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 $&gt;0$, $a$ and $b$ both real, $f$ bounded and $\int d\alpha$ is the Riemann-Stieltjes integral.

It  is trivially true if $f&gt;0$ 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.
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 $\varepsilon&gt;0$ a $\delta&gt;0$ (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 $\ge \delta$.
Consider the function $f(x) =\frac{1}{\sqrt{x}}$ and $f^2(x)$ with $\alpha(x) =x$ on the interval $(0,1) $.