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I'm currently doing a question proving that if $f$ is a monotone function and $\alpha$ is a continuous and increasing function, then $f$ is Riemann-Stieltjes integrable.

I know that when $f$ is monotone the $$\sum_{i=1}^{n}(M_i -m_i)(\alpha(x_i)-\alpha(x_{i-1}))= (\sup(f)-\inf(f))((\alpha(x_i)-\alpha(x_{i-1}))$$

My question is, are there any other cases, other than $f$ being monotone, where this is true?

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  • $\begingroup$ ??? The equation $\sum_{i=1}^{n}(M_i -m_i)(\alpha(x_i)-\alpha(x_{i-1}))= (\sup(f)-\inf(f))((\alpha(x_i)-\alpha(x_{i-1}))$ makes very little sense; there's no $\sum$ on the right side. Maybe this was a typo? I cannot quess what true equation it might have been a typo for... $\endgroup$ – David C. Ullrich May 8 '16 at 16:50
  • $\begingroup$ I guess you already know (with proof) the standard result that if $\alpha$ is monotone and $f$ is continuous then Riemann-Stieltjes integral $\int f\,d\alpha$ exists. It is a simple matter now to use integration by parts to show that $\int \alpha\,d f$ also exists. $\endgroup$ – Paramanand Singh May 10 '16 at 16:38
  • $\begingroup$ Also there is some typo in the equation you have written and I don't think this equation has any relevance to prove the result you want to prove. $\endgroup$ – Paramanand Singh May 10 '16 at 16:39
  • $\begingroup$ "Integration by parts" in my previous comment means the following result: if $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$ then $\alpha$ is also Riemann integrable with respect to $f$ on $[a, b]$ and $$\int_{a}^{b}f\,d\alpha + \int_{a}^{b}\alpha\,d f = f(b)\alpha(b) - f(a)\alpha(a)$$ $\endgroup$ – Paramanand Singh May 10 '16 at 16:43

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