Ex2-3

Solution to Chapter 2, Exercise 3 
Suppose \Gamma'is a partition of [a',b']. Then \Gamma=\Gamma'\cup\{a,b\}is a partition of [a,b]. 
P_{\Gamma}[a,b] = [f(x_0)-f(a)]^+ + \sum_{i=1}^m [f(x_i)-f(x_{i-1})]^+ + [f(b)-f(x_m)]^+ 
Since [f(x_0)-f(a)]^+\geq0and [f(b)-f(x_m)]^+\geq0, we have 
P_{\Gamma}[a,b] \geq \sum_{i=1}^m [f(x_i)-f(x_{i-1})]^+ = P_{\Gamma'}[a',b']. 
Then we have 
P_{\Gamma'}[a',b'] \leq P_\Gamma [a,b] \leq V_P[a,b]. 
Therefore, 
V_P [a',b'] = \sup_{\Gamma'} P_{\Gamma'} [a',b'] \leq V_P [a,b]. 
Similarly,
N_{\Gamma}[a,b] = [f(x_0)-f(a)]^- + \sum_{i=1}^m [f(x_i)-f(x_{i-1})]^- + [f(b)-f(x_m)]^-
Since [f(x_0)-f(a)]^-\geq0and [f(b)-f(x_m)]^-\geq0, we have 
N_{\Gamma}[a,b] \geq \sum_{i=1}^m [f(x_i)-f(x_{i-1})]^+ = N_{\Gamma'}[a',b'].
Then we have
N_{\Gamma'}[a',b'] \leq N_\Gamma [a,b] \leq V_N[a,b].
Therefore,
V_N [a',b'] = \sup_{\Gamma'} N_{\Gamma'} [a',b'] \leq V_N [a,b].