Does $\pi^{-1}(A)$ have measure zero?

Question

For the projection $\pi: \Bbb R^{n+1}\setminus \{0\}\to \Bbb {RP}^{n}$ with $n\ge 1$, if $A\subset \Bbb {RP}^{n}$ has measure zero, then $\pi^{-1}(A)$ has measure zero.

Answer 1

Yes. There exists a finite open cover $\{U_1,U_2,...,U_k\}$ of $\mathbb{R}^N-\{0\}$ such that $\pi|U_i :U_i \rightarrow \mathbb{R}P^{N-1}$ is a diffeomorphism. Since $A$ has measure 0, we get $[\pi|U_i]^{-1}(A)$ has measure 0 (as $\pi|U_i$ is diffeo).

Thus $A$ can be covered by finitely many measure $0$ subsets: $\{[\pi|U_i]^{-1}(A)\}_{i\in \{1,2,...,k\}}$. So $A$ itself has measure $0$.