Show that the function $$f(x)= \begin{cases} x, \text{ if the point } x \text{ in [0,1] is rational} \\ 0, \text{ if the point } x \text{ in [0,1] is irrational} \end{cases}$$ is not integrable.
My attempt:
Let $P_n$ be the regular partition of [0,1] with $n$ intervals. Then since rationals and irrationals are dense in $\mathbb R$ $$L(f,P_n)=\sum_{i=1}^n (x_i - x_{i-1}) \cdot \left(\inf_{x \in [x_{i-1}, x_i]} f(x)\right) = \sum_{i=1}^n 0 = 0$$ Therefore, $$\underline{\int_0^1}f=0\;$$
I'm having trouble with the upper Darboux sum $$U(f,P_n)=\sum_{i=1}^n (x_i - x_{i-1}) \cdot \left(\sup_{x \in [x_{i-1}, x_i]} f(x)\right).$$ How do I calculate this and $$\overline{\int_0^1}f?$$
