1
$\begingroup$

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 $>0$

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.