Consider a homogeneous function $m$ in $\mathbb{R}^n$ with degree zero, ie $$m(\lambda \xi) = m(\xi), \;\;\;\;\;\; \forall \lambda >0.$$ Is it true that $m \in L^\infty(\mathbb{R}^n)$ if, and only if, $m \in L^\infty(S^{n-1})$??
Attempt: [EDITED] The sufficiency isn't that trivial as I though it would be. If $m$ is in $L^\infty(\Bbb R^n) $, exists a null Lebesgue measure se $A$ in $\Bbb R$ such that, $$|m(x)| \leq \|m\|_{L^\infty(\Bbb{R}^n)}, \;\;\;\;\; \forall x \in \Bbb{R}^n\backslash A.$$ I can't find the null measureble set in $S^{n-1}$ such that the inequality above holds in the complement of this set...
We prove the necessity. Supose that $m \in L^\infty(S^{n-1})$. So there exists $A \subset S^{n-1}$ with measure zero in $S^{n-1}$ such that, $$m(\xi) \leq \|m\|_{L^\infty(S^{n-1})}, \;\;\;\; \forall\;\; \xi \in A^c.$$ So I defined a subset $B$ of $\mathbb{R}^n$ by $B:=\{ \xi \neq 0 : \xi/|\xi| \in A\}$ and I want to prove that this subset $B$ has measure zero. Supose it's proved that $|B|=0$. Then, $\xi \notin B$ implies that $\xi/|\xi| \notin A$, and using the homogeneity of $m$, we have $$|m(\xi)| = |m(\xi/|\xi|)| \leq \|m\|_{L^\infty(S^{n-1})}.$$ So, we conclude that $\|m\|_{L^\infty(\mathbb{R}^n)} < \infty$.
If you know that spherical coordinates work for the Lebesgue integral, you may write $$|B|=\int_{\mathbb{R}^n} \chi_B(x)\,dx=\int_0^\infty\int_{S^{n-1}}\chi_B(\theta,r)r^{n-1}\,d\theta\,dr=\int_0^\infty\int_{S^{n-1}}\chi_A(\theta)\,d\theta r^{n-1}\,dr=0$$ because the first integral vanishes.