TITLE: Problem 4-25 from Spivak's Calculus on Manifolds
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: This barely fit in a comment:
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 
$$
\int_{[0,1]^k} f = \int_{(0,1)^k} f.
$$