Change of variable formula for a generic measure applied to classical change of variable formula.

Question

I was reading this interesting post about changing the variables in an integral with a generic measure. I was wondering how this applies to the standard change of variable. In other words, $$\int_{F(\Omega)} f d\lambda = \int_{\Omega} f \circ F |\det DF| d\lambda $$ where $d\lambda$ is the lebesgue measure.

I think I have to show that $F_{*}(|det DF|\lambda)=\lambda$ (where $F_*$ is the pushforward of measures). In other words, for every $B$ measurable I need to show $|det DF|\lambda(F^{-1}(B))=\lambda(B)$ but I am not sure how to continue.

Answer 1

First, let us consider a special case rewriting of the theorem presented in your link.

Let $(X,\mathcal{M})$ be a measurable space, $(Y,\mathcal{N},\nu)$ a measure space, $F:X\to Y$ measurable map with measurable inverse. Also, let $F^*\nu:=(F^{-1})_{*}\nu$ be the pullback measure. Then, for any measurable function $f:Y\to [0,\infty]$, and any measurable subset $A\subset X$, we have \begin{align} \int_{F[A]}f\,d\nu&=\int_AF^*(f\,d\nu):=\int_Af\circ F\,d(F^*\nu) \end{align}

I leave it to you to see how this follows from the version in the link. I'm also going to assume you know the following measure-theory results

1. If $\phi:B\subset\Bbb{R}^n\to\Bbb{R}^m$, with $n\leq m$ is locally Lipschitz then it sends (Lebesgue) measure zero sets to (Lebesgue) measure zero sets.
2. If $T:\Bbb{R}^n\to\Bbb{R}^n$ is a linear transformation, then for every Lebesgue-measurable set $A\subset\Bbb{R}^n$, we have $\lambda(T(A))=|\det T|\lambda(A)$.
3. The Radon-Nikodym theorem
4. Lebesgue's differentiation theorem

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Our hypothesis is that $F:\Omega\to F[\Omega]\subset\Bbb{R}^n$ is a $C^1$ diffeomorphism between open subsets of $\Bbb{R}^n$, and $f:F[\Omega]\to[0,\infty]$ is Lebesgue measurable.

Then, because $F$ and its inverse are measurable, and I claim $F^*\lambda \ll \lambda$, because for any Lebesgue measurable subset $A\subset \Omega$, if $\lambda(A)=0$, then \begin{align} (F^*\lambda)(A):=((F^{-1})_{*}\lambda)(A)=\lambda(F(A))=0, \end{align} because $F$ being $C^1$ means it is locally Lipschitz, and thus sends measure zero sets to measure zero sets. So, by the abstract change of variables theorem, and the Radon-Nikodym theorem, we have \begin{align} \int_{F[\Omega]}f\,d\lambda&=\int_{\Omega}(f\circ F)\cdot d(F^*\lambda)\tag{COV}\\ &=\int_{\Omega}(f\circ F)\cdot \frac{d(F^*\lambda)}{d\lambda}\,d\lambda\tag{Radon-Nikodym} \end{align} By Lebesgue's differentiation theorem, for $\lambda$-a.e $x\in \Omega$, we have \begin{align} \frac{d(F^*\lambda)}{d\lambda}(x)&=\lim_{r\to 0^+}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}\frac{d(F^*\lambda)}{d\lambda}\,d\lambda\\ &=\lim_{r\to 0^+}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}d(F^*\lambda)\\ &=\lim_{r\to 0^+}\frac{(F^*\lambda)(B(x,r))}{\lambda(B(x,r))}\\ &=\lim_{r\to 0^+}\frac{\lambda(F(B(x,r)))}{\lambda(B(x,r))} \end{align} Our job is thus to calculate this ratio. So, fix a point $x\in \Omega$, and observe that by writing $F=DF_x\circ \underbrace{(DF_x)^{-1}\circ F}_{:=\phi}$, and using point (2) above, we have \begin{align} \frac{\lambda(F(B(x,r)))}{\lambda(B(x,r))}=|\det DF_x|\frac{\lambda(\phi(B(x,r)))}{\lambda(B(x,r))} \end{align} Now, observe that by the chain rule, $D\phi_x=DF_x^{-1}\circ DF_x=\text{id}_{\Bbb{R}^n}$. Thus, our entire problem has been reduced to proving the following lemma (which is interesting in its own right):

