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.
Best Answer
You can prove that associated to every linear transformation $T: \mathbb{R}^n \to \mathbb{R}^n$ there exists a number $\alpha(T)$ such that $$\lambda(T(E))=\alpha(T)\cdot \lambda(E)$$ for all $E \in \mathcal{B}^d$. You don't need to prove that $\alpha(T)=\mathrm{det}(T)$, and the proof of the statement above is quite direct from the basic properties of the Lebesgue measure (c.f. Rudin's Real and Complex Analysis).
Now, fix $T$ a rotation. Since the equality holds for every $E$, it holds in particular for the open unit ball $B$. Since $T(B)=B$, we have $$\lambda(B)=\alpha(T)\lambda(B),$$ and hence $$\alpha(T)=1,$$ since $\lambda(B) \neq 0$ (a quick elementary way to see this is by using the fact that it is open).
Therefore, $$\lambda(T(E))=\lambda(E)$$ for every $E \in \mathcal{B}^d$.