Prove $f$ is measurable iff $f_0$ is measurable.

Question

$f:\mathbb{R}^d\to\mathbb{R}$ called radial function if exists $f_0:\mathbb{R}_+\to\mathbb{R}$ such that for every $x\in\mathbb{R}^d$ $f(x)=f_0(||x||).$
Prove $f$ is measurable iff $f_0$ is measurable.

I've tried one way but I'm not sure about this.
Assuming that $f_0$ is measurable and defining the following sets
$f^{-1}(A)=\{x\in\mathbb{R}^d:||X||\in f_0^{-1}(A)\}=R_{f_0^{-1}(A)}$ and setting $B=f_0^{-1}(A)$
Assume that $B\subseteq\mathbb{R}_+$ is measurable and $g(x)=||x||$,$g$ is continuous function then for every open set $V(V\in \mathcal{B})$$\to$ $g^{-1}(V)\in \mathcal{B}$ is open, therefore $g$ is lebesgue measurable and then we get $R_B=\{x\in\mathbb{R}^d:||x||\in B\}=g^{-1}(B)$.
I have some doubts because $B$ is lebesgue and not borel and then $g^{-1}(B)$ is not borel measurable.

Answer 1

The composition of measurable functions is always measurable. Since you are looking at measurability with respect to the Borel $\sigma$-algebra, the continuity of the norm implies its measurability. Hence, if $f_0$ is measurable, then $f$ is measurable as composition of measurable functions.

This should answer the implication you try to show. Since these more abstract results are sometimes easier to show, I recommend proving them and applying them to this special case.