Question
Prove $f:[0,1] \rightarrow R$ defined by
$$ f(x) = \left\{\begin{aligned} & sin \left(\frac{1}{x}\right) &&, x \in (R-Q) \cap [0,1] \\ &0 &&, otherwise \end{aligned} \right.$$ is not Riemann integrable.
The author goes like this:
Consider a partition $$P=\left\{0,\frac 2 {(2n+1)\pi},\frac 2 {(2n-1)\pi},\cdots,\frac 2 {3\pi},\frac 2 {\pi},1\right\}.$$ Then the upper sum $U(f,P)=1$ as $\sup f(x)$ in every sub interval is $1$. The lower sum $L(f,P)=-\frac 2 {\pi}$;for, the contribution of the last interval $[\frac 2 {\pi},1]$ is $0$ since the minimum of the function is $0$.But in each of the other sub-intervals, the minimum is $-1$ and so the lower sum is $(-1)\times$ legth of all the sub intervals in $[0,\frac 2 {\pi}]$ which is $(-1)\times \frac 2 {\pi}=-\frac 2 > {\pi}$. This proves that $f\not\in R[0,1].$
My problem
In the book the author first takes a partition , then finds upper darboux sum and lower darboux sum then concludes that f is not riemann integrable.
Is this proof okay?
I mean there might be a partition for which $U(f,P)-L(f,P) \lt \epsilon$ is true.
EDITED on 9th December 2020
Deleted photos and added text.