So, I am taking a Several Variables Calculus course, and this problem appeared in one of my tests -
Let $\Omega \in \mathbb{R}^3$ be a bounded set, and $\Lambda = \{(x, y) \in \mathbb{R}^2 \mid (x,y,z) \in \Omega\}$ be the projection of $\Omega$ in $xy$-plane. Show that, if $\Lambda$ has content zero, $\Omega$ also has content zero.
I am aware now that defining an appropriate finite cover with an arbitrarily small volume for $\Omega$ is sufficient. But, the idea slipped my mind for that particular moment, and I applied Cavalieri's Principle. This is what my solution looked like in my paper:
As $\Omega$ is bounded, we can assume, $\exists M > 0$ such that $\forall(x,y,z) \in \Omega, z \in [-M, M]$.
Now, define $\Lambda_t = \{(x,y,t) \in \Omega \mid t \in [-M,M]\}.$ We can see, either $\Lambda_t = \varnothing$, or $\Lambda_t \subset \Lambda$.
As $\Lambda$ has content zero, then every proper subset also has volume zero. So, $v(\Lambda_t) = 0, \quad \forall t \in [-M,M]$.
So, by Cavalieri's principle, $\displaystyle v(\Omega) = \int_{-M}^{M}v(\Lambda_t)dt = \int_{-M}^{M}0dt = 0 \implies \Omega$ has volume zero.
The only problem with this solution is, that nowhere in the problem it was said that $\Omega$ is Jordan Measurable, that is, $\partial\Omega$ has to be measure zero. But I also could not find any counterexamples (that is because the statement is indeed true, you can construct a small finite cover.)
My hypothesis is, that $\Omega$ is actually Jordan-measurable. but I'm not sure how to prove that. So I'm asking a general question -
Let $\Omega \in \mathbb{R}^n$ be bounded, such that a hyperplane projection has content zero. Is $\Omega$ Jordan-measurable?
Any help would be appreciated.
EDIT: Just to clarify, proving it is content zero itself shows the set is Jordan Measurable, and I agree to that. Is there a way to directly prove it's Jordan Measurable by this information alone? That is what I wish to know.