Lebesgue measure is invariant under isometry Is it true that Lebesgue measure is invariant under isometric map? I mean standard measure of $\mathbb R^n$.
It is certainly true for interval in $\mathbb R$ (obvious). I've attempted to prove it in general by induction, and it seems that I succeed in the idea, but don't know how to prove it rigorously. (The idea is to prove that the measure only depends on the mutual distances, which are by definition are equal) Can you, please, give some hints or maybe even a solution to this problem, if it's true.
Thank you very much.
 A: Theorem. If $f\colon\mathbb R^n\to\mathbb R^n$ is a linear automorphism, and $\lambda$ is the Lebesgue measure, then
$$
    \lambda(f(B)) = |\det f|\lambda(B)\tag{1}
$$
for all Borel sets $B$ in $\mathbb R^n$.
Proof.
Decompose $f$ as a (finite) sequence of elementary operations of the following kinds

*

*Row swap: $f({\mathbf e}_i) \leftrightarrow f({\mathbf e}_j)$

*Row scaling: $f({\mathbf e}_i) \to \alpha f({\mathbf e}_i)$, $\alpha \ne 0$

*Row addition: $f({\mathbf e}_1) \to f({\mathbf e}_1) + f({\mathbf e}_2)$.


*

*Note the further restriction applied to kind 3 when compared with the usual classification.

Since the theorem is clear for operations of kinds 1 and 2, it suffices to be shown when $f$ is a row addition, i.e.,
\begin{align*}
    f\colon\mathbb R^2\times\mathbb R^{n-2}&\to\mathbb R^2\times\mathbb R^{n-2}\\
                                    (x,y,{\mathbf z})&\mapsto(x+y,y,{\mathbf z}),
\end{align*}
which takes us to the case $n=2$ with $f(x,y)=(x+y,y)$.
Now $(1)$ becomes equivalent to proving
$$
   \lambda(f([0,a)\times[0,b))) = ab  \tag{2}
$$
Suppose $a\ge b$ (the case $a < b$ follows from this one applied to $f^{-1}$). Here $f$ maps the rectangle of sides $a$ and $b$ onto the parallelogram of sides $a$ and $b$
\begin{align*}
   f([0,a)\times[0,b)) &= \{(x,y)\mid 0\le x < b,\ 0\le y \le x\}\\
        &\quad\cup [b,a)\times[0,b)\\
        &\quad\cup \{(x,y) \mid a\le x < a + b,\ x-a < y < b\},
\end{align*}
which indeed has $\lambda$-measure $ab$ because the union of the first triangle with the (horizontal) translation by $(-a,0)$ of the second is the square of side $b$.
