Let $\alpha: [a,b] \rightarrow \mathbb{R}$ be monotonically increasing and let $f \in \mathscr{R}(\alpha)$. Prove that $|f| \in \mathscr{R}(\alpha)$.

Question

So Rudin defines $|f| \in \mathscr{R}(\alpha) \iff \underline \int_{a}^{b} |f| \,d\alpha = \overline \int_{a}^{b} |f| \,d\alpha$.

There's a Theorem which states: $\underline \int_{a}^{b} f \,d\alpha \leq \overline \int_{a}^{b} f \,d\alpha$, which gives us: $\underline \int_{a}^{b} |f| \,d\alpha \leq \overline \int_{a}^{b} |f| \,d\alpha$.

So: we are left to show that $\overline \int_{a}^{b} |f| \,d\alpha \leq \underline \int_{a}^{b} |f| \,d\alpha $.

Proof: Indeed, since $f \in \mathscr{R}(\alpha)$, it satisfies the Riemann condition on $[a,b]$ so that:

$U(f,\alpha;P) \leq L(f,\alpha;P) + \epsilon \leq L(|f|,\alpha;P) + \epsilon \leq \underline \int_{a}^{b} |f| \,d\alpha + \epsilon.$

Since $-|f(x)| \leq f(x) \leq |f(x)|,$ we have that: $U(-|f|,\alpha;P) \leq U(f,\alpha;P) \leq U(|f|,\alpha;P)$. So at best, I can say:

$ - \overline \int_{a}^{b} |f| \,d\alpha \leq U(-|f|,\alpha;P) \leq U(f,\alpha;P) \leq \cdots \leq \underline \int_{a}^{b} |f| \,d\alpha + \epsilon. $

I've been stuck here for a while now. I'm not sure if I have reached a dead-end or if I'm close, but I have no idea how to proceed from here. Any and all help is very greatly appreciated. Thanks!

Answer 1

For any subinterval $I$ of a partition, we have, using $||a|-|b|| \leqslant |a-b|$,

$$\sup_{x \in I}|f(x)| - \inf_{x \in I}|f(x)| = \sup_{x,y \in I}||f(x)| - |f(y)|| \leqslant \sup_{x,y \in I}|f(x) - f(y)| = \sup_{x \in I}f(x) - \inf_{x \in I}f(x) $$

Forming Riemann-Stieltjes sums over partition intervals we get

$$U(|f|,\alpha,P) - L(|f|,\alpha,P) \leqslant U(f,\alpha,P) - L(f,\alpha,P).$$

Now you can invoke the Riemann criterion for $f$ and apply it to $|f|.$