If $f: \mathbb{R}^n \longrightarrow \mathbb{R}^n$ is isometric in a $C \subset \mathbb{R}^n$ limited subset which $\partial C$ has measure null, then $vol \ f(C) = vol (C)$.
$\textbf{My attempt:}$
I apply the Theorem of Change of Variables to obtain that $\int_{f(C)} g = > \int_C (g \circ f) |det \ f'|$ and, assuming that $g = \chi_{f(C)}$, we have $\int_{f(C)} 1 = \int_C 1 \ |det \ f'|$, i.e., $vol(f(C)) = \int_C 1 \ |det \ f'|$.
My doubts about this attempt are:
$\textbf{1.}$ I need to show that $f: int(C) \longrightarrow f(int(C))$ is a $C^1-$ isomorphism to apply the Theorem of Change of Variables, but I don't know how I can show this.
$\textbf{2.}$ I need to show that $|det f'(x)| = 1$ for each point $x$ in $C$, but I don't know how I can show this.
I think that I must use the fact that $||f(x) - f(y)|| = ||x - y||$ for each point $x, y \in C$ to solve my doubts. Someone can help me? Thanks in advance!
EDIT :
I know that alinear transformation $A$ is orthogonal when $A^T = A^{-1}$. If $f'$ is an orthogonal linear transformation, then $det \ f' = det \ f'^T = det (f')^{-1}$. Since $det f' \circ (f')^{-1} = det f * det f'^{-1}$, $det f' \circ (f')^{-1} = det Id$ and $det \ f' = det f'^T$, we have $|det f'| = 1$. I thought about this, but I don't see how the fact of $f$ be isometric, imply that $f'$ is an orthogonal matrice.
Isometries from $\mathbb{R}^n \to \mathbb{R}^n$ have determinant $\pm1$. This is due to the fact that the set of orthogonal matrices form the isometry group of $\mathbb{R}^n$. Hence, since $D_pf$ is a linear map, the underlying matrix or jacobian $J$ is orthogonal i.e $JJ^T = I$ and so $\textbf{det}(J) ^2= 1$.