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I'm having some trouble with the following question:

Let $f:[0,1] \to \mathbb R$ be defuned as: $f(x) = 0$, if $x \notin \mathbb Q$, and if $x \in \mathbb Q \cap [0,1]$ and $x = \frac pq$ is an irredicuble fraction, then $f(x) = \frac 1q$.

Prove that $f$ is integrable and find the value of $$\int_0^1 f(x) dx$$

So, first things firs, if $P$ is a partitian of $[0,1]$ I will denote the upper sum of $f$ with respect to $P$ as: $\overline \sum (f,P)$ and the lower sum as: $\underline \sum (f,P).$

What I did is:

I proved that $\sup \left\{\underline \sum(f,P),\ P\text{ is a partition} \right\} = 0$, and intuitively I think that $\inf \left\{\overline \sum(f,P),\ P\text{ is a partition} \right\}$ is also $0$, and that would mean that:

$$\sup \left\{\underline \sum(f,P),\ P\text{ is a partition} \right\} =\inf \left\{\overline \sum(f,P),\ P\text{ is a partition} \right\} = 0$$

And thus the function $f$ is integrable and: $$\int_0^1 f(x) dx = 0$$

The thing is that I'm unable to prove that $\inf \left\{\overline \sum(f,P),\ P\text{ is a partition} \right\} = 0$.

Is this even correct? If so, how can I prove this?

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1 Answers1

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For $\epsilon > 0$, there are only a finite number of rationals $r$ with $f(r)> \epsilon$. Surround each of them with a partition of width small enough that the sums of all the widths is less than $\epsilon$. In these partitions, we still know that $f \le 1$. For the remainder of $[0,1]$, the total width of all of the other partitions is less than $1$ (since the total width of all the partitions is $1$), and the maximum height of $f$ on those partitions is less than $\epsilon$.

Paul Sinclair
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