I am given a function $f:[-1,1]\times[0,1] \subset \mathbb{R}^2 \longrightarrow \mathbb{R}$ defined as it follows:
$$f(x, y)=\begin{cases} x, & \text {if } y \in \mathbb{Q}\cap[0,1] \\ 0, & \text {otherwise } \end{cases}$$
Now, regarding the Riemann integral of this function, I am asked to prove that:
- $\int_0^1 \int_{-1}^1 f(x, y) \,dx \,dy \in \mathbb{R}$
- $\nexists \int_{-1}^{1} \int_0^1 f(x, y) \,dy \,dx$
What I've worked so far:
Given a $y\in [0,1] $, I take that $y$ as "fixed" so I calculate: $$\int_{-1}^1 f(x, y) \, dx= \begin{cases} \left.\int_{-1}^1 x \, dx=\frac{x^2}{2}\right]_{-1}^1=0, & y \in \mathbb{Q}\cap[0,1] \\ \int_{-1}^1 0 \, dx=0, & \text {otherwise } \end{cases}$$ So I can conclude that $\int_0^1 \int_{-1}^{1} f(x, y) \, dx \, dy = \int_0^1 0 \, dy=0\in \mathbb{R}$
For the second case I know how to prove it, but only by this way:
Partitioning the rectangle $[-1,1]\times[0,1]$ in squares of $\frac{1}{n} \times \frac{1}{n}$, naming this partition $\mathcal{P}_{n}$ I can calculate either the lower or the upper sum and see that $$\lim _{n \rightarrow \infty} L\left(\mathcal{P}_n, f \right) \neq 0 \quad \vee \quad \lim _{n \rightarrow \infty} U(\mathcal{P}_n,f) \neq 0 $$
In case of the lower sum, I know that I only have to consider the part of the rectangle that has negative values of $x$ (which is $[-1,0] \times [0,1]$) because for the remaining part of the rectangle: $$\inf \{f(x,y): (x,y) \in [0,1] \times [0,1]\}=0$$
Therefore, $L(\mathcal{P}_n, f) =\sum_{k=0}^n \left(\frac{1}{n^2}\right) \left(-1+\frac{k}{n}\right) \cdot n=\frac{1}{n} \sum_{k=0}^n \left(-1+\frac{k}{n}\right)=-\frac{1}{2} \left( \frac{n+1}{n} \right)$
$\frac{1}{n^{2}}$ stands for the area of each subsquare; $\left(-1+\frac{k}{n}\right)=\inf \{f(x,y): (x,y) \in [-1,0] \times [0,1]\}$ and the $n$ factor takes into account that we have the same infimum for $n$ subsquares.
Since $\lim _{n \rightarrow \infty} L(\mathcal{P}_n,f)=\lim _{n \rightarrow \infty}-\frac{1}{2}\left(\frac{n+1}{n}\right)=-\frac{1}{2} \neq 0 \quad$ I can state that $\nexists \int_{-1}^1 \int_0^1 f(x,y) \, dy \, dx$.
Is there any way of proving $\nexists \int_{-1}^1 \int_0^1 f(x, y) \, dy \, dx$ without going for this method?. I would like to use the Cauchy Condition for the Riemann Integrability but I don't really know how to do it that way.