This question is related to this question (for which I still don't know the answer). There I state this result but for Riemann-Stieltjes like this:

If $f,\alpha$ are continuous and $\alpha$ increasing, $f\geq0$ and $\int_a^b f(x)d\alpha=0$ then $f=0$.

I tried to follow the same proof but I never use the continuity of $\alpha$, so I think is wrong...which makes me wonder if this is even true! Seems true because $\alpha$ is increasing, so in the sums we have a term that will be $>0$

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    $\begingroup$ I think you need $\alpha$ to be strictly increasing. $\endgroup$ – copper.hat May 8 '15 at 2:35
  • $\begingroup$ Yes, I guess it has to be because if it's constant where $f$ is positive, that part in Riemann sums is going away! :S $\endgroup$ – Lotte May 8 '15 at 2:45

I am assuming that $\alpha$ is strictly increasing.

Suppose $f(x^*) >0$, and let $x_1\le x^* \le x_2$ with $x_2-x_1 >0$ be such that $f(x) \ge \epsilon >0 $ for $x \in [x_1,x_2]$.

Then consider the partition $P=(a, x_1, x_2,b)$. Then $L(f,P,\alpha) \ge \inf_{x \in [x_1,x_2]} f(x) (\alpha(x_2)-\alpha(x_1)) \ge \epsilon (\alpha(x_2)-\alpha(x_1)) >0$.

Since $\sup_\pi L(f,\pi,\alpha) = \int_a^b f d \alpha = 0$, we obtain a contradiction.

You don't need $\alpha$ to be continuous, just strictly increasing.


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