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Given that $$\ f(x)=\begin{cases} 0, & \text{if $x$ is rational} \\ 1, & \text{if $x$ is irrational}\end{cases}$$

I want to show that the integral $$\int_0^1 f(x)\,dx$$ does not exist by using a Riemann sum and showing that it does not converge to a number.

I'm guessing that the Riemann sum that I'm searching for should look like this: $$a_n = \sum_{i=1}^n \left(f\left(\frac{i-c}{n}\right)*\frac{1}{n}\right),\ c\in [0,1]$$

and that I'm supposed to show that $$\lim_{n \to \infty} a_n$$ does not exist.

Edit: I knew that the Dirichlet function is not continuous on [0,1] and that therefore it is not integrable when I made this post. This is the reasoning that I've seen for answers to similar questions.

I was wondering if it was possible to demonstrate that it is not integrable by coming up with a Riemann sum that does not converge.

Additionally, in the form that I suggested above, the variable c does not have to be rational.

Joseph
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  • You're using a right uniform sum formula here, which is fine for computing integrals of functions we already know are Riemann integrable, but can't always tell us if the function is not Riemann integrable. In this case, $f((i - c)/n) = 0$, because $(i - c)/n$ is always rational, which makes $a_n = 0$ for all $n$, hence the limit is $0$. But, this function turns out not to be Riemann integrable. – Theo Bendit Sep 03 '23 at 15:54

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