Is there a generic change of variables formula for a measure theoretic integral that does not use the Lebesgue measure? Specifically, most references that I can find give a change of variables formula of the form:

$$ \int_{\phi(\Omega)} f d\lambda^m = \int_{\Omega} f \circ \phi |\det J_\phi| d\lambda^m $$

where $\Omega\subset\Re^m$, $\lambda^m$ denotes the $m$-dimensional Lebesgue measure, and $J_\phi$ denotes the Jacobian of $\phi$. Is it possible to replace $\lambda^m$ with a generic measure and, if so, is there a good reference for the proof? I'm also curious if a similar formula holds in infinite dimensions.

  • $\begingroup$ Maybe for measures that are absolutely continuous wrt the Lebesgue measure? $\endgroup$
    – copper.hat
    Jun 1, 2012 at 6:01
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    $\begingroup$ That's a good thought. Certainly, Radon-Nikodym could be used to generalize it to additional measures. Nonetheless, I'm still curious if there's something intrinsic to the Lebesgue measure that's required for the formula to hold. $\endgroup$
    – Oyqcb
    Jun 1, 2012 at 6:07
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    $\begingroup$ I suspect an intimate connection between the Lebesgue measure and the determinant. I would imagine any significant generalization (ie, beyond invoking Radon-Nikoydm) would need to concoct an appropriate equivalent of a determinant for the measure in question. $\endgroup$
    – copper.hat
    Jun 1, 2012 at 6:10
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    $\begingroup$ It is the general linear group action on $\mathbb{R}^n$ and the homogenity of $\mathbb{R}^n$ which makes that case so special. You may want to have a look at the Haar measure. $\endgroup$
    – user20266
    Jun 1, 2012 at 6:30
  • $\begingroup$ You may also want to have a look at Hausdorff measure, area and coarea formula. $\endgroup$
    – user20266
    Jun 1, 2012 at 6:41

5 Answers 5


Given a measure space $(X_1,M_1,\mu)$ and a measureable space $(X_2,M_2)$ you can define the pushforward measure on $M_2$ of $\mu$ by a measurable function $F:X_1\to X_2$ to be $F\mu(E)=\mu(F^{-1}(E))$. Then you have the formula

$$\int_{X_2}g\;\mathrm{d}F\mu=\int_{X_1}g\circ F\;\mathrm{d}\mu$$

which is effectively the change of variables between the measure spaces $(X_1,M_1,\mu)$ and $(X_2,M_2,F\mu)$. The change of variables with Lebesgue measure should then a special case of this (the pushforward of $|\mathrm{det} DF|\lambda$ under $F$ is $\lambda$).

  • 8
    $\begingroup$ where can I find this formula and its proof? Please help! $\endgroup$
    – le4m
    Jun 9, 2015 at 17:59
  • $\begingroup$ I'll see if I can find one, but I would think pretty much any book on measure theory. $\endgroup$
    – Dom
    Jun 17, 2015 at 2:18
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    $\begingroup$ In that last statement, don't you mean that the pushfoward of $|\det DF|\lambda$ is $\lambda$? Where are the $\mathrm{d}$'s coming from? $\endgroup$ Jul 6, 2017 at 16:01
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    $\begingroup$ what is $g$ here? $\endgroup$
    – user56834
    Feb 1, 2018 at 22:31
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    $\begingroup$ @jachilles I've found this formula as Theorem 3.6.1 in Bogachev's "Measure Theory". It's also on this Wikipedia page, which also cites Bogachev. $\endgroup$
    – Sambo
    Feb 3, 2020 at 20:46

If $(X_1,F_1)$ and $(X_2,F_2)$ are two measurable space,

$$T:X_1\to X_2$$

is a measurable function, and


is a measure on $X_1$, then the mesaure

$$\mu \circ T^{-1}:F_2\to [0,\infty]$$

defined by

$$\mu\circ T^{-1}(B)=\mu(T^{-1}(B))\qquad B\in F_2$$

is a measure on $X_2$. A measurable function $f$ on $X_2$ is integrable with respect to $\mu\circ T^{-1}$ iff $f\circ T$ is integrable with respect to $\mu$ and we have

$$\int_{X_2}f\;d(\mu\circ T^{-1})=\int_{X_1}f\circ T\;d\mu. \tag{1}$$

If $T$ is one-to-one, we have for any $A\in F_1$:

$$\int_{T(A)}f\;d(\mu\circ T^{-1})=\int_{A}f\circ T\;d\mu \tag{2}$$

Proof of (1)

It is clear that $\mu\circ T^{-1}$ is measure on $X_2$. For proof of (1) it suffices to show it holds for any nonnegative real valued function $f$ and so it is proved for every real measurable function since:


Equation (1) is proved for any complex measurable function $(f=f_{Re}+i\,f_{Im})$.

