Change of variable formula for the image of a hypercube Let $\varphi: \mathbb{R}^n\to \mathbb{R}^n$ be an injective $C^1$ map. Let $I=[0, 1]^n$. I want to show that $$m(\varphi(I))=\int_I \left|\det D\varphi(x)\right|dx.$$
This is a special case of the change of variable formula, but I think there is a simpler proof in this case (that I still can't find). Maybe one can compute $m(\varphi([0, \varepsilon]^n))$, with small $\varepsilon$, reducing to the one dimensional case and using Fubini.
What do you think?
 A: If $\varphi$ is $C^1$ and injective (not quite enough, since the derivative may be 0, otherwise the measure falls apart. You need $\varphi$ to be a diffeomorphism). We want to compute
$$ m(\varphi(I)) =\int_{\varphi(I)} dy $$
Since $\varphi$ is $C^1$ and injective, we have the inverse function theorem. I.e. we have $\varphi^{-1} : \mathbb{R}^n \to \mathbb{R}^n$. Thus it makes sense to consider $x = \varphi^{-1}(y)$. Since everything matches in one-to-one correspondence, we know that $\varphi^{-1}(\varphi(I))=I$, which means
$$m(\varphi(I)) =\int_{\varphi(I)} dy = \int_{I} d ( \varphi(x)) $$
Now we just have to understand what $d ( \varphi(x))$ means. This is the statement of the change of variables formula...i.e. we see this isn't any simpler to prove. In fact, it's the same proof since if you show if for a small cube, you can always build up more complicated regions in cubes. In term of differential forms we have
$$ dy = dy_1 \wedge \ldots \wedge dy_n 
\to d ( \varphi(x)) =d \varphi_1(x) \wedge \ldots d \varphi_n (x) = (\partial_{x_1}\varphi_1 dx_1 \wedge \ldots \wedge \partial _{x_n} \varphi_1  dx_n) \wedge \ldots \wedge((\partial_{x_1}\varphi_n dx_1 \wedge \ldots \wedge \partial _{x_n} \varphi_n  dx_n)$$
which works out to $|\det D \varphi | dx_1 \wedge \ldots dx_n = |\det D \varphi | dx$ when all is said and done. Otherwise you'll have to go by the the definition you're comfortable with.
