I assume that the integral using in this discussion is the Riemannian integral.
The Change of variable theorem states that :
Let $U\subset\mathbb R^n$ be an open, rectifiable set. Let $\varphi:\ U\longrightarrow\varphi(U)\subset\mathbb R^n$ be a $C^1$-diffeomorphism and $f:\ \varphi(U)\longrightarrow\mathbb R$ be integrable (i don't know if it is assumed to be continuous). Then $(f\circ\varphi)\big|\det J_\varphi\big| $ is integrable over $U$, and $$\intop_{\varphi(U)}f=\intop_U(f\circ\varphi)\big|\det J_\varphi\big|. $$
So to be able to apply this theorem, one need to have the map $\varphi$ to be diffeomorphic. But then, why I saw so many people apply this theorem for the cases where $U$ and $\varphi(U)$ are not diffeomorphic ? For instance, when calculate the volume of the 2-torus, which is parametrized by \begin{align} \varphi:\ \ [0,a]\times [0,2\pi]\times[0,2\pi]\,&\longrightarrow\,\mathbb R^3 \\ (r,u,v)\,&\longmapsto\,\varphi(r,u,v)\,=\,\begin{bmatrix} \big(R+r\cos v\big)\cos u \\ \big(R+r\cos v\big)\sin u \\ r\sin v \end{bmatrix} \end{align} Here, $\varphi$ is clearly can't be a diffeomorphism from $[0,a]\times [0,2\pi]\times[0,2\pi]$ to the solid torus, even when only the interior of both is considered, since we know that a 3-ball will never be diffeomorphic to a solid torus which has a hole. But somehow, people keep applying the Change of variable theorem for $\varphi$ to evaluate the volume of torus.
Could anyone enlight me this issue ? Thank you
tl;dr: Conclusions of the change of variables theorem remain true if the hypotheses hold up to sets of measure zero.
$\newcommand{\eps}{\varepsilon}$A complete general explanation is liable to be lengthy. Relevant search terms for the curious and motivated include partitions of unity and geometric measure theory.
Here, the important (sufficient) points are that
Consequently, one way to make things rigorous here is to "shrink" the cube by an arbitrarily real $\eps$ with $0 < \eps < \min(a, \pi)$. That is, restrict to $$ [\eps, a - \eps] \times [\eps, 2\pi - \eps] \times [\eps, 2\pi - \eps], $$ integrate using change of variables, and take the limit as $\eps \to 0$.
Because the limiting value of the integral agrees with the formal result of integrating over the closed box, it's customary to pretend the change of variables theorem applies even if Strictly Speaking a bit more care was required.
Since we're on the subject, it would be a problem (i.e., could change the value of the integral) if our box were, say, $$ [0, a] \times [0, 4\pi] \times [-2\pi, 2\pi], $$ or $$ [0, a] \times [0, 2\pi + 10^{-12}] \times [0, 2\pi]. $$ In either case, we'd be multiply-covering part of the image in a way that removing sets of measure zero doesn't repair.