Prove $f(x)= 1$ for $x \in \mathbb{Q}$ and $f(x)=0$ for $x \notin \mathbb{Q}$ is not integrable.

Question

I want to prove that

$$f(x) = \begin{cases} 1 \text{ for } x \in \mathbb{Q}\\ 0 \text{ for } x \notin \mathbb{Q} \end{cases}$$
is not integrable on $[0,1]$.

Now I'm at the point in the book where we have just learned about Riemann integrability, with partitions. So far, this is the definition we have used:

A function $f: [a,b] \rightarrow \mathbb{R}$ is integrable if $$\forall \epsilon > 0, \exists P: \overline{S}-\underline{S} < \epsilon,$$ where $P$ is a partition of the interval and $\overline{S}$ and $\underline{S}$ are upper and lower sums of the function.

Now if I'm right, I need to prove $$\exists \epsilon > 0, \forall P: \overline{S}-\underline{S} > \epsilon,$$ in order to establish that the function is indeed not integrable. And this is where I'm struggling. I have the idea that, although the interval itself is bounded, the number of $x \in \mathbb{Q}$ in $[0,1]$ is not bounded. Then how could I ever choose a $\epsilon$ such that all partitions result in a larger difference? If on the other hand the number would have been finitie, say $n$, I would let $\epsilon > n \cdot 1 = n$ and then all partitions would meet this criteria (at least, I think).

Any suggestions?

Answer 1

For that function, each superior sum is equal to $1$, and each inferior sum is equal to $0$. Therefore, take any $\varepsilon\in(0,1]$.

Answer 2

On the interval $[a,b] $,all the upper Riemann sums are $(b-a)$ and the lower Riemann sums are zero. There are no sequences of upper and lower sum converging to a common limit.

Thus your function is not integrable on [a,b]