if the projection of $E\subseteq \mathbb{R}^n$ onto $\mathbb{R}^{n-1}$ has (n-1) dim. measure zero, then the set $E$ has $n$-dim. measure zero

Question

In the book of Mathematical Analysis by Zorich, at page 121, it is asked that

Prove that if the projection of a bounded set $E\subseteq \mathbb{R}^n$ onto a hyperplane $\mathbb{R}^{n-1}$ has (n-1) dimensional measure zero, then the set $E$ itself has $n$-dimensional measure zero.

I even cannot see the result intuitively, so I'm hoping you guys can help me to prove it, or give some hints about how to prove it.

Answer 1

You have $E \subset P(E) \times [a,b]$, where $P$ is the projection and $a,b$ are some real numbers. This follows from the fact that $E$ is bounded. Therefore $0 \leq m_n(E) \leq (b-a) m_{n-1}(P(E))=0$. (Also, $E$ is Lebesgue measurable because it is contained in a known measurable set which has zero measure, and the Lebesgue measure is complete.)