Change of Variables in Multiple Integrals III

Today we finish up the proof of the change of variables formula for multiple integrals:

$\displaystyle\int\limits_Xf(x^1,\dots,x^n)\,d(x^1,\dots,x^n)=\int\limits_{g^{-1}(X)}f(g(u^1,\dots,u^n))\left\lvert\frac{\partial(x^1,\dots,x^n)}{\partial(u^1,\dots,u^n)}\right\rvert\,d(u^1,\dots,u^n)$

So far, we’ve shown that we can chop $X$ up into a collection of nonoverlapping regions $T_{(k)}$ and $g^{-1}(X)$ into their preimages $A_{(k)}=g^{-1}(T_{(k)})$ . Further, within each $A_{(k)}$ we can factor $g(u)=\theta_{(k)}(\phi_{(k)}(u))$ , where $\phi_{(k)}$ fixes each component of $u$ except the last, and $\theta_{(k)}$ fixes that one. If we can show the formula holds for each such region, then it will hold for arbitrary (compact, Jordan measurable) $X$ .

From here we’ll just drop the subscripts to simplify our notation, since we’re just concerned with one of these regions at a time. We’ve got $T$ and its preimage $A=g^{-1}(T)$ . We’ll also define $B=\phi(A)$ , so that $T=\theta(B)$ . For each real $\xi$ we define

$\displaystyle\begin{aligned}T(\xi)&=\{(x^1,\dots,x^{n-1}\vert(x^1,\dots,x^{n-1},\xi)\in T\}\\B(\xi)&=\{(t^1,\dots,t^{n-1})\vert(t^1,\dots,t^{n-1},\xi)\in B\}\end{aligned}$

Then $(T(\xi),\xi)=\theta(B(\xi),\xi)$ , since $\theta$ preserves the last component of the vector. We also define

$\displaystyle\begin{aligned}c&=\inf\{\phi^n(u)\vert u\in A\}\\d&=\sup\{\phi^n(u)\vert u\in A\}\end{aligned}$

The lowest and highest points in $T$ along the $n$ th coordinate direction. Now we can again define $F(x)=f(x)\chi_T(x)$ and set up the iterated integral

$\displaystyle\int\limits_TF(x)dx=\int\limits_c^d\int\limits_{T(x^n)}F(x^1,\dots,x^{n-1},x^n)\,d(x^1,\dots,x^{n-1})\,dx^n$

We can apply the inductive hypothesis to the inner integral using $x=\theta(t)$ , which only involves the first $n-1$ coordinates anyway. If we also rename $x^n$ to $t^n$ , this gives

$\displaystyle\int\limits_TF(x)dx=\int\limits_c^d\int\limits_{B(t^n)}F(\theta(t^1,\dots,t^{n-1},t^n))\lvert J_\theta(t)\rvert\,d(t^1,\dots,t^{n-1})\,dt^n$

Which effectively integrates as $t$ runs over $B=\phi(A)$ . But now we see that $B(t^n)$ lies within the projection $A_n$ of $A$ , as we defined when we first discussed iterated integrals. We want to swap the order of integration here, so we have to rewrite the limits. To this end, we write $A=[a^1,b^1]\times\dots\times[a^n,b^n]$ , $A_n=[a^1,b^1]\times\dots\times[a^{n-1},b^{n-1}]$ , and define

$B^*(u^1,\dots,u^{n-1})=\{\phi^n(u^1,\dots,u^{n-1},u^n)\vert a^n\leq u^n\leq b^n\}$

which runs over the part of $B$ above some fixed point in $A_n$ . Then we can reverse the order of integration to write

$\displaystyle\int\limits_TF(x)dx=\int\limits_{A_n}\int\limits_{B^*(u^1,\dots,u^{n-1})}F(\theta(t^1,\dots,t^{n-1},t^n))\lvert J_\theta(t)\rvert\,dt^n\,d(t^1,\dots,t^{n-1})$

Now we can perform the one-dimensional change of variables on the inner integral and swap out the variables $u^1=t^1$ through $u^{n-1}=t^{n-1}$ to write

$\displaystyle\int\limits_TF(x)dx=\int\limits_{A_n}\int\limits_{a^n}^{b^n}F(\theta(\phi(u)))\lvert J_\theta(\phi(u))\rvert\lvert J_\phi(u)\rvert\,du^n\,d(u^1,\dots,u^{n-1})$

But now we recognize the product of the two Jacobian determinants as the Jacobian of the composition:

$\displaystyle J_\theta(\phi(u))J_\phi(u)=J_{\theta\circ\phi}(u)$

and so we can recombine the iterated integral into the $n$ -dimensional integral

$\displaystyle\int\limits_TF(x)dx=\int\limits_AF([\theta\circ\phi](u))\lvert J_{\theta\circ\phi}(u)\rvert\,du$

Finally, since $g=\theta\circ\phi$ we can replace the composition. We can also replace $F$ by $f$ since there’s no chance anymore for any evaluation of $f$ to go outside $T$ . We’re left with

$\displaystyle\int\limits_Tf(x)dx=\int\limits_{g^{-1}(T)}f(g(u))\lvert J_g(u)\rvert\,du$

Essentially, what we’ve shown is that we can always arrange to first change $n-1$ variables (which we handle by induction) and then by the last variable. The overall scaling factor is the $n-1$ -dimensional scaling factor from the first transformation times the one-dimensional scaling factor from the second, and this works because of how the Jacobian works with compositions.

This establishes the change of variables formula for regions within which we can write $g$ as the composition of two functions, one of which fixes all but the last coordinate, and the other of which fixes that one. Since we established that we can always cut up a compact, Jordan measurable set $X$ into a finite number of such pieces, this establishes the change of variables formula in general.

January 7, 2010 - Posted by John Armstrong | Analysis, Calculus

4 Comments »

John,

What are the assumptions on f? I missed them if you stated them.

Can the assumptions on g be relaxed?

–rich

Comment by rich | January 2, 2011 | Reply
$f$ should only need to be integrable, while $g$ needs to be differentiable (in the higher-dimensional sense) to even define the Jacobian determinant. I don’t think I had to strengthen that assumption, but it certainly can’t be weakened.

Comment by John Armstrong | January 2, 2011 | Reply
I think g differentiable is a sufficient condition. I wonder whether g can be differentiable in a some sort weak sense since all takes place under and integral.

Comment by Rich Lehoucq | January 4, 2011 | Reply
It wouldn’t entirely surprise me if there were some give to the results, but whatever improvement is possible is likely not general. That is, we can weaken one condition by strengthening some other condition. Analytic results often have complicated and fussy “boundaries”, especially as compared with algebraic or categorical ones.

Comment by John Armstrong | January 4, 2011 | Reply

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