$f(x) = f_0(||x||)$, show that $f$ is measurable iff $f_0$ is measurable Let $f_0:[0,\infty) \to \mathbb{R}$, and let $f:\mathbb{R}^m \to \mathbb{R}$. say that $f(x) = f_0(||x||)$
Prove: $f$ is measurable if and only if $f_0$ is measurable.
I tried to solve it myself and got down to this problem:
$\{x \in \mathbb{R}^m: ||x|| \in B\} $ is measurable if and only if $B$ is measurable.
But I have no idea how to prove it
 A: I assume here that measurable means "Borel measurable". If $f_0$ is measurable, it is clear that $f$ is measurable since $||*||$ is continuous, hence measurable. Conversely, if $f$ is measurable and $S$ is a measurable set of $\mathbb R_+$, then we have to prove that $f_0^{-1}(S)$ is a measurable set.
Let us note $\rho(x)=||x||$, hence $f^{-1}(x) = \rho^{-1}(f_0^{-1}(x))$.
So, with $C=f_0^{-1}(B)$, you have to prove that if $C$ is a subset of $\mathbb R_+$ such that $\rho^{-1}(C)$ is measurable, then so is $C$. But $C = \rho(\rho^{-1}(C))$ (since $\rho$ is surjective). So, you have to show that if $T=\rho^{-1}(C)$ is a measurable subset of $\mathbb R^n$, $\rho(T)$ is a measurable subset of $\mathbb R_+$. Since $\rho$ is surjective, it transports boolean set operations from $\mathbb R^n$ to $\mathbb R_+$, and since the open balls generate the Borel sigma algebra of $\mathbb R^n$, it suffice to observe that the image of an open ball of $\mathbb R^n$ by $\rho$ is an open interval of $\mathbb R_+$ (if the "open ball" are convex, e.g. for euclidean distance, which can be assumed a priori).  
Regarding your second question, the answer is in fact included inside the end of what I wrote above (of course, one direction is evident: if $B$ is measurable, then so is $\rho^{-1}(B)$ since $\rho$ is continuous). 
A: Hints:


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*Suppose that $f_0$ is (Borel-)measurable. Show that $x \mapsto p(x) := \|x\|$ is Borel-measurable and conclude that the composition $f = f_0 \circ p$ is measurable.

*Suppose that $f$ is (Borel-)measurale. Show that $\mathbb{R}^m \ni x \mapsto x_1 \in \mathbb{R}$ is Borel-measurable. Deduce that $x_1 \mapsto f_0(|x_1|)$ is measurable. Conclude.

