# How to show that a function is not Riemann integrable by using Cauchy criterion?

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

• Can you explain in your question what you mean for a partition $P$ by $S(f, P)$? Nov 17, 2021 at 9:48
• $S(f;\dot{P})$ is the Riemann sum with $\dot{P}$ is the tag partition. Nov 17, 2021 at 10:11
• Can you please write it precisely in the question? In particular, at which point is $f$ evaluated in each interval? That is key to provide a meaningful answer. Nov 17, 2021 at 10:21

Since $$x \mapsto x+1$$ is Riemann integrable, it follows that as $$\|P\| \to 0$$ for partitions $$P$$ with rational tags,

$$S(f,P) \to \int_{-2}^2(x+1) \, dx = \left.\frac{1}{2}x^2 \right|_{-2}^2 + \left. x \right|_{-2}^2 = 4$$

Similarly, since $$x \mapsto -x$$ is Riemann integrable, it follows that as $$\|Q\| \to 0$$ for partitions $$Q$$ with irrational tags,

$$S(f,Q) \to \int_{-2}^2(-x) \, dx = -\left.\frac{1}{2}x^2 \right|_{-2}^2 = 0$$

Hence, there exist $$\delta_1,\delta_2 > 0$$ such that if $$P$$ has rational tags and $$\| P\| < \delta_1$$ and if $$Q$$ has irrational tags and $$\|Q\| < \delta_2$$, then

$$|S(f,P) - 4| < 1,\quad |S(f,Q)| = |S(f,Q) - 0| < 1$$

This implies that if $$\|P\| < \delta = \min(\delta_1,\delta_2)$$, then $$S(f,P) - 4 > -1$$, and, hence $$|S(f,P)|= S(f,P) > 3$$. Also, if $$\|Q\| < \delta = \min(\delta_1,\delta_2)$$, then $$-|S(f,Q)| > -1$$.

Take $$\varepsilon_0$$ = 1. For any $$\eta > 0$$ we can choose partitions $$P$$ (with rational tags) and $$Q$$ (with irrational tags) such that

$$\|P\|, \|Q\| < \min(\delta, \eta) \leqslant \eta,$$

but, by the reverse triangle inequality,

$$|S(f,P) - S(f,Q)|\geqslant |S(f,P)| - |S(f,Q)|>3 -1 = 2 > \varepsilon_0$$

Therefore, the Cauchy criterion is violated and $$f$$ is not Riemann integrable.

• Can we say for each part of function, namely $x+1$ and $-x$ are Riemann integrable and then use the definition of Riemann integral that is $|S(f;P)-L|<\varepsilon$ ? Because the example in the book calculate the Riemann sum first. Nov 18, 2021 at 13:20
• @ccmatyn: Yes - and that is what I used. If the tags are rational then the Riemann sum $S(f,P)$ can be arbitrarily close to the integral of $x+1$ which is $L = 4$. If the tags are irrational then the Riemann sum $S(f,Q)$ can be arbitrarily close to the integral of $-x$ which is $L = 0$.
– RRL
Nov 18, 2021 at 15:47