We have a borel set $A\subset \mathbb{R}^n$ such that for every $v\in \mathbb{R}^n$, $\left \| v\right \|=1$, the set $A_v=\left \{t\in \mathbb{R}:tv\in A\right \}$ has measure zero. Prove that $A$ has measure zero.
I am supposed to make use of the Change of Variables Theorem, but I cannot see how to use it.
What I could prove is that if we extend the definition of $A_v$ for every nonzero $v\in \mathbb{R}^n$ in the obvious way, then for $s\in \mathbb{R}\setminus \left \{0\right \}$, we have $sA_v=A_{\frac{1}{s}v}$, therefore $\chi _{A_v}\left (\frac{x}{\left \|v\right \|}\right )=\chi _{A_{\frac{v}{\left \|v\right \|}}}\left (x\right )$ and we can use the Change of Variables Theorem to prove that $A_v$ has measure zero for every nonzero $v$, not only for those which have norm $1$.
Let $\mathbb S$ be the unit sphere of $\mathbb R^n$. Then $n$-dimensional Lebesgue measure can be written in "spherical coordinates" as $dm(x) = r^{n-1} dr\; d\mu(v)$ where $\mu$ is a measure on $\mathbb S$ and $x = r v$, $r \in [0,\infty)$, $v \in \mathbb S$. Thus
$$|A| = \int_{\mathbb S} d\mu(v) \int_0^\infty dr\; r^{n-1} \chi_A(r v) $$
But $|A_v| = 0$ says $\chi_A(r v) = 0$ for almost every $r \in \mathbb R$, and thus for every $v$ we have $\int_0^\infty dr \; r^{n-1} \chi_A(r v) = 0$.