Let $f$ be a nonnegative real-valued measurable function. Then there exists a sequence of simple measurable functions $\phi_n=\sum_{i=1}^{m_n} c_{n,i}\,\chi_{B_{n,i}}$ such that

$$\phi_n\nearrow f$$

and since the integral of left side of (1) equals to:

$$ \begin{align*} &\int_{X_2}f\;\;d(\mu\circ T^{-1}) \\ &=\lim_{n\to \infty}\int_{X_2}\phi_n \;d(\mu\circ T^{-1}) \\ &=\lim_{n\to \infty}\int_{X_2}\sum_{i=1}^{m_n}c_{n,i}\,\chi_{B_{n,i}} \;d(\mu\circ T^{-1}) \\ &=\lim_{n\to\infty}\sum_{i=1}^{m_n}c_{n,i}\, \int_{X_2}\chi_{B_{n,i}} \;d(\mu\circ T^{-1})\\ &=\lim_{n\to\infty}\sum_{i=1}^{m_n}c_{n,i}\;\mu(T^{-1}(B_{n,i}))\\ &=\lim_{n\to\infty}\sum_{i=1}^{m_n}c_{n,i}\;\int_{X_1}\chi_{T^{-1}(B_{n,i})}\;d\mu\\ &=\lim_{n\to\infty}\sum_{i=1}^{m_n}c_{n,i}\;\int_{X_1}\chi_{B_{n,i}}\circ T\;\;d\mu \\ &=\lim_{n\to \infty}\int_{X_1}\left[\sum_{i=1}^{m_n}c_{n,i}\,\chi_{B_{n,i}}\right]\circ T\;\; d\mu \\ &=\lim_{n\to \infty}\int_{X_1}\phi_n\circ T\;\; d\mu \\ &=\int_{X_1}f\circ T\;\; d\mu. \end{align*} $$

The last paragraph is established because $\phi_n\circ T$ is a simple measurable function such that

$$\phi_n\circ T\;\nearrow\; f\circ T.$$

Proof of (2)

Now for proof of (2), we have:

$$ \begin{align*} \int_{T(A)}f\;\;d(\mu\circ T^{-1}) &=\int_{X_2}f\;\chi_{T(A)}\;\;d(\mu\circ T^{-1}) \\ &=\int_{X_1}(f\;\chi_{T(A)})\circ T\quad d\mu \\ &\color{magenta}{=}\int_{X_1}(f\circ T)\,.\,\chi_{A}\quad d\mu \\ &=\int_{A}f\circ T\quad d\mu. \end{align*} $$

The equality in pink holds because

$$ \begin{align*} &(f\;\chi_{T(A)})\circ T\,(x) \\ &= f(T(x))\;.\chi_{T(A)}(T(x)) \\ &= \begin{cases} f(T(x)) & \text{ $T(x)\in T(A)$} \\ 0 & \text{ $T(x)\notin T(A)$} \end{cases} \\ &\color{red}{=}\begin{cases} f(T(x)) & \text{ $x\in A$} \\ 0 & \text{ $x\notin A$} \end{cases} \\ &=[\,(f\circ T)\,.\,\chi_A\,](x) \end{align*} $$

where the equality in red holds because $T$ is one to one.


Also you can have look on V.I. Bogachev. "Measure Theory."

In the case you are interested in probability theory, see R. Durrett, "Probability: Theory and Examples", 4th ed, 2010, pp 30-31.

  • $\begingroup$ It's called image measure, so together with change of variable, the measure is also changed. Interesting thing happens when you need to compute the integral, because you have to change back to Lebesgue integral. In this case, Jacobian comes out again. Can you derive the Jacobian in image measure settings? It's not included in Durrett's text perhaps because it is a measure theoretic problem per se. $\endgroup$ Jun 29, 2020 at 11:43

My upcoming book "Measure-Theoretic Calculus in Abstract Spaces: on the playground of infinite-dimensional spaces"$^\dagger$ has a proof for this.

Essentially, the usual Change of Variable formula involves two parts. First, transfer the integral to the second measurable space with the induced measure. Then, on the second measurable space, if one has a desired measure to integrate w.r.t, then use Radon-Nikodym Theorem to convert the integral w.r.t the induced measure to an integral w.r.t the desired measure.


$\dagger$ The book has been released. Link to the (Springer) bio.

  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ Oct 27, 2023 at 6:00
  • $\begingroup$ @CraniumClamp There is no link. $\endgroup$ Oct 27, 2023 at 6:31
  • $\begingroup$ @AnneBauval A link needn't be an html hyperlink. A reference to a particular text is a link (particularly in terms of the limited number of "canned" responses that one is able to post with a delete vote). You are free to argue that the answer here is more than "just" a link/reference, but you should understand the word "link" more broadly. $\endgroup$
    – Xander Henderson
    Dec 8, 2023 at 2:46
  • $\begingroup$ @XanderHenderson "Link-only answers can become invalid if the linked page changes" cannot refer to a book. When none of the "canned responses" is appropriate, one can always vote to delete with "no comment needed", and post a personal comment separately. $\endgroup$ Dec 8, 2023 at 6:26
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    $\begingroup$ @AnneBauval Yes, but books go out of print, journal articles are hard to access for people without educational access to those journals, etc. The canned response is not perfect, but it is not entirely wrong. Math SE is not seeking to be a list of bibliographic resources, but rather, to be a self-contained resource. The canned responses are based on the needs of SO, and there isn't much we can do about them. But I do think that the "link only" response is better than none at all, because of the bit about including essential bits of the argument. $\endgroup$
    – Xander Henderson
    Dec 8, 2023 at 13:07

please have a look at the monograph by Patric Muldowney theory of Random variation John Wiley and sons. it suggests a formuala and proves using Henstock-kurzweil apparoach


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