How to prove Lebesgue Measure is dilation-invariant for any B borel set

Question

Let $\rho\cdot B = \{\rho\cdot b: b \in B\}$ a Borel set. I want to prove that $\forall B \in \mathbb{B}(\mathbb{R}^n)$ and $\rho > 0$ fixed we have: $$ \mathcal{L}(\rho\cdot B) = \rho^n\cdot\mathcal{L}(B) $$ where $\mathcal{L}(B)$ is the Lebesgue Measure.

Answer 1

The functions $A\mapsto\mathcal L(\rho\cdot A)$ and $A\mapsto\rho^n\mathcal L(A)$ are measures on the Borel sets of $\Bbb R^n$ which coincide on the $\pi$-system of the hyperrectangles. The set on which two measures coincide is a $\lambda$-system, therefore by Dynkin's $\pi$-$\lambda$ lemma the two measures coincide on the $\sigma$-algebra generated by the hyperrectangles. Which is the Borel sets.

Answer 2

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Theorem. Let $M\in GL_n(R)$ and $A$ be a measurable subset of $\R^n$. Then $\lambda(M(A))=\det(M)\lambda(A)$, where $\lambda$ is the Lebesgue measure on $\R^n$.

Proof. We know that for each $T\in GL_n(\R)$ the assignment $\mc B\ni B\mapsto \lambda(T(B))\in [0, \infty]$ is a translation invariant measure, call it $\lambda_T$, on the Borel $\sigma$-algebra $\mc B$ on $\R^n$. There there is $c_T\in \R$ such that $\lambda_T=c_T\lambda$. This gives a map $c:GL_n(\R)\to \R^*$ defined as $c(T)=c_T$. Note that this is a group homomorphism, and since $\R^*$ is an abelian group, $c$ factors through the abelianization of $GL_n(\R)$. But the commutator subgroup of $GL_n(\R)$ is $SL_n(\R)$, and thus the abelianization of $GL_n(\R)$ is $SL_n(\R)$.

What this tells us is that if $T, S\in GL_n(\R)$ are in the same $SL_n(\R)$-coset, then $c_T=c_S$. But $T$ and $S$ are in the same $SL_n(\R)$-coset if and only if $\det(T)=\det(S)$. Thus, for a given $T\in GL_n(\R)$, to find $c_T$ we just need to find an $S\in GL_n(\R)$ having the same determinant as $T$, and for which one can easily calculate the value of $c$. Such an $S$ can in fact be constructed by considering the diagonal-matrix whose first diagonal entry is $\det(T)$, and all the other diagonal entries are $1$. $\blacksquare$