# Problem 4-25 from Spivak's Calculus on Manifolds

I am reading through Spivak's Calculus on Manifolds and have come across a technicality in one of the problems that is annoying me. It is Problem 4-25, the statement of which is

Let $c$ be a singular $k$-cube and $p:[0,1]^k\to[0,1]^k$ be a 1-1 function such that $p([0,1]^k)=[0,1]^k$ and $\det p'(x)\geq 0$ for $x\in[0,1]^k$. If $\omega$ is a $k$-form, show that $$\int_c \omega = \int_{c\circ p} \omega.$$

I think Spivak intended this to be a straightforward exercise in the use of the change of variables theorem (his Theorem 3-13, which I copy below, making the change that Problem 3-39 allows, namely removing the assumption that $\det g'(x)\neq 0$).

3-13 Theorem. Let $A\subset\mathbf{R}^n$ be an open set and $p:A\to\mathbf{R}^n$ a 1-1, continuously differentiable function. If $f:p(A)\to\mathbf{R}$ is integrable, then $$\int_{p(A)} f = \int_A (f\circ p)\lvert\det p'\rvert.$$

Indeed if we unwind the definitions of integrating forms over cubes, Problem 4-25 is solved as long as we can say that $$\int_{[0,1]^k} (f\circ p)\lvert \det p' \rvert = \int_{[0,1]^k} f.$$ for a (say smooth) function $f:[0,1]^k\to\mathbf{R}$. There's one problem: $[0,1]^k$ is not open so I can't apply the theorem above directly. What I would like to do is replace the integrals over $[0,1]^k$ by integrals over $(0,1)^k$ but I don't really see how this will work. I don't know that the corresponding integrals would then be equal. I don't know that $p$ restricts to a bijection from $(0,1)^k$ to $(0,1)^k$. I don't know that $p((0,1)^k)$ is open because I can't use the inverse function theorem. Any way I try to approach this I seem to run into technical problems. I am happy to assume that everything is smooth.

Note: What it means in Spivak for a function to be differentiable on a non-open set is that the function extends to a differentiable function on an open set.

• You don't need $g((0,1)^k)$ to be open in order to use the theorem (though this is indeed true, but it's a hard theorem called the theorem of invariance of domain). The two integrals will be the same because their difference will be over a set of measure $0$ (the boundary of the cube). – i like xkcd Aug 31 '13 at 1:34
• I understand that $\int_{[0,1]^k} (f\circ p)\lvert\det p'\rvert=\int_{(0,1)^k} (f\circ p)\lvert\det p'\rvert$ and $\int_{[0,1]^k} f = \int_{(0,1)^k} f$. But then why is it true that $\int_{(0,1)^k} (f\circ p)\lvert\det p'\rvert = \int_{(0,1)^k} f$? – frakbak Aug 31 '13 at 1:51

You are correct, I should have made that more precise. The thing is that, by the theorem you have $$\int_{(0,1)^k} (f \circ p )|\det p'| = \int_{g((0,1)^k)} f.$$ But observe that $p((0,1)^k)) = p([0,1]^k \setminus \partial[0,1]^k)$. And since $p$ is $1-1$ and onto, this is $$p((0,1)^k) = p([0,1]^k) \setminus p(\partial[0,1]^k).$$ What you are subtracting has measure $0$, so you can just write $$\int_{(0,1)^k} (f \circ p )|\det p'| = \int_{[0,1]^k} f,$$ and now you can complete the proof using that $$\int_{[0,1]^k} f = \int_{(0,1)^k} f.$$
• It seems you are using the fact that $p((0,1)^k)$ is open, then, to apply the theorem? So we do need invariance of domain. – frakbak Aug 31 '13 at 2:45
• @frakbak Where am I using that $p((0,1)^k) is open? I am just applying Theorem 3-13 to the open set$(0,1)^k$... right? – i like xkcd Aug 31 '13 at 3:04 • Ah. Yes, you are applying the theorem as I stated it. I included in the theorem the revision made by Problem 3-39, which removed the assumption that$\det p'>0$everywhere. In the original theorem you got that$p(A)$was open for free because of the inverse function theorem. But with the possibility that$\det p'=0$you no longer get that for free. When I was thinking about Problem 3-39 I might have included the openness of$p(A)\$ as an extra hypothesis, so I probably should have included it in my statement above. But it follows from invariance of domain...which as you say is a hard theorem. – frakbak Aug 31 '13 at 3:12