Here is Theorem 6.12 (d) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
If $f \in \mathscr{R}(\alpha)$ on $[a, b]$ and if $\lvert f(x) \rvert \leq M$ on $[a, b]$, then $$ \left\lvert \int_a^b f d\alpha \right\rvert \leq M \left[ \alpha(b) - \alpha(a) \right]. $$
Here is my proof of this assertion.
As $\lvert f(x) \rvert \leq M$ on $[a, b]$, so $-M \leq f(x) \leq M$ on $[a, b]$ and for every partition $P$ of $[a, b]$, we have $$ -M \left[ \alpha(b) - \alpha(a) \right] = - M \sum_{i=1}^n \left[ \alpha \left( x_i \right) - \alpha \left( x_{i-1} \right) \right] \leq L(P, f, \alpha ) \leq U(P, f, \alpha) \leq M \sum_{i=1}^n \left[ \alpha \left( x_i \right) - \alpha \left( x_{i-1} \right) \right] = M \left[ \alpha(b) - \alpha(a) \right]. \tag{1} $$ And, as $f \in \mathscr{R}(\alpha)$ on $[a, b]$, so for every partition $P$ of $[a, b]$, we also have $$ L(P, f, \alpha ) \leq \int_a^b f d \alpha \leq U(P, f, \alpha). \tag{2} $$ From (1) and (2) we obtain $$-M \left[ \alpha(b) - \alpha(a) \right] \leq L(P, f, \alpha ) \leq \int_a^b f d \alpha \leq U(P, f, \alpha) \leq M \left[ \alpha(b) - \alpha(a) \right], $$ and so $$-M \left[ \alpha(b) - \alpha(a) \right] \leq \int_a^b f d \alpha \leq M \left[ \alpha(b) - \alpha(a) \right], $$ which implies that $$ \left\lvert \int_a^b f d \alpha \right\rvert \leq M \left[ \alpha(b) - \alpha(a) \right], $$ as required.
Is this proof lacking in logic, rigor, or presentation?
