Proof collecting: A short proof of chage of variable formula

Question

I've been searching for a proof of the change-of-variable formula for multi-variable Riemann integration, but most of them seem to be too long and cumbersome. I'd like to see one proof that uses clever approximation skills or some advanced techniques, and of course,the shorter, the better. Thanks in advance :).

By the way, any proof that involves interesting ideas are also welcomed.

Answer 1

Change of formula reads \begin{align*} \int_{a}^{b}f(\varphi(x))\varphi'(x)dx=\int_{a}^{b}f(u)du, \end{align*} say, for $\varphi'>0$ sufficiently good.

1) Using integration by parts, one can do it for $f(x)=x^{n}$, $n=0,1,2,...$

2) Using Stone-Weierstrass, the formula holds for continuous functions $f$.

3) Using the fact that $C[a,b]$ is $L^{1}$ dense in $R[a,b]$, the space of Riemann integrable functions, the result follows.

Answer 2

Since, you asked for ideas, one interesting derivation of change of variables is from Radon-Nikodyn theorem about derivative of measures. First, let us state a precise formulation. Let $U \subset \mathbb{R}^n$ be an open set. If $f:U \to \mathbb{R}^n$ is continuously differentiable, 1-to-1, and $\det Df(x) \neq 0$ for any $x \in U$, then for any integrable $w:f(U) \to \mathbb{R}^n$, $$ \int_{f(U)} w(y) \, d\mathcal{L}^n(y) = \int_U w(f(x)) \, |\det Df(x)| d\mathcal{L}^n(x)\, . \qquad \text{(C.of.V. 1)} $$

Proof Sketch:

1. It suffices to prove C.of.V for characteristic functions of measurable sets, i.e. for any measurable $E\subset U$, $$ \mathcal{L}^n({f(E)}) = \int_E |\det Df(x)| \, d\mathcal{L}^n(x)\, . \qquad \text{(C.of.V. 2)} $$ Then the general case c.of.V. 1 follows from approximation by characteristic functions.

2. I will assume $U=\mathbb{R}^n$ for simplicity. Define a new measure on $\mathbb{R}^n$ by $$ \mu(E):=\mathcal{L}^n({f(E)}) \, . $$

3. Show stuff about this measure. Like it is well-defined, Borel, etc. You'll probably will also need to show $f(E)$ is Lebesgue measurable iff $E$ is.

3.5) Prove that $\mu$ is absolutely continuous w.r.t $\mathcal{L}^n$, i.e. if $\mathcal{L}^n({E})=0$ then $\mu(E)=0$.

4. Next compute the Radon-Nikodyn derivative: $$ \frac{d\mu}{d\mathcal{L}^n}(x) = \lim_{r \to 0} \frac{\mu(f(B(x,r))}{\mathcal{L}^n(B(x,r))} = |\det Df(x)| \, . $$ This follows from approximating $f(B(x,r))$ by $Df(x)(B(x,r))$. The approximation comes from differentiability of $f$, and gets arbitrarily close as $r\to 0$. since $Df(x)$ is a linear map, even before having $r\to 0$, we have $\mathcal{L}^n(Df(x)(B(x,r))) = |\det Df(x)|\mathcal{L}^n(B(x,r))$.
5. By Radon-Nikodyn, for every $\mathcal{L}^n$-measurable $E$, $$ \mu(E) = \int_E \frac{d\mu}{d\mathcal{L}^n}(x) \, d\mathcal{L}^n(x) = \int_E |\det Df(x)| \, d\mathcal{L}^n(x), $$ Which is just C.of.V 2.

Apparently this is done in Rudin's book, but I do not have access to them to verify this! Will hopefully update once I find out.

Note: As one matures in math, what is elementary, easy, natural, elegant, etc keeps changing. So, feel free to criticize this proof! :)