TITLE: Rotation invariance of the Lebesgue measure
QUESTION [6 upvotes]: Denote the Borel sets in $\mathbb R^d$ as $\mathcal B^d$.
Is there a proof for the rotation invariance of the Lebesgue measure that doesn't use already that one has 
$$ \lambda(A^{-1}(B)) = \vert \operatorname{det} A \vert ^{-1} \lambda (B) \qquad \text{for all } A \in \operatorname{GL}(\mathbb R^d), \ B \in \mathcal B^d?$$
For example one can show easily that $\lambda$ is invariant under translation just using that intervals are invariant under translation and a $\cap$-closed generator of the Borel sets. This the first step in the proof of the above statement.
Hence I am interested to see a proof for the rotation invariance likewise without the above statement.

REPLY [2 votes]: One quick proof of this is using the fact that Lebesgue measure on $\mathbb{R}^n$ is $n$-dimensional Hausdorff measure for the Euclidean metric (up to a scale factor).  Since the definition of Hausdorff-measure uses only the metric, any isometry automatically preserves it.
Another way to prove it is using the fact that Lebesgue measure is a Haar measure and Haar measures are unique up to scaling.  Since the composition of Lebesgue measure with a rotation is then also a Haar measure, it differs from Lebesgue measure by a constant factor.  You can then find the constant factor is $1$ because a rotation fixes the unit ball, which has positive measure.