Show that $|\int_a^b f dg | \le \|f\|_{\infty} \|g\|$

Question

The following is from Bruckner's Real Analysis :

I could make a rigorous argument for items (a) and (b), but although intuitively item (c) seems to be reasonable but I cannot make a rigorous proof for that. Please help, thanks!

Answer 1

Note that in exercise 12:8.2, $f \in \mathcal{C}[a,b]$.

Given any partition of $[a,b]$, say $ a= x_0 <x_1 < \cdots <x_n =b$, and let $t_1,t_2, \dots t_n $ satisfy $x_{i-1} \leq t_i \leq x_i$, for all $i =1, \dots, n$. Then, we have \begin{align*} \left | \sum_{i=1}^n f(t_i) (g(x_i) -g(x_{i-1}) \right | &\leq \sum_{i=1}^n |f(t_i)| | (g(x_i) -g(x_{i-1})| \leq \\ &\leq \|f\|_\infty\sum_{i=1}^n |(g(x_i) -g(x_{i-1})|\leq \|f\|_\infty V(g,[a,b])= \|f\|_\infty \|g\| \end{align*}

So $$ \left | \int_a^b f dg \right|= \left | \lim_{\Delta(P)\to 0} \sum_{i=1}^n f(t_i) (g(x_i) -g(x_{i-1}) \right| = \lim_{\Delta(P)\to 0} \left | \sum_{i=1}^n f(t_i) (g(x_i) -g(x_{i-1}) \right| \leq \|f\|_\infty \|g\|$$