Proving that $\int_a^bf^pd\alpha=0$ implies $\int_a^bfd\alpha=0$ for $f \in \mathcal{R}(\alpha)$ and $f\ge0$ on $[a,b]$ The question posed may seem trivial to many. But I couldn't find a trivial solution to above within the theory upto Riemann-Stieltjes integral. I woudn't be considering arguments from Lebesgue theory as I haven't studied it so far. Here I would try to prove it from within the theory of Riemann-Stieltjes integral.
No assumption is made regarding continuity of $f$. 
$\alpha$ is monotonically increasing on $[a,b]$  
Let $A=\alpha(b)-\alpha(a)$
Let $M=sup\{f(x)|x \in [a,b]\}$
$f \in \mathcal{R}(\alpha)$ on $[a,b]$  
$f \ge 0$ on $[a,b]$
Given $\int_a^bf^pd\alpha=0$  
For any $\epsilon > 0$ there exists a partition $P$ of $[a,b]$, $P=\{a=x_0<x_1...<x_i<...<x_n=b\}$ such that $U(P,f^p,\alpha)=\sum_1^nM_i(\alpha(x_i)-\alpha(x_{i-1}))< \epsilon$ where $M_i=sup\{(f(x))^p|x \in [x_i,x_{i-1}]\}$. For any $k>0$ let $K=\{i|M_i>k^p\}$. Let $l(K)=\sum_{i \in K}(\alpha(x_i)-\alpha(x_{i-1}))$.  
Now $$k^p.l(K)<\epsilon \implies l(K)<\frac{\epsilon}{k^p}$$
Choosing $\epsilon< k^p\delta \implies l(K)<\delta$ for any $\delta > 0$  
For the same partition $P, K'=\{i|M'_i>k\}=K$ where $M'_i=sup\{f(x)|x \in [x_i,x_{i-1}]\}$  
$U(P,f,\alpha)\le M.l(K)+k.A<M\delta+kA$.  
Both $\delta$ and $k$ can be chosen to be arbitrarily small. Hence $U(P,f,\alpha)$ can be made to be as close to $0$ as we wish.
This proves $\int_a^bfd\alpha=0$
Any trivial proof would be welcome.
 A: Take a step function $0\le s\le f$. Then $0\le s^p\le f^p$ and so $0\le \int s^p dx\le\int f^pdx=0$. Hence $\int s^pdx=0$ but since s is piecewise constant, this implies that $s=0$. In particular, $\int s dx=0$ and since this is true for every step function below $f$ then the lower Riemann integral of $f$ must be zero. 
A: I believe, some constraints must be imposed on $p$, at least it must not be $0$. But, for a large class of values of $p$, specifically $p>1$, Hölder's inequality can be used. E.g.
$$0\leq \int_{a}^{b}fd\alpha = \int_{a}^{b}f\cdot 1d\alpha \leq \left(\int_{a}^{b}f^pd\alpha \right)^{\frac{1}{p}}\cdot \left(\int_{a}^{b}1^qd\alpha \right)^{\frac{1}{q}}=0$$
A: You do not need p>1. My previous answer works. If you don't like step functions, take a partition $a=t_0<\cdots<t_n=b$ and consider $\inf_{[t_{i-1},t_i]}f$. Then $$0\le \sum_i(\alpha(t_i)-\alpha(t_{i-1}))\inf_{[t_{i-1},t_i]}f^p\le \int_a^b f^pd\alpha=0$$ and so $(\alpha(t_i)-\alpha(t_{i-1}))\inf_{[t_{i-1},t_i]}f^p=0$ for every $i$, which implies that $\alpha(t_i)-\alpha(t_{i-1})=0$ or $\inf_{[t_{i-1},t_i]}f^p=0$. If $\inf_{[t_{i-1},t_i]}f^p=0$, then $\inf_{[t_{i-1},t_i]}f=0$ (just play with the definition of inf). So you have $\sum_i(\alpha(t_i)-\alpha(t_{i-1}))\inf_{[t_{i-1},t_i]}f=0$ for every lower sum and thus $\int_a^b f d\alpha=0$. This is really the same idea behind the other proof.
