Let $\| \dot\ \|:\mathbb{R}^n \to \mathbb{R}$ be the euclidean norm.
By continuity $\| \dot\ \|^{-1}$ preserves Borel measurability so it suffices to check that it preserves null sets. In the case of $\| \dot\ \|$ i have nothing.
I basically trying to prove that the euclidean norm preserves measurability and is Lesbegue-Lesbegue measurable (measurable from the Lebesgue $\sigma$-algebra to the Lebesgue $\sigma$-algebra)
Let $f$ be the norm function on $\mathbb{R}^n$. We can re-write $\mathbb{R}^n$ as $S^{n-1} \times (0, \infty) \cup \{0\}$. The restriction of $f$ to $\{0\}$ is definitely measurable. And, the restriction of $f$ to $S^{n-1} \times (0, \infty)$ is measurable because $f$ is a projection onto the second factor of $S^{n-1} \times (0, \infty)$. This is because, if $A$ is a measurable subset of $Y$, then $X \times A$ is a measurable subset of $X \times Y$.
If $X = A \cup B$ where $A$ and $B$ are measurable, and $g$ is some function defined on $X$, then $g$ is measurable if and only if it is measurable on $A$ and $B$. Since $\{0\}$ and $\mathbb{R}^n-\{0\}$ are measurable subsets of $\mathbb{R}^n$, and $f$ is measurable on both, $f$ is a measurable function.
Note that the above works for both Lebesgue-Lebesgue measurability and Borel-Borel measurability.