I mainly work in statistics and I know only basic measure theory. I was trying to understand the Co-Area formula by Federer.

If $f:\mathbb{R}^M\to \mathbb{R}^N$ is a Lipschitz function with $M \geq N$ then $$ \int_A J_N f(x) d\mathcal{L}^M x = \int_{R^N} \mathcal{H}^{M-N} (A\cap f^{-1}(y)) d\mathcal{L}^N y $$

Could anyone please give me an intuition of what the formula implies and what it means practically? In particular, this is a simple example I am trying to wrap my head around: suppose we are in 2 dimensions and there is a curve defined by a function

manifold picture

What does the co-area formula tell us in this context?

Here are some practical questions:

  • My understanding is that we have some space of dimension $M$ and there a manifold in it of dimension $M-N$ and we would like to compute the area of this manifold using the Hausdorff measure. Is this what is happening?
  • I am completely lost as to why we need $A\cap f^{-1}(y))$ what's the intuition behind this?
  • On the right hand side we have $\mathcal{H}^{M-N}$. I am familiar with the Lebesgue measure $d\lambda$ however I am lost as to why we seem to have both the Lebesgue and the Hausdorff measure on the right hand side. I thought we should only have the Hausdorff measure on the RHS?
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    $\begingroup$ Can you add more detail so that I may not need to purchase a book to understand the question? $\endgroup$ Feb 15, 2021 at 18:40
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    $\begingroup$ It is a generalization of say, integration using spherical shells - i.e. the volume of a balls is an integral over a one parameter family of spheres, with an appropriate choice of measure when integrating the spheres ($n-1$ dimensional Lebesgue measure, or Hausdorff measure, if your object for example has some kind of fractal structure). The formula is trivial when integrating smooth functions, but the formulation for Lipschitz functions is a bit more subtle, and requires some knowledge of the technical details of geometric measure theory. $\endgroup$ Feb 15, 2021 at 18:43
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    $\begingroup$ @leslietownes apologies, I published the question too early! I just added some questions. I will try to formulate them more concretely. $\endgroup$ Feb 15, 2021 at 18:43
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    $\begingroup$ If you have Lebesgue measure on the plane, how will you use just that to talk about length of a curve in the plane? Think about slicing a region into curves and trying to compute the area of the region from information about the curves. $\endgroup$ Feb 15, 2021 at 18:47
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    $\begingroup$ Reading recommendations: especially if you're a beginner, Federer's book is very hard to follow, I'd rather recommend Geometry of Sets and Measures in Euclidean Spaces by Mattila and Measure Theory and Fine Properties of Functions by Evans-Gariepy. $\endgroup$ Feb 24, 2021 at 23:03

2 Answers 2

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This is most easily understood when $N = 1$. Suppose $A \subset \mathbb{R}^m$ is a bounded domain and $f: A \rightarrow \mathbb{R}$ is a smooth function whose gradient is everywhere nonzero.

Its level sets are non-intersecting hypersurfaces. Suppose first that you want to compute the volume of the $A$ in terms of the surface areas of the hypersurfaces. Suppose $a$ is the inf of $f$ on $A$ and $b$ is the sup. You can divide the interval $[a,b]$ into equal sized subintervals $I_1 = [t_0,t_1], \dots, I_N = [t_{N-1},t_N]$ of size $\delta = (b-a)/N$. The volume of $A$ is the sum of the volumes of $f^{-1}(I_k)$. On the other hand, each $f^{-1}(I_k)$ is a shell with varying thickness. At each point, the thickness is roughly $\Delta t/|\nabla f|$. So the volume of the shell is roughly $$ V(f^{-1}(I_k)) \simeq \Delta t\int_{f^{-1}(t_k)}\frac{dA}{|\nabla f|}, $$ where $dA$ is the $(M-1)$-dimensional Hausdorff measure on $f^{-1}(t_k)$. Adding up the volumes of the shells and taking the limit $N \rightarrow \infty$, we see that the volume of $A$ is $$ V(A) = \int_{-\infty}^{\infty} \int_{f^{-1}(t)} \frac{dA_t}{|\nabla f|}\,dt, $$ where $dA_t$ is the $(M-1)$-dimensional Hausdorff measure of $f^{-1}(t)$.

If, on the other hand, you integrate $|\nabla f|$ on $A$ using the same approach, you get $$ \int_{f^{-1}(I_k)} |\nabla f|\,dx \simeq \Delta t\int_{f^{-1}(t_k)}|\nabla f|\frac{dA}{|\nabla f|}, $$ and therefore $$ \int_A |\nabla f|\,dx = \int_{-\infty}^{\infty} \int_{f^{-1}(t)}\,dA_t\,dt $$ The $N>1$ case is similar, except you chop $\mathbb{R}^N$ into small rectangular pieces and use the fact that the cross section of the inverse image of each piece is roughly a parallelogram whose volume is $(J_Nf)^{-1}$. and therefore $$ \int_A J_Nf\,dx = \int_{\mathbb{R}^N} \int_{f^{-1}(y)}\,dA_y\,dy, $$ where $dA_y$ is the $(M-N)$-dimensional Hausdorff measure on $f^{-1}(y)$.

The general formula is $$ \int_A \phi(x)\,dx = \int_{\mathbb{R}^N} \int_{f^{-1}(y)} \phi(x)\frac{dA_y}{J_Nf}\,dy $$

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First, it follows from approximation by characteristic functions that the formula in the OP generalizes to $$ \int_{R^M} u(x) J_N(Df(x)) \, d\mathcal{L}^{M} = \int_{R^N} \Big(\int_{f^{-1}(z)}u\, d\mathcal{H}^{M-N}\Big)d\mathcal{L}^N(z)\, , $$ for every measurable $u\colon R^M \to [0,+\infty]$ . It is part of the claim that the different components are measurable and that the integrals are meaningful.

This means that to integrate a function $u$ we may first integrate along fibers coming from the function $f$. The most familiar example of this is the Fubini's theorem: If you take $f(x_1,\cdots,x_M) = (x_1,\cdots,x_N)$.

The next example is integration in spherical coordinates: take $f:R^M \to R^1$ be $f(x)=|x|$.

So, that must be a good start for intuition.

Important delicate issues:

  1. Even for smooth $f$ we cannot guarantee that all fibers $f^{-1}(z)$ will be manifolds (let alone of dimension $M-N$). Example: $f(x,y)=x^2-y^2$, and level set of zero. In Lipschitz case, you can draw any shape and let $f(x)$ be distance of $x$ to that set. Obviously, $f$ is Lipschitz and level set of zero is the shape you started with. The reason coarea formula survives this is that if $f$ is Lipschitz, then for a.e. $z$ the sets $f^{-1}(z)$ will be countably $\mathcal{H}^{M-N}$-rectifiable and integration against Hausdorff measure will be meaningful.

  2. Integrability of $z \to \mathcal{H}^{M-N}(f^{-1}(z)\cap A)$ is not easy! It requires the so-called coarea inequality.

  3. The coarea formula is not obvious even for sets $A$ of measure zero!

  4. There is a proof of the coarea formula that uses the implicit function theorem to "straighten" the fibers so that the proof reduces to Fubini's theorem.

Pictorially, if $U$ is a (measurable) set in $R^M$ (here in my picture $R^2$) then $F^{-1}(z)$ (here sample $z=a_1,a_2,\cdots$ drawn) "foliate" $U$. The coarea formula says that integration over $U$ can be done first by integration along these fibers/level sets.

enter image description here

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