Let $f$ be the characteristic function of the set $S=\{1,\frac{1}{2},\frac{1}{3},\dots \}$; thus $f(x)=1$ when $x\in S$ and $f(x)=0$ otherwise. If $0< \epsilon < 1$ and $(P,T)$ is a partition pair for $[0,1]$ such that mesh$(P)<\epsilon$, show that $0\le R(f,P,T)<\epsilon + 2\sqrt{\epsilon}$.
$P,T \subset [a,b]; P = \{x_0,\dots,x_n\}, T=\{t_1,\dots,t_n\} \text{ such that } a=x_0\le t_1 \le x_1 \le t_2 \le x_2 \le \cdots \le t_n \le x_n = b$
$R(f,P,T)=\sum_{i=1}^nf(t_i)\Delta x_i=f(t_1)\Delta x_1+f(t_2)\Delta x_2 + \cdots +f(t_n)\Delta x_n \text{ where } \Delta x_i = x_i-x_{i-1}$
mesh$(P)$ is the largest subinterval $[x_{i-1},x_i]$
So I know that if $\epsilon$ is small then the largest interval will be small and since the function is a collection of points most of them will be missed. So $R(f,P,T)$ must be approaching zero. \
I know that if I let $N$ be the largest integer such that $1/N>\sqrt{\epsilon}$ and $k$ be the largest index such that $x_k>\sqrt{\epsilon}$ that $\sum_{j=1}^kf(t_j)(x_j-x_{j-1})\le \sum_{j=1}^k 1\cdot (x_j-x_{j-1}) = x_k$. So $R(f,P,T)\le x_k+N\epsilon$. I'm not sure how of if this gets me where I need though