# The Unapologetic Mathematician

## Change of Variables in Multiple Integrals II

Okay, let’s get to actually proving the change of variables formula for multiple integrals. To be explicit:

Let $g=(g^1,\dots,g^n)$ be a continuously differentiable function defined on an open region $X\subseteq\mathbb{R}^n$. Further, assume that $g$ is injective and that the Jacobian determinant $J_g$ is everywhere nonzero on $S$. The inverse function theorem tells us that we can define a continuously differentiable inverse $g^{-1}$ on all of the image $g(S)$.

Further, let $X$ be a Jordan measurable subset of $g(S)$, and let $f$ be defined and continuous on $X$. Then we have the change of variables formula

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

We will proceed by induction on the dimension $n$. For $n=1$, this is exactly the one-dimensional change of variables formula, which we already know to be true. And so we’ll assume that it holds in all $n-1$-dimensional cases, and prove that it then holds for $n$-dimensional cases as well.

Since the Jacobian determinant $J_g(v)$ is nonzero, we cannot have $\left[D_ng^k\right](v)=0$ for all $k$; at least one of the components must have a nonzero partial derivative with respect to $v^n$ at any given point $v\in S$. Let’s say, to be definite, that $\left[D_ng^n\right](v)\neq0$. We will now (locally) factor $g$ into the composite of two functions $\theta$ and $\phi$, which will each have their own useful properties. First, we will define

$\displaystyle\phi(u)=(u^1,\dots,u^{n-1},g^n(u))$

This is clearly continuously differentiable, and it’s even injective on some neighborhood of $v$, by our assumption that $\left[D_ng^n\right](v)\neq0$. Further, the Jacobian determinant $J_\phi$ is exactly the partial derivative $D_ng^n$, and so the inverse function theorem tells us that in some neighborhood of $v$ we have a local inverse $\psi$, with $\psi(\phi(u))=u$ in some neighborhood of $v$. We can now define

$\displaystyle\theta^k(t)=g^k(t^1,\dots,t^{n-1},\psi^n(t))$

for $1\leq k\leq n-1$, and define

$\displaystyle\theta^n(t)=t^n$

Then for each $u$ in a small enough neighborhood of $v$ we have $\theta(\phi(u))=g(u)$. The first function $\phi$ leaves all components of $u$ fixed except for $u^n$, while the second function $\theta$ leaves $u^n$ fixed. Of course, if we used a different partial derivative $D_ng^k$, we could do the same thing, replacing $u^k$ instead with $g^k$ in $\phi$, and so on.

Now if $X$ is a Jordan measurable compact subset of $g(S)$, then its inverse image $g^{-1}(X)$ will also be compact since $g^{-1}$ is continuous. For every point in $g^{-1}(X)\subseteq S$, we can find some neighborhood — which we can take to be an $n$-dimensional interval — and a factorization of $g$ into two functions as above. As we move around, these neighborhoods form an open cover of $g^{-1}(X)$. And since $g^{-1}(X)$ is compact, we can take an open subcover.

That is, we can cover $g^{-1}(X)$ by a finite collection of open intervals $A_{(k)}$, and within each one we can write $g(u)=\theta_{(k)}(\phi_{(k)}(u))$, where the function $\phi_{(k)}$ leaves all the components of $u$ fixed except for the last, while $\theta_{(k)}$ leaves that last one fixed. By subdividing these intervals, we can assume that they’re nonoverlapping. Of course, if we subdivided into open sets we’d miss the shared boundary between two subintervals. So we’ll include that boundary in each subinterval and still have $p$ nonoverlapping intervals $A_{(k)}$.

Then we can define $T_{(k)}=g(A_{(k)})$. Since $g$ is injective, these regions will also be nonoverlapping. And they’ll cover $X$, just as $A_{(k)}$ covered $g^{-1}(X)$. So we can define $F(x)=f(x)\chi_X(x)$ and write

$\displaystyle\int\limits_Xf(x^1,\dots,x^n)\,d(x^1,\dots,x^n)=\sum\limits_{k=1}^p\int\limits_{T_{(k)}}F(x^1,\dots,x^n)\,d(x^1,\dots,x^n)$

Our proof will thus be complete if we can show that the change of variables formula holds for these regions, and we will pick up with this final step next time.

January 6, 2010 - Posted by | Analysis, Calculus

1. […] far, we’ve shown that we can chop up into a collection of nonoverlapping regions and into their preimages . […]

Pingback by Change of Variables in Multiple Integrals III « The Unapologetic Mathematician | January 7, 2010 | Reply

2. Don’t you mean $T_{(k)}=g(A_{(k)})$?

Comment by Relheun | January 25, 2010 | Reply

3. Yes, thanks. Fixed.

Comment by John Armstrong | January 25, 2010 | Reply

4. “replacing $u^n$ instead with $D_ng^k$ in $\phi$, and so on.” Did you mean $g^k$ ? Also, it would be nice if you could explain how to construct $\phi$ and $\theta$ in the case when k \neq n .
Thanks.

Comment by sswarnendu | September 9, 2014 | Reply

5. I think that’s correct; it may even be clearer to replace $u^k$ with $g^k(u)$, though replacing $u^n$ still works. Now we have $\phi(u)=\left(u^1,\dots,u^{k-1},g^k(u),u^{k+1},\dots,u^n\right)$. The construction of $\theta$ from this point should also be clearer.

Comment by John Armstrong | September 9, 2014 | Reply

6. Thanks a lot….

Comment by sswarnendu | September 10, 2014 | Reply