# Does this set have area in $\mathbb{R}^2$?

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.

• Btw I'm somewhat new to this community so I'm not 100% sure if those are the "right" tags. If you believe there are other tags that are more suited to this question, please tell me! Commented Jul 19, 2021 at 15:23
• The Lebesgue integral exists and is equal to zero, and that will be the case whenever $\mathcal A$ has measure zero. Not sure about whether the Riemann integral exists for this particular set. Commented Jul 19, 2021 at 15:26
• You can always replace a degenerate rectangle with a rectangle of arbitrarily small area, the "arbitrary smallness" not depending on any other parameters that might exist. So, given any $\epsilon > 0,$ if you have $n$ degenerate rectangles, then cover each of them with a rectangle of area $\frac{\epsilon}{n}.$ This contributes an additional $\epsilon$ to the sum of the areas of all rectangles you're using, so if you want the final result to be less than some given positive number, make the sum of the other areas be less than half this number and choose $\epsilon$ to be half this number. Commented Jul 19, 2021 at 15:29
• Also, both the Riemann and Lebesgue integral of the characteristic function of your set exist (and both integrals are equal to $0)$, because the function is discontinuous precisely on the 4 segments of the square, and their union is a closed measure zero set in ${\mathbb R}^2$ (see this answer for the real-valued function version for Riemann integrability). Commented Jul 19, 2021 at 15:35
• Now what do I do next with this question? --- It's fine with me if someone else wants to write an answer, and if no one does, then I might when I have time (but I'm too busy today). Actually, it's OK for YOU to write an answer (and pick it, if no one else writes one you think is better; if you write something soon, or someone else does, maybe wait a day or two before deciding on the best answer), and this might be the best thing for you, as others can then help out with your "math writing". By the way, you should mention in the question that you're dealing with Riemann integration. Commented Jul 19, 2021 at 16:31

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.