Is f(x) = 1 if x is rational, 0 if x is irrational Riemann integrable on [0,1]?

[missavvy]

Homework Statement

Let A={1/n, n =natural number}
f: [0,1] -> Reals
f(x) = {1, x in E, 0 otherwise
Prove f is riemann integrable on [0,1]

Homework Equations

The Attempt at a Solution

Not quite sure, but I think supf = 1 and inf f= 0 no matter what partition you take, then
Spf - spf = 1
so it is not r-integrable..?
(obv i skipped lots of steps but I am not sure if it is actually r-int or not, i said its not)

 

[LeonhardEuler]

That's not right; consider this:
You have a function that is 1 for x=1, but 0 everywhere else. You integrate in an interval including x=1.
The value of the function in the upper sum in the part of the partition containing x=1 is always 1, while the lower is 0. But the width of the interval in the partition containing x=1 diminishes to 0 as the mesh of the partition approaches 0. So the function is integrable, and the integral on every interval is 0.

This should reveal the mistake you were making. Can you apply this kind of reasoning to your example?