Let $f: [0;1] \to \mathbb R$
$$f(x) = \begin{cases} 0 & x = 0 \\ \dfrac{1}{x}-\left[\dfrac{1}{x}\right] & \text {otherwise} \end{cases}$$ Prove that f is Riemann integrable. I have a solution in my textbook however i dont understand it. Could anyone explain intuition why such partiton was chosen? Why right after $U(P,f)-L(P,f)$ we have $\frac{1}{n_0+1}$?
Solution: Let $\varepsilon>0$, there exist natural number $n_0$ such that $\frac{1}{n}<\varepsilon$ for all $n>n_0$> Let P be partition such that:
$0=x_0<x_1=\dfrac{1}{n_0+1}<x_2< \dots <x_{n'_0}=\dfrac{1}{n_0}<x_{n'_0+1}< \dots <x_{n'_1}=\dfrac{1}{n_0-1}< \dots <x_{n'_{n_0-1}}=1$
and $x_i-x_{i-1}<\dfrac{\varepsilon}{4n_0}$ for $i \geq 2$.
Then: $$U(P,f)-L(P,f)=\frac{1}{n_0+1}+\sum\limits_{i=2}^{n'_{0}}(M_i-m_i)(x_i-x_{i-1})+\sum\limits_{k=0}^{n_0-2}\sum\limits_{i=n'_k+1}^{n'_{k+1}}(M_i-m_i)(x_i-x_{i-1})<\frac{\varepsilon}{2}+2n_0\frac{\varepsilon}{4n_0}=\varepsilon$$ So $f$ is Riemann integrable.
The idea behind the chosen partition is that the map $f$ is
Base on those fact, you can take for the two first points of the partition zero itself and $1/(n_0+1)$. The difference between the upper sum and lower sum on such interval is less than $1/(n_0+1)$. Which can be as small as we want providing that $n_0$ is chosen large enough.