Is a bijective projection function measure preserving?

Question

A subspace with dimension strictly less than the dimension of vector space has (Lebesgue) $measure=0$. Let $V$ be a vector space with $dimension=n$. To show that some set $S$ in V is zero-measure, is it enough to show the existence of bijective projection between $S$ and a subset of a subspace of $V$ with $dimension < n$?

Answer 1

No. As shown in Lebesgue Measure of the Graph of a Function, there exists a function $f : \mathbb{R} \to \mathbb{R}$ whose graph $G = \{(x,f(x)) : x \in \mathbb{R}\}$ is a Lebesgue non-measurable subset of $\mathbb{R}^2$. But if we let $\pi(x,y) = (x,0)$ be projection onto the $x$ axis, then $\pi|_G : G \to \mathbb{R}$ is bijective.

If you can first show that your set $S$ is measurable, then it will have measure zero. This can be shown using Fubini's theorem.

Let $E$ be a proper subspace of $V$ and let $\pi : V \to E$ be orthogonal projection onto $E$. Suppose that $\pi|_S$ is injective. Let $F$ be the orthogonal complement of $F$. Then Lebesgue measure $m$ on $V$ is the product measure of the Lebesgue measures $m_{E}$, $m_{F}$ on $E,F$ respectively. Now by Fubini's theorem, we can write $$m(S) = \int 1_S\,dm = \int_E \int_F 1_S(x+y) \,m_F(dx) \,m_E(dy).$$ But by assumption, for each $y \in E$, there is at most one $x\in F$ such that $x+y \in S$. (If $x+y, x'+y \in S$, then $\pi(x+y)=\pi(x'+y)=y$. Since $\pi$ is injective on $S$, we have $x+y=x'+y$, hence $x=x'$.) Hence the function $x \mapsto 1_S(x+y)$ is zero almost everywhere, so $\int_F 1_S(x+y)\,m_F(dx) = 0$ for every $y \in E$. Thus the double integral on the right side is 0 and we conclude $m(S) =0 $.

If you don't want to require $\pi$ to be orthogonal projection, then the result is still true but you have to include a Jacobian factor in the double integral. The details are slightly tedious.