Say $f\ge 0$ and $f \in R(\alpha)$ over a finite interval, is $f^p$ integrable in the same interval? The title pretty much says it, $p$ is some real $>0$, $a$ and $b$ both real, $f$ bounded and $\int d\alpha$ is the Riemann-Stieltjes integral.
It  is trivially true if $f>0$ since then $f^p = e^{p \log f}$ over the whole interval but I'm not sure how to start if we allow points with $f(x)=0$. In this case $f^p$ is not the composition of a continuous function with $f$ so the trivial argument valid before is useless.
My analysis background is Rudin's PoMA ch1-6 so a answer without measure theory would be welcome.
 A: The theorem you want is Theorem 6.11 in Rudin POMA. The function $f^p$ is given by the composition of f with the continuous function $h(t)=|t|^p$. I won't reproduce the proof of the theorem here, but basically, you first use uniform continuity of the function $h$ in the interval $[0,M]$, where $M=\sup f$, to find for any given $\varepsilon>0$ a $\delta>0$ (that you can always take less than $\varepsilon$) such that $|a^p-b^p|\le \varepsilon$ for all $a,b\in [0,M]$ with $|a-b|\le \delta$. Then you apply the definition of Riemann-Stieltjes integral to find a lower sum and an upper sum for $f$ which differ by $\delta^2$. Then you study the sup and inf of $f^p$ in each of the partitions considering the cases in which $\sup_{I_i} f-\inf_{I_i}\le \delta$ or $\ge \delta$. 
A: Your problem is not that complicated as you are trying to think of it. The essence of the problem can be reduced to a simpler question

What is $a^{p}$ when $a = 0$ and $p$ is a positive real number?

If $p$ is a positive rational number then the usual algebraic definition of $a^{p}$ says that it is $0$ when $a = 0$. The real issue here is the case when $p$ is a positive irrational number like $\sqrt{2}$. So our question is equivalent to

What is the value of $0^{\sqrt{2}}$?

One option is to keep the symbol $0^{\sqrt{2}}$ as undefined and mention that $a^{p}$ for irrational positive $p$ makes sense only when $a > 0$. Another option (which works only when $p > 0$) is to define $0^{p}$ as $\lim_{a \to 0^{+}}a^{p} = 0$. Thus $0^{\sqrt{2}} = 0$.
This appears to make sense because of another way of looking at irrational powers. When $p$ is irrational and positive then we have a sequence of positive rational numbers $p_{n}$ such that $p_{n} \to p$ as $n \to \infty$. And hence we should have $$0^{p} = \lim_{n \to \infty}0^{p_{n}} = 0$$ This option thus in effect makes $g(x) = x^{p}$ continuous in $[0, \infty)$ and your problem is easily solved.
The other option to keep $0^{p}$ as undefined for positive irrational $p$ requires a bit more care to handle. In order to discuss the integrability of $f^{p}$ we need to find out points where $f$ vanishes and at these points assume that $f^{p}$ can be given any value. If there are only a finite number of such points then $f^{p}$ is integrable. Otherwise we can't guarantee the integrability of $f^{p}$. A trivial example is when $f$ is identically $0$ in $[a, b]$ and then $f^{p}$ is undefined for all points of $[a, b]$ and hence it does not really make sense to talk of integrability of $f^{p}$.
A: Consider the function $f(x) =\frac{1}{\sqrt{x}}$ and $f^2(x)$ with $\alpha(x) =x$ on the interval $(0,1) $. 
