Maybe the notation varies so let me just clarify this: if $F\subset\mathbb{R}^n$, we define:
$\overline{\text{Vol}}(F) := inf_{G \text{ is a grid on a rectangle $R\supset F$}}V(F,G)$
(the outer sum $V(F,G)$ is the sum of volumes of rectangles in the grid $G$ which intersect $\overline{E}$)
and if is a Jordan set (boundary has volume zero, that is, $\forall \epsilon > 0$ there is a grid $G$ on a rectangle $R\supset F$ s.t. the outer sum $V(\partial F,G)$ is less than $\epsilon$), then we define:
Vol$(F) := $ $\overline{\text{Vol}}(F)$.
Here is my question:
(this is exercise 12.3.8 from Wade's $\textit{Analysis}$)
Is there a fairly quick solution to this that does not invoke the fact that $f$ is continuous except on a measure zero set:
Let $E\subset \mathbb{R}^2$ be nonempty and Jordan and let $f: E\to [0,\infty)$ be integrable (on $E$). Prove that the volume of $\Omega:= \{(x,y,z)\in \mathbb{R}^3 \mid (x,y)\in E, 0\leq z\leq f(x,y)\}$ satifsifes: Vol$(\Omega) = \int\int_EfdA$.
Once I can show that (i) $\Omega$ is Jordan, (ii) $A:= \{(x,y,z)\in \mathbb{R}^3 \mid (x,y)\in \overline{E}, 0\leq z\leq f(x,y)\}$ is Jordan and equal to $\overline{\Omega}$, and (iii) Vol$(\Omega) = $ Vol$(A)$,
and I assume WLOG that f is continuous on $E$ (*) (the set of points of discontinuity has measure zero and I think this implies volume zero and I think that implies Jordan measure or Vol$(-)$ of zero)
then I am really done ($A$ is certainly projectiable under the assumption (*), so by theorem 12.39 we will have: $\int_E f = \int\int_E(\int_0^{f(x,y)}1dz)dydx = \int_A f = \text{Vol}(A) = $ Vol$(\overline{\Omega}) = $ Vol$(\Omega) + $ Vol$(\partial \Omega) = $ Vol$(\Omega) + 0 = $Vol $(\Omega)$ (the first equality holds due to the FTC from single variable calculus).
I guess I can prove (i), (ii), (iii) assuming (*) (maybe some of them, especially (i) without) but I feel like this is kind of sloppy and making the argument more convoluted than it needs to be (we haven't proved my class for instance that if the set differences between $U$ and $V$ in $\mathbb{R}^2$ have measure zero (definition 12.27), then if one of $U$ or $V$ is Jordan then I guess they both are and if $f:C\supset U,V\to \mathbb{R}$ is integrable then $\int_Uf = \int_Vf$, but I feel like this is all getting really convoluted and long, and there is probably an easier way. (I think I have more or less proven that $A = \overline{\Omega}$ if we assume $f$ is continuous but still...). It just seems like I am doing it wrong just because it so complicated at this point (I am up to like 6 pages of scratch work now).
Also, no I cannot use the change of variables theorem (**) here (idk if that would be helpful anyway but let's just assume not, that isn't until the section in the book after this exercise anyway, so apparently it is feasible without it I just don't know how).
It really kind of looks like I should be using Fubini somewhere but I don't think $\Omega$ needs to be a product of rectangles (it seems like far from it, possibly besides using that theorem (**) I mentioned that I can't use for this). Maybe I should be using Caravelli Principle somewhere but I am not really sure how either (I was trying to use that corollary of Caravelli that involves projective sets as you can see, the problem is really just that $f$ need not be continuous and/or I don't know how to rigorously justify saying WLOG assume f is continuous on $E$)