Why isn't the Dirichlet Function Riemann Integrable?

Question

Can we find out the upper and lower sums based solely on the domain [a,b] ?

Answer 1

On the interval $[0,1]$, let $$f(x)=\begin{cases}1&\text{if }x\in\mathbb Q\\0&\text{otherwise}\end{cases}$$ For any partition $0=x_0<x_1<\ldots <x_n=1$, the Riemann upper sum is $$U_{\mathbf x}=\sum_{i=1}^n(x_i-x_{i-1})\sup_{x_{i-1}<x<x_i}f(x)=\sum_{i=1}^n(x_i-x_{i-1})=1$$ because in each interval $(x_{i-1},x_i)$ there is some rational $q\in(x_{i-1},x_i)$ and hence the supremum is $\sup_{x_{i-1}<x<x_i}f(x)\ge f(q)=1$ (and of course it is not larger than $1$). Likewise, the lower sum is $$L_{\mathbf x}=\sum_{i=1}^n(x_i-x_{i-1})\inf_{x_{i-1}<x<x_i}f(x)=\sum_{i=1}^n(x_i-x_{i-1})\cdot 0=0$$ beacuse the existence of irrationals $\alpha\in(x_{i-1},x_i)$ cause $\inf_{x_{i-1}<x<x_i}f(x)\le f(\alpha)=0$ (and of course it is not below $0$).

If we consider $$g(x)=\begin{cases}x&\text{if }x\in\mathbb Q\\0&\text{otherwise}\end{cases}$$ instead, the situation hardly changes: Since there are rationals arbitrarily close to the right ends of the intervals, we have in fact that $\sup_{x_{i-1}<x<x_i}=x_i$; thus the upper sum is the same as for the identity function itself and therefore tends to $\frac12$ as the partition gets finer. But the lower sum is still $0$.