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 $\mathbf{P}=\{x_0<\dots<x_n\}$ 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$(L(f,P))$.
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.