Understanding the solution of the problem from book Kaczor ,Nowak 1.1.8

Question

Here I cannot understand the 2nd part proof given in the ans from the book of W.J.Kaczor and M.T.Nowak book volume 3 can anyone help me to understand that or any other equivalent proof.

Answer 1

If you choose tags $\ t_k=-1+\frac{k-1}{n}\ $ for $\ 1\le k\le n\ $ and $\ t_k=\frac{k-n}{n}\ $ for $\ n+1\le k\le 2n-1\ $ with the partitions $\ P_n\ $ given by RRL in his or her comment, then these values of $\ t_i\ $ are admissible (i.e. $\ x_{i-1}\le t_i\le x_i\ $), and \begin{align} S(P_n,f,\alpha)&=\sum_{k=1}^{2n-1}f(t_i)\big(\alpha(x_i)-\alpha(x_{i-1})\big)\\ &=0 \end{align} because $\ f(t_i)=0\ $ for $\ 1\le k\le n\ $ and $\ \alpha(x_i)-\alpha(x_{i-1})=0\ $ for $\ n+1\le k\le 2n-1\ $. If you take $\ t_n=\frac{1}{2n} $ instead of $\ t_n={-}\frac{1}{n}\ $, however, and leave all the other values of $\ t_i\ $ the same, then they're still admissible, but now $$ S(P_n,f,\alpha)=1 $$ because $\ f(t_n)=1\ $ and $\ \alpha(x_n)-\alpha(x_{n-1})=1\ $ while all the other values of $\ f(t_i)\ $ remain the same and none of the values of $\ x_i\ $ have changed.

What's happening here is that $\ n\ $ is the only value of $\ i\ $ for which $\ \alpha(x_i)-\alpha(x_{i-1})\ne0\ $ and when you shift the tag $\ t_n\ $ from the interval $\ \left[{-}\frac{1}{n},0\,\right]\ $ to $\ \left(0,\frac{1}{n}\,\right]\ $ the value of $\ f(t_n)\ $ changes from $\ 0\ $ to $\ 1\ $.