Let $n\in\mathbb{N},M\subset\mathbb{R}^n$ with $\mu^*(M)$ being finite and $A$ a linear mapping, which is diagonal to the standard basis of $\mathbb{R}^n$.
Show that: $\mid\det(A)\mid\mu^*(M)=\mu^*(A(M)) $.
To be honest, I have no idea how to begin and I doubt that using the definition of $\mu^*$ will not be particularly useful.
I'd appreciate any help.
Define $\mu_{1}(M)=\mu^{*}(A(M))$, and $\mu_{2}(M)=|\det(A)|\mu^{*}(M)$. Consider an open box $B\subset \mathbb{R}^{n}$, whose edge vectors are given by $\lambda_{1}e_{1},\ldots,\lambda_{n}e_{n}$, where $\lambda_{i}\in\mathbb{R}$ and $\{e_{1},\ldots,e_{n}\}$ is the standard basis of $\mathbb{R^{n}}$.
Notice that the volume of $B$ is $\mu^{*}(B)=|\lambda_{1}\lambda_{2}\ldots\lambda_{n}|$, i.e., $\mu_{2}(B)=|\det(A)||\lambda_{1}\lambda_{2}\ldots\lambda_{n}|$. It is given that $A$ is a diagonal matrix. Let $A=diag[d_{1},\ldots,d_{n}]$. After applying $A$ to $B$, edge vectors of the distorted version $A(B)$ will be $d_{1}\lambda_{1}e_{1},\ldots,d_{n}\lambda_{n}e_{n}$. Therefore volume of $A(B)$ is $\mu^{*}(A(B))=|d_{1}\lambda_{1}d_{2}\lambda_{2}\ldots d_{n}\lambda_{n}|=|d_{1}\ldots d_{n}||\lambda_{1}\ldots\lambda_{n}|=|\det(A)|\mu^{*}(B)=\mu_{1}(B)$. Therefore $\mu_{1}(B)=\mu_{2}(B)$.
Since the Borel sigma algebra of $\mathbb{R^{n}}$ is generated by such sets $B$, the statement $\mu_{1}(M)=\mu_{2}(M)$ is true for all $M\subset \mathbb{R^{n}}$.