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According to https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral, if $g$ is a real-valued function of bounded variation and $f$ is another real-valued function, then one can define the Riemann--Stieltjes integral $\int_s^t f d g = \lim_{\lvert \pi \rvert \to 0} \sum_{x,y\in \pi} f(x)(g(y)-g(x))$, where the limit is over all sequences of partitions $\pi$ of $[s,t]$ with vanishing mesh $\lvert \pi \rvert = \sup_{x,y \in \pi} \lvert v - u\rvert$. Is the converse true? I.e., if the Riemann--Stieltjes integral exists, must $g$ be of bounded variation?

xy z
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    Are you asking if $g$ must be of bounded variation when the integral $\int_a^b f , dg$ exists for every continuous integrand $f$? The answer to that is yes. On the other hand, if a Riemann-Stieltjes integral $\int_a^b f , dg$ exists then it is not necessary that $g$ be of bounded variation -- even with restrictions on $f$ (non-constant , continuous, etc.) – RRL Feb 03 '24 at 18:49
  • @RRL Ah, I think I found a reference for this in (2.21) Exercise of Revuz and Yor's "Continuous Martingales and Brownian Motion". So this is only true for continuous integrands then. If $f$ were ''nicer'', e.g., Hölder continuous, then $g$ might be of unbounded variation? – xy z Feb 03 '24 at 19:56
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    See here for a proof that the existence of $\int_a^bf ,dg$ for all continuous $f$ implies $g \in BV$. I also think you can find functions $f$ and $g$ both of unbounded variation such that $\int_a^b f,dg$ exists. I have to look at that reference. I'm still not exactly sure what you are asking. – RRL Feb 03 '24 at 20:06
  • Thanks, it is a nice proof, I think I understand it. As far as I can see, it is essentially the same argument as outlined in my earlier reference, i.e., using the uniform boundedness principle. – xy z Feb 03 '24 at 20:44

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