This question is inspired by several others on a similar topic: see e.g. this one and a sequence of linked questions. Let us so far focus on $\Bbb R^n$ endowed with standard Borel structure. For any map $\phi\in \mathcal C^1(\Bbb R^n,\Bbb R^n)$ we have the following formula for the change of variables in the integral $$ \int_{\phi(\Omega)}f\;\mathrm d\lambda = \int_\Omega (f\circ\phi)\;|\phi'|\mathrm d\lambda \tag{1} $$ where $\Omega$ is any Borel set, $f$ is any Borel and bounded, $\phi'$ is the Jacobian of $\phi$ and $\lambda$ is the $n$-dimensional Lebesgue measure on $\Bbb R^n$.

On the other hand, for any Borel measure $\mu$ the following formula applies $$ \int_\Omega (f\circ\phi)\;\mathrm d\mu = \int_{\phi(\Omega)}f\;\mathrm d(\phi_*\mu) \tag{2}. $$ where $(\phi_*\mu)(A) = \mu(\phi^{-1}(A))$ is the pushforward measure. As a result, provided the fact that $ \frac{\mathrm d\mu}{\mathrm d\lambda} = |\phi'| $ we obtain for any admissible $\Omega$ and $f$ that $$ \int_{\phi(\Omega)}f\;\mathrm d\lambda = \int_{\phi(\Omega)}f\;\mathrm d(\phi_*\mu) $$ which implies that $\lambda|_{\phi(\Omega)} = (\phi_*\mu)|_{\phi(\Omega)}$. Moreover, if $\phi(\Omega) = \Bbb R^n$ then $\lambda = \phi_*\mu$.

Let $\mathcal P(\Bbb R^n)$ be the set of all Borel measures on $\Bbb R^n$, and for any $\phi\in \mathcal C^1(\Bbb R^n,\Bbb R^n)$ with the range $\Bbb R^n$ let us define an operator $\phi'$ on $\mathcal P(\Bbb R^n)$ given by $\mathrm d\phi'(\mu) := |\phi'|\mathrm d\mu.$ As a result, from the discussion above we obtain that $\lambda$ solves the equation $$ \phi_*(\phi'(\mu)) = \mu \tag{3}. $$ for any such $\phi$. Clearly, for some $\phi$ there may be multiple solution: e.g. if $\phi =\mathrm{id}_{\Bbb R^n}$ then $\phi_*\circ \phi' = \mathrm{id}_{\mathcal P(\Bbb R^n)}$ so that any $\mu$ satisfies $(3)$ in this case.

Q1: Is that true, that $\lambda$ is the only positive measure (up to scaling) that satisfies $(3)$ for all $\phi$ that range over $\Bbb R^n$?

Q2: If such a measure is not unique, what similar properties do they have? Perhaps, it has to be equivalent to the Lebesgue measure.

Q3: Is there a "characteristic" map $\hat\phi$ such that if $\mu$ satisfies $(3)$ for $\hat\phi$, then it satisfies it for all $\phi$?

Q4: Is there any intuitive reason, why the Lebesgue measure satisfies this equation?

P.S. Feel free to fix my notation (especially $\phi'$ as an operator), or comment for which $\Omega$, $f$ and $\phi$ formulas above are well-defined - I didn't have a chance to study geometric measure theory.

I hope that the current formulation may shed some light on the similarities between two different formulas for the change of coordinates, which are not clear to me at the moment. Also, I know that similar formulas for the change of coordinates are available for smooth manifolds endowed e.g. with Hausdorff measures, so perhaps this feature applies not only to the Lebesgue measure, but to $n$-dimensional Hausdorff measures as well.

Another consequence of $(1)$ and $(2)$ is that for any $\mathcal C^1$-bijection $\phi$ it holds that $$ \frac{\mathrm d(\phi_*\lambda)}{\mathrm d\lambda} = |(\phi^{-1})'|\circ\phi^{-1}. $$

  • $\begingroup$ One small remark: if a measure satisfies (3), then it is translation invariant. So the answer to Q1 is Yes if you consider only measures which are finite on compact sets. $\endgroup$
    – Etienne
    Jun 11, 2013 at 20:56

1 Answer 1


Before I start I'd like to point out that your change of variables formula only works for invertible functions (technically injective functions as well, but let me get back on that later). There is a more general version involving differential forms that works for arbitrary differentiable functions, but let's not complicate things.

