Is this riemann integrable on [0,1]?

$\displaystyle f(x)= 0$ when x is rational and $\displaystyle f(x)= 1$ when x is irrational

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- Apr 22nd 2010, 01:57 PMCrazyCat87Riemann Integrability
Is this riemann integrable on [0,1]?

$\displaystyle f(x)= 0$ when x is rational and $\displaystyle f(x)= 1$ when x is irrational - Apr 22nd 2010, 02:07 PMDrexel28
- Apr 22nd 2010, 02:24 PMCrazyCat87
I guess I'm having trouble with this material since I'm unsure of what you're asking here....

- Apr 22nd 2010, 03:21 PMDrexel28
- Apr 23rd 2010, 06:42 AMCrazyCat87
I understand that it's at most 1 and 0, I just don't know how to prove this using the definition, which I think is what I'm supposed to do

- Apr 23rd 2010, 10:12 AMDrexel28
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 .