Ex2-14

Let [a,b]=[-1,1] and takef(x)=\begin{cases}0&\text{ if } -1\le x<0,\\1&\text{ if } 0 \le x \le 1. \end{cases} \, \, \, \, \, \, \, \, \,\, \, \,
\phi(x)=\begin{cases}0&\text{ if } -1\le x \le0,\\1&\text{ if } 0 < x\le 1. \end{cases}
and we can calculate\int_{-1}^{0}f d\phi=\lim_{\left\vert \Gamma \right\vert \to 0 } \sum f(\xi_i)[\phi(x_i)-\phi(x_{i-1})]
=\lim_{\left\vert \Gamma \right\vert \to 0 } \sum f(\xi_i)(0-0)=0
\int_{0}^{1}f d\phi=\lim_{\left\vert \Gamma \right\vert \to 0 } \sum f(\xi_i)[\phi(x_i)-\phi(x_{i-1})]
=f(\xi_i)\cdot(1-0)=1
The second equation exist at the point 0, also in this situation we can't get the\int_{-1}^{1}f d\phi
, because these two functions are both discontinuous (jump discontinuity of first kind) at point 0. In this condition, the different choice of partitions give different bound for R_\Gamma 
, as it will give value of 0 or 1 conditionally, so we can't find the limit\int_{-1}^{1}f d\phi
. 
For the Theorem 2.17, we can use contrapositive of the theorem, which is:If \int_{a}^{c}f d\phi 
 or \int_{c}^{b}f d\phi 
 does not exist, then \int_{a}^{b}f d\phi 
 does not exist for a<c.Then the statement above yields\int_{a}^{c}f d\phi 
 and \int_{c}^{b}f d\phi 
 may both exist for a<c, but \int_{a}^{b}f d\phi 
 does not exist.