Prove that $\int_{[0,1]\times [0,1]} f(x,y) \,dA$ doesn't exist. I cannot prove this.

Question

I am watching a lecture on multivariable mathematics by Theodore Shifrin on YouTube.

Let $R:=[0,1]\times[0,1]$.
Let $f:R\to\mathbb{R}$ be a function such that $f(x,y) = \begin{cases} \frac{1}{q} & (x=\frac{p}{q}, y=\frac{s}{q}, \frac{p}{q}\text{ and }\frac{s}{q}\text{ are irreducible fractions}),\\ 0 & (\text{otherwise}). \end{cases}$
Prove that $\int_{R} f(x,y) \,dA$ doesn't exist.

I must prove that for some $\epsilon_0 >0$, there doesn't exist a partition $\mathcal{P}$ of $R$ such that $U(f,\mathcal{P})-L(f,\mathcal{P}) < \epsilon_0$.
In other words, I must prove that there exists some $\epsilon_0 >0$ such that for any partition $\mathcal{P}$ of $R$, $U(f,\mathcal{P})-L(f,\mathcal{P}) \geq \epsilon_0$.
But I cannot prove this.

If $f$ were defined as follows, I could prove that $\int_{R} f(x,y) \,dA$ doesn't exist.
I quess that Professor Shifrin intended the following $f$ since this $f$ is much easier for his purpose.
But I am interested in my question itself now.

Let $f:R\to\mathbb{R}$ be a function such that $f(x,y) = \begin{cases} 1 & (x=\frac{p}{q}, y=\frac{s}{q}, \frac{p}{q}\text{ and }\frac{s}{q}\text{ are irreducible fractions}),\\ 0 & (\text{otherwise}). \end{cases}$
Prove that $\int_{R} f(x,y) \,dA$ doesn't exist.

Answer 1

You can't prove it because the integral exists.

Fix $\varepsilon > 0$ and pick $N \in \mathbb{N}$ such that $\frac{1}{N+1} \leqslant \varepsilon$. The set $$F = \left\{ \frac{p}{q} : 0 \leqslant p \leqslant q \leqslant N \ \& \ \frac{p}{q} \text{ is an irreducible fraction} \right\}$$ has some finite cardinality $s$. Take $M \in \mathbb{N}$ such that $\frac{s^2}{M^2} \leqslant \varepsilon$ and consider the partition $\mathcal{P}$ of $R$ into squares of side length $\frac{1}{M}$. It suffices to prove that $U(f, \mathcal{P}) \leqslant 2 \varepsilon$.

We have that $$U(f, \mathcal{P}) = \sum_{S \in \mathcal{P}} \frac{1}{M^2} \cdot \sup_{z \in S} f(z).$$

Divide $\mathcal{P}$ into subsets $\mathcal{P}_1$ and $\mathcal{P}_2$ where $\mathcal{P}_1$ consists of squares which intersect $F \times F$ and $\mathcal{P}_2$ of the remaining squares. Then $|\mathcal{P}_1| \leqslant s^2$, so

$$\sum_{S \in \mathcal{P}_1} \frac{1}{M^2} \cdot \sup_{z \in S} f(z) \leqslant \frac{s^2}{M^2} \cdot 1 \leqslant \varepsilon.$$

For $S \in \mathcal{P}_2$ we have that $\sup \limits_{z \in S} f(z) \leqslant \frac{1}{N+1} \leqslant \varepsilon$, hence

$$\sum_{S \in \mathcal{P}_2} \frac{1}{M^2} \cdot \sup_{z \in S} f(z) \leqslant \frac{M^2}{M^2} \cdot \varepsilon = \varepsilon.$$

Therefore

$$U(f, \mathcal{P}) = \sum_{S \in \mathcal{P}_1} \frac{1}{M^2} \cdot \sup_{z \in S} f(z) + \sum_{S \in \mathcal{P}_2} \frac{1}{M^2} \cdot \sup_{z \in S} f(z) \leqslant \varepsilon + \varepsilon = 2 \varepsilon.$$

Answer 2

Given any partition of the unit square you have $$U(f,\mathcal{P})=1$$ and $$L(f,\mathcal{P}) =0$$ For $U(f,\mathcal{P})$ you pick rational points and for $ L(f,\mathcal{P})$ you pick irrational points so you have $$U(f,\mathcal{P})-L(f,\mathcal{P}) =1 $$