# Theorem 6.12 (d) in Baby Rudin: If $\lvert f(x) \rvert \leq M$ on $[a, b]$, then $\lvert \int_a^b f d\alpha \rvert \leq \ldots$

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?

• Sorry but I have to ask you something I do not understand. Why have you discarded the following simpler and more clear (for me) proof? $$\bigg|\int_{a}^{b}{f(\alpha)\,d\alpha}\bigg|\leq \int_{a}^{b}{|f(\alpha)|\,d\alpha} \leq M\int_{a}^{b}{\,d\alpha} = M[\alpha(b)-\alpha(a)]$$. Sorry if this I present is not evident, I am not a matematician and I would like to know why it is not this way. Thank you – HBR Jun 18 '17 at 18:10
• @HBR the reason why I haven't used (the first one of) your inequalities is because in a rigorous treatment of a mathematical discipline one can only use the results that have been established in a particular discussion in the proof of any assertion. In this particular case, Rudin has not (directly) proved the first one of your inequalities. However, of course we can prove it using the machinary developed up to this point in the book! – Saaqib Mahmood Jun 19 '17 at 10:56
• Thank you for your answer!! Kindly appreciate! – HBR Jun 19 '17 at 11:02
• @HBR you're welcome. – Saaqib Mahmood Jun 19 '17 at 11:21