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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.

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    $\begingroup$ Hint: an isometry defined on a subset of $\mathbb R^n$ into $\mathbb R^n$ extends to an isometry of the whole space onto itself. Then it has the form of an orthogonal map followed by a translation. $\endgroup$ – GEdgar Nov 22 '12 at 21:42
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    $\begingroup$ If I remember right, there is a theorem one usually proves beforehand, saying that the Lebesgue measure $\lambda$ is the only translation invariant measure on $\Bbb R^n$ satisfying some extra hypothesis (IIRC that the measure of the cube $[0,1]^n$ equals one.) You then take a linear isomorphism $A$ and consider the measure $\mu=A^*\lambda$ defined on measurable sets $X$ via $\mu(X)=\lambda(A^{-1}(X))$, show that it is translation invariant and finite on the cube, so that it is proportional to Lebesgue measure. The constant of proportionality is the absolute value of the determinant of $A$. $\endgroup$ – Olivier Bégassat Nov 22 '12 at 21:58
  • $\begingroup$ Thank you guys very much. I just want to supply your beautiful answers with some links. proof that translation invariant measure is Lebesgue measure up to a constant factor $\endgroup$ – user197284 Nov 23 '12 at 5:52
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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

  1. Row swap: $f({\mathbf e}_i) \leftrightarrow f({\mathbf e}_j)$
  2. Row scaling: $f({\mathbf e}_i) \to \alpha f({\mathbf e}_i)$, $\alpha \ne 0$
  3. 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 $b$.

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