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Given this modified Dirichlet function: f(x) = 0 if x is in Q, else f(x) = x. I am wondering if this function is Darboux integrable on the interval [0, 2].

I managed to show that every lower Darboux sum is equal to zero, therefore the lower Darboux integral is 0. My intuition tells me that the upper Darboux integral is also 0, but I can't think of how to show that.

I thought of using the fact that f(x) is Darboux integrable if and only if: for every epsilon there exists a partition of [0, 2] such that $S(P) - s(P) < \epsilon$ where S, s are the lower Darboux sums. In our case $s(P)=0$ for every P so we have to show that for every epsilon there exists a partition such that $S(P) < \epsilon$, but didn't manage to come close to anything.

PS. If possible, please don't give a solution that uses Reimann's integral, I haven't studied it.

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up vote 1 down vote accepted

The function is not integrable by Darboux (and equivalently by Riemann) as any point in (0,2] is a discontinuity point. (take a rational sequence approaching $x$, the values of $f(x_n)$ are constantly $0$. If $x$ is irrational, then $\lim f(x_n) \neq f(x)$)

If you want to stick to the Darboux integral definition, since the lower Darboux sum is always 0, and the upper Darboux sum of a given finite partition is $\sum_{i = 1}^{n-1} (x_{i+1}-x_i)\cdot x_{i+1}$.

In this case, the upper sum is just the sum of the rectangles created by the partition, so if we reduce each rectangle by the parts it exceeded the function $f(x)=x$

Take some finite partition $P = \langle 0 = x_0,\ldots,x_{n+1}=2 \rangle$ of $[0,2]$.

$\sum_{i=0}^n (x_{i+1} - x_i)\cdot x_{i+1} \ge $

$\sum_{i=0}^n (x_{i+1} - x_i)\cdot x_{i+1} - \frac{1}{2}\sum_{i=0}^n(x_{i+1} - x_i)^2 =$

$\frac{1}{2}\sum_{i=0}^n (2x_{i+1}^2 - 2x_{i+1}x_i - x_{i+1}^2 + 2x_{i+1}x_i - x_i^2) =$ $\frac{1}{2}\sum_{i=0}^n(x_{i+1}^2 - x_i^2) = 2$

and therefore we have that every upper sum is $\ge2$ for every partition.

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