Let $\mathcal{A} \subset \mathbb{R}^2$ the following set: $\mathcal{A} = A_1 \cup A_2 \cup A_3 \cup A_4$, where:
$$A_1 = \{(q_1, 0) \in \mathbb{R}^2 : q_1 \in [0,1] \cap \mathbb{Q}\}$$
$$A_2 = \{(1, q_2) \in \mathbb{R}^2 : q_2 \in [0,1] \cap \mathbb{Q}\}$$
$$A_3 = \{(q_1, 1) \in \mathbb{R}^2 : q_1 \in [0,1] \cap \mathbb{Q}\}$$
$$A_4 = \{(0, q_2) \in \mathbb{R}^2 : q_2 \in [0,1] \cap \mathbb{Q}\}$$
That is, $\mathcal{A}$ are the points over the edges of the unitary square $[0,1] \times [0,1]$ having only rational coordinates. Now I know this set has measure zero because $\mathbb{Q} \cap [0,1]$ is a countable set and therefore $A_1, A_2, A_3$ and $A_4$ are countable, and $\mathcal{A}$ is therefore a finite union of countable sets, which makes it countable too. However, I'd like to see whether or not the set has area in the sense that
$ \int_\mathcal{A} 1_\mathcal{A} $ exists, where $1_\mathcal{A} (x,y) = 1$ if $(x,y) \in
\mathcal{A}$ and $1_\mathcal{A} (x,y) = 0$ otherwise. I built this set based on the fact that $\int_0^1 1_\mathbb{Q} (x)\ dx$ doesn't exist.
I suspect $\mathcal{A}$ has no area, but I'm not quite sure. Being the devil's advocate, we can cover this set by a finite collection of degenerate rectangles such as $\{0\} \times [0,1]$ that have volume (area) zero, but as far as I know the rectangle covering definition of volume (area) doesn't take into account degenerate rectangles.
I'm trying to see if this set is an example of measure zero sets which have no volume (in this case, area). Thank you for reading me!
EDIT: All integrals here are Riemann integrals.
Best Answer
Ok so I'm writing this answer based on a comment by @DaveL.Renfro (thanks!). Spoiler alert: my assumption turned out to be false, $\mathcal{A}$ has volume zero.
As I stated in the question we can cover $\mathcal{A}$ by 4 degenerate rectangles $\Omega_1, \Omega_2, \Omega_3$ and $\Omega_4$, where: $$ \Omega_1 = [0,1] \times \{0\}$$ $$ \Omega_2 = \{1\} \times [0,1]$$ $$ \Omega_3 = [0,1] \times \{1\}$$ $$ \Omega_4 = \{0\} \times [0,1]$$
These are of course the 4 edges of the unit square. Now, in order the finish the proof that $\mathcal{A}$ has indeed volume let $\epsilon > 0$ arbitrary. We take $\Omega'_1, \Omega'_2, \Omega'_3$ and $\Omega'_4$ given by: $$\Omega'_1 = [0,1] \times \left[-\frac{\epsilon}{8},\frac{\epsilon}{8}\right]$$ $$\Omega'_2 = \left[1-\frac{\epsilon}{8},1+\frac{\epsilon}{8}\right] \times [0,1]$$ $$\Omega'_3 = [0,1] \times \left[1-\frac{\epsilon}{8},1+\frac{\epsilon}{8}\right]$$ $$\Omega'_4 = \left[-\frac{\epsilon}{8},\frac{\epsilon}{8}\right]\times [0,1]$$
It's not hard to check that $\mathcal{A} \subset \bigcup_{i=1}^4 \Omega'_i$ and also $V(\Omega'_1) + V(\Omega'_2) + V(\Omega'_3) + V(\Omega'_4) = \epsilon/2 < \epsilon$, so for each $\epsilon > 0$ we can find a finite rectangle cover for $\mathcal{A}$ with a total area less than $\epsilon$, so $\mathcal{A}$ has volume zero.