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Let $f,\phi:[a,b] \rightarrow \mathbb{R}$ be a continuous map, and the function of bounded variation respectively. And, $g$ is a continuous map on $[a,b]$. Then, following results hold.

(i) A map $ \psi:[a,b] \rightarrow \mathbb{R}$ s.t. $\psi(x)=\int_a^xfd\phi$ is of bounded variation.

(ii) $\int_a ^b gd\psi =\int_a^bfgd\phi$.

(i) is clearly proved from the definition of the bounded variation. But, I think the proof of the part (ii) is difficult.

How can I solve this? I tried the integration by parts - Riemann-Stieltjes version. But, there is no gain.

Chris kim
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1 Answers1

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Recall that the Riemann-Stieltjes integral $\int_a^b g d\psi$ is defined as the limit of $$\sum_{i=1}^N g(\xi_i) (\psi(t_i)-\psi(t_{i-1}))$$

where $a=t_0\le t_1\le \cdots\le t_N = b$, $t_{i-1}\le \xi_i\le t_i$ as $\max_i |t_i - t_{i-1}|\to 0$.

We have $$\psi(t_i)-\psi(t_{i-1})=\int_{t_{i-1}}^{t_i} f d\phi=f(\zeta_i)(\phi(t_i)-\phi(t_{i-1}))$$

for some $\zeta_i\in [t_{i-1},t_i]$ by the mean value theorem.

Thus,

$$\sum_{i=1}^N g(\xi_i) (\psi(t_i)-\psi(t_{i-1}))=\sum_{i=1}^N g(\xi_i) f(\zeta_i) (\phi(t_i) - \phi(t_{i-1}))$$

As $\max_i |t_i - t_{i-1}|\to 0$ the right hand side converges to $\int_a^b fg d\phi$.

J.R.
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  • Does the mean value theorem holds well? I think this proof assumes that $\phi$ is monotonic. – Chris kim Mar 26 '16 at 14:44
  • Yes, the mean value theorem holds. Our $phi$ is of bounded variation, so it equals the difference of two monotonic functions. – J.R. Mar 26 '16 at 14:53
  • Oh, I had some misunderstanding. It's clear because any bounded variation can be decomposed to the difference of increasing functions. :) – Chris kim Mar 26 '16 at 14:53
  • Ah.... I didn't know that. ;( Where is the "accept" button? – Chris kim Mar 26 '16 at 16:00
  • As you said, I have used this site. But, I didn't know that rule and how to accept the answer( manual) because I used this site sometimes. (Furthermore, I didn't use this site about two years(2014). ;) ) – Chris kim Mar 26 '16 at 16:10