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 $$[f(x_0)-f(a)]^+\geq0$$and $$[f(b)-f(x_m)]^+\geq0$$, 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 $$[f(x_0)-f(a)]^-\geq0$$and $$[f(b)-f(x_m)]^-\geq0$$, 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,

$$V_N [a',b'] = \sup_{\Gamma'} N_{\Gamma'} [a',b'] \leq V_N [a,b].$$