Is this riemann integrable on [0,1]?
$\displaystyle f(x)= 0$ when x is rational and $\displaystyle f(x)= 1$ when x is irrational
The point is this. If you give me any partition of $\displaystyle [0,1]$ into subintervals, let's just call them $\displaystyle I_1,\cdots,I_n$, we find that $\displaystyle U(P,f)=1$. Why? Well, let's look at something. These intervals are presumably non-degenerate (not one point) and so in any of these intervals there are both irrational and rational numbers. So, $\displaystyle \sup_{x\in I_k}f(x)=1$ for any subinterval since it must contain an irrational. So, $\displaystyle U(P,f)=\sum_{j=0}^{n}\sup_{x\in I_j}f(x)\Delta I_j=\sum_{j=1}^{n}\delta I_j=1-0=1$. But, why does this last sum equal one? The way I wrote it should give you a clue. Start with that .