Let $B \in \mathcal{B}^{d}$ and $\alpha\neq 0$
Any ideas on how to show:
$\lambda^{d}(\alpha B)=|\alpha|^d\lambda^{d}(B)$
My idea using "simple sets":
Let $B:=[a_{1},b_{1}[\times...\times[a_{d},b_{d}[$ this means:
Note without loss of generality $\alpha > 0$ otherwise switch intervals $[a,b[$ to $[\alpha b,\alpha a[$
$\alpha B=[\alpha a_{1},\alpha b_{1}[\times...\times[\alpha a_{d},\alpha b_{d}[$
Therefore $\lambda^{d}(\alpha B)=\prod_{i=1}^{d}(\alpha b_{i}-\alpha a_{i})=\alpha^{d}\prod_{i=1}^{d}( b_{i}- a_{i})=\alpha\lambda^{d}(B)$
Since $\{[\alpha a_{1},\alpha b_{1}[\times...\times[\alpha a_{d},\alpha b_{d}[\}$ is a $\cap-$stable generator of $\mathcal{B}^{d}$ it follows:
$\lambda^{d}(\alpha B)=|\alpha|^d\lambda^{d}(B)$ for any $B \in \mathcal{B}^{d}$