Let $f:[-2,2] \to \mathbb{R}$ is defined as follows: $f(x) = \begin{cases} x+1, x \in \mathbb{Q} \\ -x, x \in \mathbb{Q^c} \end{cases}$.
Is $f \in \mathcal{R}[-2, 2]$? Explain your answer!
I think this function is not Riemann integrable. I need to show this by using the negation of Cauchy criterion for integral.
Negation of Cauchy criterion for integral: $f \notin \mathcal{R}[a, b]$ if and only if there exists $\varepsilon_0 > 0$ such that for all $\eta > 0$ there exists tag partition $\dot{P}$ and $\dot{Q}$ where $||\dot{P}||<\eta$ and $||\dot{Q}||<\eta$ such that $|S(f; \dot{P}) - S(f; \dot{Q})| \geq \varepsilon_0$.
I'm stuck in choosing the tag partition. I've tried the tag partition $\dot{P}$ with the tag are rational numbers and tag partition $\dot{Q}$ with the tag are irrational numbers and didn't get good answer. Can you help me?
Note: please do not use upper and lower integral