Let $Q\subseteq \mathbb{R}^n$ be a rectangle and let $f: Q \rightarrow \mathbb{R}$ be a positive function.
Let $\Gamma=\left \{ (x_1, …, x_n, x_{n+1}): 0\leq x_{n+1}\leq f(x_1, … , x_n) \right \}$.
Prove that $f$ is Riemann integrable $\Leftrightarrow$ $\Gamma$ is Jordan-measurable.
The $\Rightarrow$ direction is quite easy: it's enough to observe that
$\partial\Gamma \subseteq G \cup (B_f\times[0, M]) \cup (\partial Q\times [0, M])$
where $G=\left \{ (x_1, …, x_n, x_{n+1}): x_{n+1}= f(x_1, … , x_n) \right \}$, $B_f$ is the set of discontinuities of $f$ and $M$ is a bound for $f$.
All three have measure 0 if $f$ is integrable.
But the other direction is harder. Does anyone has an idea?
Best Answer
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Bd}{\partial}\newcommand{\eps}{\varepsilon}$Hint: If your definition of Jordan measurable coincides with Wikipedia's (i.e., "inner" and "outer" approximations have arbitrarily close area), then each lower sum for the Riemann integral is an "inner" approximation and every upper sum is an "outer" approximation.
Edit: The strategy is to cover the graph of $f$ by finitely many rectangles $R_{i}$ of small area, then to subdivide $Q$ into rectangles $Q_{j}$ so that each $R_{i}$ is partitioned (modulo overlapping boundaries) by its intersections with the "columns" $Q_{j} \times [0, M]$. Finally, over each $Q_{j}$, choose an interval $[m_{j}, M_{j}]$ so that
$m_{j} \leq f(x) \leq M_{j}$ for all $x$ in $Q_{j}$, and
$Q_{j} \times [m_{j}, M_{j}]$ is contained in the union of the $R_{i}$.
The corresponding approximations to the Riemann integral are arbitrarily close.
Since the details are potentially delicate, here's a sketch of the one-dimensional case, phrased in a way to facilitate generalization to arbitrarily many variables.
Let $Q$ be a closed interval, $f:Q \to \Reals$ a bounded function satisfying $0 \leq f(x) \leq M$ for all $x$, and $$ \Gamma = \{(x, y) : 0 \leq f(x) \leq y,\ x \in Q\} $$ the Jordan-measurable region under the graph of $f$.
Fix $\eps > 0$, and pick finitely many rectangles $R_{i} = [a_{i}, b_{i}] \times [c_{i}, d_{i}]$ (with $1 \leq i \leq I$, say) of total area at most $\eps/2$ whose union contains the graph $G$ of $f$. (Use the remaining $\eps/2$ of the "total error budget" to cover $(Q \times \{0\}) \cup (\Bd Q \times [0, M])$.) There exists a partition $P = (x_{j})_{j=0}^{J}$ of $Q$ such that for every $j = 1, \dots, J$ and for every $i = 1, \dots, I$, either $Q_{j} = [x_{j-1}, x_{j}]$ is contained in $[a_{i}, b_{i}]$, or else $(x_{j-1}, x_{j}) \cap (a_{i}, b_{i}) = \varnothing$. [Note 1]
For each $j$, let $m_{j}$ be the largest of the $c_{i}$ such that
$Q_{j} \subset [a_{i}, b_{i}]$, and
$c_{i} \leq f(x)$ for all $x$ in $Q_{j}$.
In words, $m_{j}$ is the height of the highest lower edge among rectangles lying over $Q_{j}$ that also lies on or below the graph of $f$ over $Q_{j}$.
Similarly, let $M_{j}$ be the smallest of the $d_{i}$ such that $Q_{j} \subset [a_{i}, b_{i}]$ and $f(x) \leq d_{i}$ for all $x$ in $Q_{j}$. [Note 2]
Consequently, [Note 3] $$ U(f, P) - L(f, P) = \sum_{j=1}^{N} (M_{j} - m_{j})(x_{j} - x_{j-1}) \leq \sum_{i=1}^{I} (d_{i} - c_{i})(b_{i} - a_{i}) < \eps/2. \tag{*} $$ In words, the Darboux sums can be made closer than $\eps/2$ for arbitrary $\eps > 0$, so $f$ is Riemann integrable over $Q$.
Notes:
In the one-dimensional case, take $P = (x_{j})$ to be the sorted union of the division points $a_{i}$ and $b_{i}$. Generally, subdivide this way "in each variable". Geometrically, cover the graph of $f$ with finitely many boxes $R_{i}$ of total volume at most $\eps/2$. Project away the $(n+1)$th coordinate, and take the smallest grid subdivision of $Q$ that contains the shadows of the $R_{i}$.
Each $Q_{j}$ is contained in at least one $[a_{i}, b_{i}]$, and by construction is completely contained in every $[a_{i}, b_{i}]$ whose interior intersects $Q_{j}$. Because the $R_{i} = [a_{i}, b_{i}] \times [c_{i}, d_{i}]$ cover the graph of $f$, for each $j$, there exists an $i$ such that $c_{i} \leq f(x)$ for all $x$ in $Q_{j}$, and there exists an $i'$ such that $f(x) \leq d_{i'}$ for all $x$ in $Q_{j}$.
By construction, for each $j = 1, \dots, J$ we have $m_{j} \leq f(x) \leq M_{j}$ for all $x$ in $Q_{j}$. The first equality in (*) follows since the $Q_{j}$ have disjoint interiors.
By construction, $Q_{j} \times [m_{j}, M_{j}] \subset \bigcup_{i=1}^{I} [a_{i}, b_{i}] \times [c_{i}, d_{i}]$, so $$ \bigcup_{j=1}^{J} Q_{j} \times [m_{j}, M_{j}] \subset \bigcup_{i=1}^{I} [a_{i}, b_{i}] \times [c_{i}, d_{i}], $$ which gives the first inequality in (*).