I am teaching a course on Multivariable Calculus for Graduate students. I came across this nice article by Lax where a special case of the change of variables theorem is proved:

Theorem. Let $f:\mathbb{R}^n \to \mathbb{R}$ be a continuous function with compact support and let $\phi:\mathbb{R}^n \to \mathbb{R}^n$ be a continuously differentiable function that is identity outside of some ball. Then $$\int_{\mathbb{R}^n} f = \int_{\mathbb{R}^n} (f \circ \phi)J_\phi,$$ where $J_\phi$ denotes the determinant of the derivative matrix of $\phi$.

I presented a differential forms version of Lax's proof given by Ivanov in class. Michael Taylor has also given a differential forms version of Lax's proof and writes that the standard version of the change of variables theorem is easily established using Lax's version. Lax himself has a follow-up article in which a more or less standard version of the change of variables theorem is proved but the proof is quite long. One thing that is not clear to me is how to give a simple proof of the following standard version of change of variables theorem using Lax's theorem:

Theorem. Let $\phi:U \to V$ be a $\mathcal{C}^1$-smooth diffeomorphism of open sets in $\mathbb{R}^n$ and let $f:V \to \mathbb{R}$ be a continuous function with compact support. Then $$ \int_V f = \int_U (f \circ \phi)|J_\phi|.$$

I found an article by Kumaresan and Santhanam that claims to deduce the above version from Lax's version but the statement given by them omits the absolute value on $J_\phi$ and their proof is incorrect at the last step.

Is it possible to deduce the standard change of variables theorem from Lax's theorem in an easy way?


1 Answer 1


We may assume in Th.2 that $U$ is connected, so the (never vanishing) $J_\phi$ has constant sign, that we can assume positive w.l.o.g. We first establish the formula locally, for special $f$ (with "small enough supports"): precisely, assume that $\phi^{-1}(\mathrm{supp}\, f)$ is included in a ball $B\Subset U$ such that

(B) there exists a diffeomorphism $\psi:\mathbb{R}^n\to\mathbb{R}^n$ such that $\psi =\phi$ on $B$, and $\psi$ is the identity outside some large ball.

In this case, $ \phi( x)= \psi(x)$ for all $x\in B$, and $f(\phi( x))=f(\psi(x))=0$ for all $x\notin B$ (because $\psi$ is injectve) so that by Th1 $$\int_U(f\circ\phi)|J_\phi|=\int_{\mathbb{R}^n}(f\circ\psi)J_\psi=\int_V f.$$

To extend the formula to any $f$, first observe that, given the diffeo $\phi:U\to V$ with $J_\phi>0$, and $x\in U$, there is a ball $B\Subset U$ satisfying (B) (this point is elementary though maybe technical: one can e.g. use the radius as a deformation parameter, and homotop $\phi$ on its differential, and then to the identity map within the linear group). Therefore, for any $y\in V$ there is a nbd $W$ such that the formula of thm2 holds for functions with support in $W$. Since all integrals depend linearly on $f$, the formula holds then for any function with support in $V$, by partition of unity.

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    $\begingroup$ This argument is essentially the one given by Kumaresan and Santhanam but I am not sure it is correct. Let $A := \phi^{-1}(\textsf{supp}(f))$. Then $A$ is compact. Then by a standard partition of unity argument, we can construct a function $\psi:\mathbb{R}^n \to \mathbb{R}^n$ that agrees with $\phi$ on $A$ and is identity outside a neighbourhood that contains $A$. Lax's theorem applies to this $\psi$ (irrespective of the behaviour of $J_{\psi}$ which could even be $0$ at some points). But it is not clear that $$ \int_U (f \circ \phi)|J_\phi| = \int_U (f \circ \psi) J_\psi.$$ $\endgroup$ Nov 6, 2016 at 11:39
  • $\begingroup$ Right! I see: as it is formulated, Th1 does not guarantee that $$\int_{U\setminus A}(f\circ\psi)J_\psi=0$$ (a suitable version of it would ensure this too.) To bypass this point and use thm1, I think the simplest thing is to prove first the formula for $f$ with small support, so that the corresponding $\psi$ can be made bijective. Then for general $f$ a partition argument can reduce to the local case (all integrals depend linearly on $f$). Is it clear? In case I'll edit and add some details. $\endgroup$ Nov 6, 2016 at 13:13
  • $\begingroup$ I hope this question is not to naive, but why are you allowed to assume that the sign of $ J_{\phi} $ is positive? $\endgroup$ Dec 26, 2017 at 11:45
  • $\begingroup$ @DavidWallauch Because the sign of $J_\phi$ is only incidental. E.g. you can reverse it changing the basis of the target space to $(-e_1,e_2,\dots,e_n)$. $\endgroup$ Dec 26, 2017 at 22:17

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