Prove $f$ is not Riemann Integrable?

Let $$f:[0,1]\rightarrow\mathbb{R}, f(x)=\begin{cases} x & \text{when x is rational} \\ 1-x & \text{when x is irrational} \end{cases}$$ Show that $$f$$ is not Riemann Integrable.

Can you show provide an argument with upper sum and lower sum? Any hint will be appreciated.

My attempt: $$f\in \mathcal{R}[0,1] \Rightarrow f\in\mathcal{R}[0,0.5]$$. Let $$\mathbf{P}=\{x_0<\dots be a partition, then $$U(f,P)\leq\sum_{i=1}^n(1-x_{i-1})(x_i-x_{i-1})$$ and $$L(f,P)=\sum_{i=1}^n(x_{i-1})(x_i-x_{i-1})$$.

But how to show that $$f$$ is not Riemann, or there does not exist any inf$$(U(f,P))$$ or sup$$(L(f,P))$$.

• Note that the measure of the set of rational numbers is $0$ – Fareed Abi Farraj Feb 26 '20 at 14:19
• You'll have to use the fact that the rationals are dense in the reals. This implies that - no matter how fine your partition is gonna be - the upper sum will be always greater than the Riemann-integral of $u(x) = \max\{x,1-x\}$ and your lower sum will be smaller than the integral of $l(x)=\min\{x,1-x\}$. Therefore the lower and upper sum cannot converge to the same number. – frog Feb 26 '20 at 14:20

Let $$x\in[x_{i-1},x_i]$$. As $$x\leq \frac12$$, we have $$1-x\geq \frac12$$

If $$x$$ is irrational then $$f(x)=1-x \geq \frac12$$

If $$x$$ is rational then $$f(x)=x \leq \frac12$$

So $$\inf_{x\in [x_{i-1},x_i]} f(x) = \inf_{x\in [x_{i-1},x_i] \cap \mathbb{Q}} f(x)$$

Hence $$\inf_{x\in [x_{i-1},x_i]} f(x) = \inf_{x\in [x_{i-1},x_i] \cap \mathbb{Q}} x = x_{i-1}$$

Similarly,

$$\sup_{x\in [x_{i-1},x_i]} f(x) = \sup_{x\in [x_{i-1},x_i] \cap \mathbb{R}\setminus\mathbb{Q}} f(x)$$

Hence $$\sup_{x\in [x_{i-1},x_i]} f(x) = \sup_{x\in [x_{i-1},x_i] \cap \mathbb{R}\setminus\mathbb{Q}} 1-x =1- x_{i-1}$$

We deduce that

$$L(f,P)=\sum_{i=1}^n (x_{i-1})(x_i-x_{i-1})$$

$$U(f,P)=\sum_{i=1}^n (1-x_{i-1})(x_i-x_{i-1})$$

Let's denote $$\Pi$$ the set of all the partitions of $$\left[0,\frac12\right]$$, we have

$$L(f)=\sup_{P\in\Pi}L(f,P)=\int_0^{1/2} x dx=\frac18$$

and $$U(f)=\inf_{P\in\Pi}L(f,P)=\int_0^{1/2} (1-x) dx=\frac38$$

$$L(f) \neq U(f)$$ so $$f$$ is not Riemann-integrable.