Prove that the bounded function $f$ defined by $f(x)=0$ if $x$ is irrational and $f(x)=1$ if $x$ is rational is not Riemann integrable on $[0,1]$.
I was given the hint to use the inverse definition of Riemann integrable and consider the cases of the partition being all rationals, and all irrationals between $[0,1]$, but I'm not too sure how to go about it.
Best Answer
Hint: Every subinterval of $[0,1]$ contains both rational numbers and irrational numbers.
Look at the definition of a function being Riemann integrable on the interval $[a,b]$ (in this case $[0,1]$). This definition must fail for this function.
Why?
Well, since every subinterval contains irrational numbers, the infimum of the function values over any subinterval is $0$. Similarly, since every sub-interval contains rational numbers, the supremum of the function values over any subinterval is $1$.
What does this tell you about the upper and lower sums over any partition?