Is there a change of variables formula for a measure theoretic integral that does not use the Lebesgue measure

Question

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.

Answer 1

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 $\lambda$).

Answer 2

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

$$\mu:F_1\to[0,\infty]$$

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:

$$f=f^+-f^-.$$

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.

Answer 3

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.

Answer 4

My upcoming book "Measure-Theoretic Calculus in Abstract Spaces: on the playground of infinite-dimensional spaces" 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.

Answer 5

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