Q:

Prove that: $\lambda^{d}(s_{a}(M))=a^{d}\lambda^{d}(M)$ for any lebesgue-measurable set where $s_{a}(M)=aM$

Let $s_{a}$ be the scaling by a certain factor $a >0$, then show that $\lambda^{d}(s_{a}(M))=a^{d}\lambda^{d}(M)$ for any lebesgue-measurable set.
Define the generator of the borel$-\sigma-$algebra, namely, $E:=\{ [c_{1},b_{1}[\times...\times [c_{d}, b_{d}[:c,b \in \mathbb  R^{d}\}$
Note that for any $C:=[c_{1},b_{1}[\times...\times [c_{d}, b_{d}[$ we get: 
$\lambda^{d}(s_{a}(C))=\lambda^{d}([ac_{1},ab_{1}[\times...\times [ac_{d}, ab_{d}[)=\prod\limits_{i=1}^{d} a\vert b_{i}-c_{i}\vert=a^{d}\prod\limits_{i=1}^{d} \vert b_{i}-c_{i}\vert=a^{d} \lambda^{d}(C)$
So I have shown that $\lambda^{d}(s_{a}(M))=a^{d}\lambda^{d}(M)$ holds on a generator $E$ of the borel measurable sets. 
Am I allowed to then say it would indeed hold for any set $M\in \mathcal{B} = \sigma(\tau_{d})$ where $\mathcal{B}$ is the borel$-\sigma-$algebra? $(**)$
If this is the case then I am nearly done. Since the Lebesgue sigma algebra is merely the completion of the borel sigma algebra, any $A \in \mathcal{L}^{d}$ can be written as $A=B\cup N$ where $B$ is a borel set and $N$ is a lebesgue set with measure zero, hence $\lambda^{d}(s_{a}(A))=\lambda^{d}(s_{a}(B\cup N))= \lambda^{d}(s_{a}(B))+ \lambda^{d}(s_{a}(N))=\lambda^{d}(s_{a}(B))=a^{d}\lambda^{d}(B)=a^{d}\lambda^{d}(B\cup N)=a^{d}\lambda^{d}(A)$ 
It was used that $s_{a}$ is lipschitz continuous and thus maps Lebesgue Null sets to Lebesgue Null sets. 
Is my proof ok? I am particularly not sure about $(**)$

A:

It is not sufficient that the boxes form a generating set. It's also important that they form a $\pi$-system in order to invoke the Uniqueness Theorem for $\sigma$-finite measures (see statslab.cam.ac.uk/~james/Lectures/pm3.pdf). In this case you're comparing the measures $\mu(A)=\lambda^d(s_a(A))$ and $\nu(A)=a\lambda^d(A)$.

