Riemann sum: Let the function $f$ be defined by $f=\cases{0, when&$x=0$\cr 1, when&$0<x\leq 1.$\cr}$

Question

Let the function $f$ be defined by

$f=\cases{0, when&$x=0$\cr 1, when&$0<x\leq 1.$\cr}$

The task is to show that for every $\epsilon>0$ there exists a partition $P_{{\epsilon }}$ of the interval $[0, 1]$ such that the lower Riemann $L \left( f,P_{{\epsilon }} \right)$ sum would be $L \left( f,P_{{\epsilon }} \right)>1-\epsilon.$

I need help.

Answer 1

Let $\epsilon>0$ be arbitrary. Choose $n\in \mathbb{N}$ such that $n> \frac{1}{\epsilon}$ and consider the partition $P_n=\{0,\frac{1}{n},\frac{2}{n},\dots ,1-\frac{1}{n},1\}$.

Then $L(f,P_n)=1-\frac{1}{n}$ and $U(f,P_n)=1$.

So $L(f,P_n)=1-\frac{1}{n}>1-\epsilon $.

Therefore for each $\epsilon>0$, there exists a partition $P_\epsilon$ such that $L(f,P_\epsilon)>1-\epsilon $.