change of variable formula

Question

Let $\chi:\Omega \to \Bbb{R}^n$ be smooth function where $\Omega \subset \Bbb{R}^n$, assume the derivatie $D\chi$ being negative definite matrix.Is possible to let the following change of variable formula holds?for $S$ a bounded set and $g\in L^{1}_{loc}(\Bbb{R}^n)$. If $\chi$ is only negative semidefinite does this hold?

$$\int_{\chi(S)} g(x)dx = \int_{S}g(\chi(y)) |\det(D\chi)|dy$$

Typically I assume $\chi$ be a diffeomorphism, however, I recently came across this version of change of variable formula when proving the Alexandroff maximum principle, I don't know if this type of change of variable formula holds, is there some reference? I can't find this on wiki.

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As shown in the post below, the condition that $\chi$ is "injective" is necessary to let the change of variable formula hold.

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Althong if $\chi$ for arbitary smooth function this version change of variable formula does not hold.If we assume or restrict the domain to be convex, and If $\chi = Du$ is derivative of some function $u:\Omega \to \Bbb{R}$, then under the assumption $D\chi = D^2 u <0$ we can prove $D\chi$ is in fact a injective function, therefore the change of variable formula holds .(that is we add two additional condition: first assume $\chi = Du$ for some $u$, second we assume or restrict $\Omega$ be the convex set.)

To see this since $D^2 u <0$ we deduce that $u$ is concave, therefore

$$u(x) \le u(y) + Du(y)\cdot(y-x)\\ u(y) \le u(x) + Du(x)\cdot(x -y)$$

If we assume $Du(x) = Du(y)$ then we deduce that $u(x) = u(y) + Du(y)\cdot(y-x)$ therefore:

$$u(y) = u(x) + Du(x)\cdot (y- x) + \int_0^1t<D^2u(x + (1-t) y )(y-x),(y-x)> dt$$.(where we use the convexity of the domain $\Omega$)

therefore the integrand must be zero everywhere,which implies $x = y$.

Answer 1

Here's the 'vanilla' version of the theorem (which is still a pain to prove):

Let $\Omega\subset\Bbb{R}^n$ be open and $\chi:\Omega\to\chi[\Omega]\subset\Bbb{R}^n$ a $C^1$ diffeomorphism between open subsets of $\Bbb{R}^n$. Then, for any Lebesgue-measurable function $f:\chi[\Omega]\to[0,\infty]$, we have \begin{align} \int_{\chi[\Omega]}f(x)\,dx &=\int_{\Omega}(f\circ\chi)(y)\cdot |\det D\chi(y)|\,dy. \end{align}

By dealing with positive and negative parts, and real/complex parts, you can deal with the case of $f\in L^1(\chi[\Omega])$. Now, one can strengthen the theorem by weakening slightly the hypotheses on $\chi$ using Sard's theorem (see for example Sternberg's Lectures on Differential Geometry). With the aid of this theorem, we can prove:

Let $\Omega\subset\Bbb{R}^n$ be open, and $\chi:\Omega\to\Bbb{R}^n$ a $C^1$ map, and $C=\{x\in\Omega\,:\, \text{$D\chi(x)$ is not surjective}\}$ be the critical set of $\chi$, and suppose $\chi$ is injective on $\Omega\setminus C$. Then, for any Lebesgue-measurable function $f:\chi[\Omega]\to[0,\infty]$, we have \begin{align} \int_{\chi[\Omega]}f(x)\,dx &=\int_{\Omega}(f\circ\chi)(y)\cdot |\det D\chi(y)|\,dy. \end{align}

To prove this version, note that Sard's theorem tells us $\chi[C]$ has Lebesgue measure zero. Also, $C$ is a closed set (since it's the set of points where $\det D\chi(x)=0$, i.e the preimage of the closed set $\{0\}$ under a continuous function), and the restriction $\chi: \Omega\setminus C\to \chi[\Omega\setminus C]$ is a $C^1$ diffeomorphism here (by the inverse function theorem and bijectivity). So, the previous version of the theorem can be applied here. Now, we have: \begin{align} \int_{\chi[\Omega]}f(x)\,dx&=\int_{\chi[\Omega\setminus C]} f(x)\,dx\\ &=\int_{\Omega\setminus C}(f\circ \chi)(y)\cdot|\det D\chi(y)|\,dy\\ &=\int_{\Omega}(f\circ \chi)(y)\cdot|\det D\chi(y)|\,dy. \end{align} To emphasize again, the first equality is because $\chi[\Omega]\setminus\chi[C]\subset \chi[\Omega\setminus C]\subset\chi[\Omega]$ and $\chi[C]$ has Lebesgue measure zero. The second equality is because of the previous version of the theorem. The final equality is because on the set $C$, we have $\det D\chi(y)=0$, so the integrals are equal (you're just adding zero).

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Your statement with the set $S$, assuming it is measurable (boundedness is unnecessary), is a special case; just apply it to $f$ multiplied by an appropriate indicator function.

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Finally, if $\chi$ is not injective, then you have to modify the theorem. Either you have to take into account the 'multiplicity' somehow, or what is more commonly done, you just restrict the domain to where the map is injective, and apply the previous version of the theorem. To see an example where the theorem as stated fails without any form of injectivity, consider $\Omega= (0,1)\times (0,4\pi)\subset\Bbb{R}^2$, and $\chi:\Omega\to\Bbb{R}^2$ the polar coordinate map $\chi(r,\theta)=(r\cos\theta,r\sin\theta)$. Then, $\chi[\Omega]$ is the open unit disc minus the origin. Take $f=1$ the constant function. Then, $\int_{\chi[\Omega]}f(x)\,dx=\pi$, since it's the measure (i.e area) of $\chi[\Omega]$. On the other hand, $|\det D\chi(r,\theta)|=r$, so $\int_{\Omega}r\,d(r,\theta)=\int_0^{4\pi}\int_0^1r\,dr\,d\theta=2\pi$ by Fubini. So, we're getting twice the answer we expect, and this is precisely because our map is going over points twice (not injective).