Here is the link to my Math SE post on Probs. 10 (a) through (c), Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Probs. 10 (a), (b), and (c), Chap. 6, in Baby Rudin: Holder's Inequality for Integrals
Here is the link to my first Math SE post on Prob. 10 (d), Chap. 6, in Baby Rudin:
Prob. 10 (d), Chap. 6, in Baby Rudin: Holder's Inequality for Improper Integrals
And, here is the link to my Math SE post on Prob. 8, Chap. 6, in Baby Rudin:
Prob. 8, Chap. 6, in Baby Rudin: The Integral Test for Convergence of Series
Here I'll be attempting a proof of the Holder's inequality for the improper integrals defined in Prob. 8, Chap. 6, in Rudin.
My Attempt:
Let $p$ and $q$ be positive real numbers such that $1/p + 1/q = 1$.
Suppose that $f$ and $g$ are complex functions which are Riemann-Stieltjes integrable with respect to a monotonically increasing function $\alpha$ on $[a, b]$ for every $b > a$, where $a$ is a fixed real number. Then the holder's inequality gives $$ \left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert \leq \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q}. \tag{0} $$
Suppose that the integrals $\int_a^\infty f g \ \mathrm{d} \alpha$, $\int_a^\infty \lvert f \rvert^p \ \mathrm{d} \alpha $, and $ \int_a^\infty \lvert g \rvert^q \ \mathrm{d} \alpha $ all converge.
Then by the definition in Prob. 8, Chap. 6, in Rudin, we have $$ \int_a^\infty f g \ \mathrm{d} \alpha = \lim_{b \to \infty} \int_a^b f g \ \mathrm{d} \alpha, \tag{1} $$ $$ \int_a^\infty \lvert f \rvert^p \ \mathrm{d} \alpha = \lim_{b \to \infty} \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha , \tag{2} $$ and $$ \int_a^\infty \lvert g \rvert^q \ \mathrm{d} \alpha = \lim_{b \to \infty} \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha . \tag{3} $$
So,
$$ \begin{align} \left\lvert \int_a^\infty f g \ \mathrm{d} \alpha \right\rvert &= \left\lvert \lim_{b \to \infty} \int_a^b f g \ \mathrm{d} \alpha \right\rvert \qquad \mbox{ [ by (1) above ] } \\ &= \lim_{b \to \infty} \left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert \qquad \mbox{ [ using the continuity of the map $t \mapsto \lvert t \rvert$ ] } \\ &\leq \lim_{b \to \infty} \left[ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \right] \\ & \qquad \qquad \mbox{ [ using (0) and a property of the limits ] } \\ &= \lim_{b \to \infty} \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \lim_{b \to \infty} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \qquad \mbox{ [ by Theorem 4.4 (b) in Rudin ] } \\ &= \left( \lim_{b \to \infty} \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \lim_{b \to \infty} \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \\ & \ \ \ \mbox{ [ using the continuity of the map $y \mapsto y^r$ for $y \geq 0$, $r$ being a given real number ] } \\ &= \left( \int_a^\infty \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^\infty \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q}. \qquad \mbox{ [ by (2) and (3) above ] } \end{align} $$
Is this proof correct? If so, then is it rigorous enough too for Rudin?
If incorrect, then where?
How to show that the map $y \to y^r$ for $y \geq 0$ is continuous, for any given real number $r$, especially when $r$ is irrational?