3-21) "Let $A \subseteq \mathbb{R}^n$ be a closed rectangle, and $C \subseteq A$. Show that $C$ is Jordan measurable if and only if for every $\varepsilon > 0$, there is a partition $P$ of $A$ such that
$$\sum_{S \in \sigma_1}v(S) - \sum_{S \in \sigma_2}v(S) < \varepsilon$$
where $\sigma_1$ is the collection of subrectangles $S$ determined by the partition $P$ such that they intersect $C$, and $\sigma_2$ consists of those which are contained in $C$."
So I think I have an outline of a solution but there are a few steps which I can't justify even though they seem to be true intuitively.
proof of $\implies$ direction:
Suppose $C$ is Jordan measurable and let $\varepsilon > 0$ be given.
Then we can choose a collection of closed rectangles $\{U_i\}_{i \in \mathbb{N}}$ to cover the boundary of $C$ such that $\sum_{i=1}^{\infty} v(U_i) < \varepsilon$. Since bd($C$) is closed and bounded, it's compact thus it has a finite subcover $\{U_{\omega}\}_{\omega \in \Omega}$.
Next, we construct a partition $P$ of $A$ using the endpoints of the closed intervals which make up the closed rectangles $U_{\omega}$; so that each $U_{\omega}$ is a union of subrectangles determined by $P$.
Now, because $\sigma_2 $ is contained in $\sigma_1$, we have that $$\sum_{S \in \sigma_1}v(S) - \sum_{S \in \sigma_2}v(S) = \sum_{S \in \sigma_1 - \sigma_2}v(S) $$
Now here is where I can't properly justify my next step: I want to claim that the RHS is $\leq \sum_{\omega \in \Omega} v(U_{\omega})$, because from the pictures I drew that's what it seems like...
Assuming that step is true, we have that \begin{align} \sum_{\omega \in \Omega} v(U_{\omega}) &\leq \sum_{i=1}^{\infty} v(U_i) \\ &< \varepsilon \end{align}
Since $\varepsilon > 0$ was arbitrary, this completes the "proof" of this direction.
For the other direction, for any given $\varepsilon > 0$, we can choose a partition $P$ such that
$$ \sum_{S \in \sigma_1}v(S) - \sum_{S \in \sigma_2}v(S) < \varepsilon$$
Once again since $\sigma_2 \subseteq \sigma_1$, we have that
$$\sum_{S \in \sigma_1 - \sigma_2}v(S) < \varepsilon$$
Here, I would like to claim that $\sigma_1 - \sigma_2$ covers bd($C$), which once again seems like it should be true from the pictures I've drawn, but I can't prove it. If this is indeed true then we've just shown that bd($C$) has content $0$, which implies it has measure $0$; thus completing the proof.
I'd appreciate any help in justifying those two steps of my proof, and also any comments on any other mistakes which are present. Thanks!
If $x \in \partial C$ ($x$ is a boundary point) then every open rectangle containing $x$ contains points both in $\text{int} \, C$ and $\text{ext} \, C$.
For the second part, the rectangles in $\sigma_1 - \sigma_2$ must cover $\partial C$.
Suppose $x \in \partial C \cap C$ but is not contained in a rectangle in $\sigma_1 - \sigma_2$. It must then belong to a rectangle $S_x$ in $\sigma_2$ which is contained in $C$. This requires $x$ to be on the boundary of that rectangle. Otherwise $x$ is contained in the interior of $S_x$, which is an open rectangle that contains no points in $\text{ext} \, C$, a contradiction. Since $x$ is on the boundary of $S_x$, it belongs to at least one other rectangle which is in $\sigma_1-\sigma_2$. This follows because if $x$ did not belong to a rectangle in $\sigma_1-\sigma_2$, it lies in the interior of a union of rectangles in $\sigma_2$ and there is a subrectangle containing $x$ that contains no points in $\text{ext} \, C$.
On the other hand suppose $x \in \partial C$ and $x \notin C$, but is not contained in a rectangle in $\sigma_1 - \sigma_2$. It must then belong to a rectangle $S_x \notin \sigma_1$ of the partition $P$. Again, this requires $x$ to be on the boundary of that rectangle. Otherwise $x$ is contained in the interior of $S_x$, which is an open rectangle that contains no points in $\text{int} \, C$, a contradiction. By similar reasoning above, $x$ belongs to at least one other rectangle in $\sigma_1- \sigma_2$.
For the first part, the boundary $\partial C$ is covered by a countable collection of open rectangles $U_j$ that contain points both in $\text{int} \, C$ and $\text{ext} \, C$ (since every open rectangle containing a boundary point must have this property) such that $\sum_j v(U_j) < \epsilon$. Since $\partial C$ is compact, there is a finite subcover by closed rectangles $\{\bar{U}_\omega\}_{\omega \in \Omega}$, where each closed rectangle contains both points in $\text{int} \, C$ and $\text{ext} \, C$. Proceed from here.