Let $\phi:\Omega\to\Bbb{R}^n$ be a $C^1$ function, where $\Omega\subset\Bbb{R}^n$ is open, such that at a point $x\in \Omega$, we have $D\phi_x=\text{id}_{\Bbb{R}^n}$. Then, $\lim\limits_{r\to 0^+}\frac{\lambda(\phi(B(x,r)))}{\lambda(B(x,r))}=1$.

For the proof, fix $0<\epsilon<1$. Then, by definition of $D\phi_x=\text{id}_{\Bbb{R}^n}$, there exists a $\delta_1>0$ such that if $\|h\|\leq \delta_1$ then \begin{align} \|\phi(x+h)-\phi(x)-h\|\leq \epsilon\|h\| \end{align} So (triangle inequality), for any $r<\delta_1$, we have $\phi(B(x,r))\subset B(\phi(x),(1+\epsilon)r)$.

Next, by the inverse function theorem, $\phi$ is a local $C^1$ diffeomorphism with $D(\phi^{-1})_{\phi(x)}=\text{id}_{\Bbb{R}^n}$ as well, so we can apply the same reasoning as above to deduce there exists $\delta_2>0$ such that for any $r<\delta_2$, we have $\phi^{-1}(B(\phi(x),r))\subset B(x,(1+\epsilon),r)$, or equivalently, $B(\phi(x),r)\subset \phi(B(x,(1+\epsilon)r))$. Thus, if $0<r<\min(\delta_1,\delta_2)$ then \begin{align} B(\phi(x),(1-\epsilon)r)&\subset \phi\bigg(B(x, (1+\epsilon)(1-\epsilon)r)\bigg)\\ &\subset\phi(B(x,r))\\ &\subset B(\phi(x),(1+\epsilon)r) \end{align} Thus, taking measures and dividing by the measure of $B(x,r)$, we get the inequality \begin{align} \frac{\lambda(B(\phi(x),(1-\epsilon)r))}{\lambda(B(x,r))}\leq \frac{\lambda(\phi(B(x,r)))}{\lambda(B(x,r))} \leq \frac{\lambda(B(\phi(x),(1+\epsilon)r))}{\lambda(B(x,r))} \end{align} Now, recall that the Lebesgue measure is translation-invariant and that the measure of balls scales as the $n^{th}$ power of the radius. Thus, \begin{align} (1-\epsilon)^n\leq \frac{\lambda(\phi(B(x,r)))}{\lambda(B(x,r))} \leq (1+\epsilon)^n \end{align} Since $0<\epsilon<1$ was arbitrary, this shows $\lim\limits_{r\to 0^+}\frac{\lambda(\phi(B(x,r)))}{\lambda(B(x,r))}$ exists and equals $1$. Thus, the entire proof is complete.

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To recap:

1. We first invoked the general change of variables theorem.
2. Next, we used the Radon-Nikodym theorem to express the integral with respect to $F^*\lambda$ as an integral with respect to $\lambda$, so our problem is now to calculate the Radon-Nikodym derivative $\frac{d(F^*\lambda)}{d\lambda}$.
3. Lebesgue's differentiation theorem tells us that the Radon-Nikodym derivative can actually be calculated as a limit of a quotient of the two measures applied to balls (or really any other nicely shrinking set).
4. We then used the fact that a linear transformation $T$ (in our case $DF_x$ for a fixed $x\in\Omega$) distort Lebesgue measure by a factor of $|\det T|$, in order to reduce to the case where $DF_x=\text{id}_{\Bbb{R}^n}$.
5. Finally, we used that lemma to show the limit is $1$.