Also I'd like to change the notation $\phi'(\mu)$ to $|\phi'| \mu$.

Anyway, using the fact that $\phi'$ is invertible and that $\phi_*^{-1} \phi_*(\mu) = \mu$ we can show that for any $\mu$ satisfying $ \phi_*(|\phi'| \mu) = \mu $ we also have that

$$ |\phi'| \mu = \phi_*^{-1}(\mu) $$

or equivalently

$$ \tag{1}\label{*} \def\d{\mathrm{d}} \frac{\d\phi_*^{-1}(\mu)}{\d \mu} = |\phi'| $$

Note that when $\eqref{*}$ holds for all translations then $\mu$ must be the Lebesgue measure. Since that is the only (complete) translation invariant measure.

Now let $\mu$ be any measure for which $\frac{\d \mu}{\d \lambda}$ exists (i.e. $\mu \ll \lambda$), then for any $f$

$$ \int_{\Omega} f \, \d \phi_*^{-1}(\mu) = \int_{\phi(\Omega)} (f \circ \phi^{-1}) \, \d \mu = \int_{\phi(\Omega)} (f \circ \phi^{-1}) \frac{\d \mu}{\d \lambda} \d \lambda = \int_{\Omega} f |\phi'| \, \frac{\d \mu}{\d \lambda}\d \lambda. $$

So, using the properties of the Radon–Nikodym we find that ($\lambda$-almost everywhere)

$$ \frac{\d \phi_*^{-1}(\mu)}{\d \lambda} = \frac{\d \phi_*^{-1}(\mu)}{\d \mu} \frac{\d \mu}{\d \lambda} = |\phi'| \, \frac{\d \mu}{\d \lambda} $$

which, provided $\frac{\d \mu}{\d \lambda} \neq 0$ almost everywhere, implies $\eqref{*}$. Therefore $\eqref{*}$ holds for all measures equivalent to the Lebesgue measure (i.e. $\mu \ll \lambda$ and $\lambda \ll \mu$).

As for what happens when $\mu \not\ll \lambda$ it's hard to say. It seems to depend quite a lot on the properties of $\phi$, but for any particular $\phi$ you can construct a pathological measure by starting with a measure $\mu$ and considering the sum

$$ \sum_{k=-\infty}^\infty (\phi_*|\phi'|)^k(\mu) $$

where for $k \geq 0$

$$ \begin{align} (\phi_*|\phi'|)^0(\mu) &= \mu \\ (\phi_*|\phi'|)^k(\mu) &= (\phi_*|\phi'|)^{k-1} \phi_*(|\phi'|\mu))\\ (\phi_*|\phi'|)^{-k}(\mu) &= (\phi_*|\phi'|)^{-k+1} (|\phi'|^{-1}\phi_*^{-1}(\mu)) \end{align} $$

This sum won't always behave, but letting $\mu$ be $\delta_x$, the dirac delta measure at $x$, then as long as $x$ isn't a fixed point and $|\phi'(\phi^k(x))|$ doesn't become $0$ or undefined for any $k$ the resulting measure remains finite and well defined at least for all sets that only contain finitely many of the points in $\{ \phi^k(x) : k \in \mathbb{Z} \}$. Since this only excludes countably many $x$ there must be an $x$ for which the pathological construction works and is reasonably well behaved.


To get back on the injective functions $f : \mathbb{R}^n \to U \subset \mathbb{R}^n$ , note that those are still invertible when considered as functions onto their image $U$ rather than all of $\mathbb{R}^n$. This complicates things slightly but all the steps in first part of the proof still work. The main issue is that the pathological construction becomes somewhat more difficult, in case $\phi$ doesn't induce an isomorphism on any non-trivial subspace of $\mathbb{R}^n$ you're forced to let $x$ be a fixpoint which results in the measure $\infty \delta_x$, technically this is still a counter example but it's not that interesting.


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