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A Riemann-Stieltjes integral $\int_a^b fdg$ may not exist if $f$ and $g$ are both discontinuous at the same point. With common discontinuity and assuming $g$ is increasing it can be shown that upper and lower sum difference $U(f,g,P) – L(f,g,P) \not\to 0$ as the norm of the partition $\|P\| \to 0$.

References I’ve seen always prove existence with assumptions that $g$ is increasing or bounded variation. Is this just for convenience or is it necessary? Can we say $\int_a^b fdg$ always exists if both $f$ and $g$ are continuous without further hypotheses for $g$?

WoodWorker
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Consider the case $f(x) = 1$ and let $g$ be some continous function with unbounded variation then you can find partitions $(x^n)_n$ of $[a, b]$ for which $\sum_{i = 0}^{n} (g(x^n_{i+1}) - g(x^n_i))$ becomes arbitrarily large (as $n \rightarrow \infty$) so $$ \int_{a}^{b} 1 dg(x) $$ doesn't exist despite the fact that f and g are both continous. Continous increasing functions always have bounded variation over a compact set (it's less than $(b-a) \cdot (g(b) - g(a))$).

OneAndOnlyDaniel
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  • Thanks. I'm aware that a constant function is integrable for any integrator $g$. My question is about existence for general continuous $f$. – WoodWorker Jun 05 '19 at 00:01
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    Also, if $g$ has unbounded variation, then $\sum |g(x^n_{i+1}) - g(x^n_{i})|$ becomes arbitrarily large, not $\sum (g(x^n_{i+1}) - g(x^n_{i})) = g(b) - g(a)$. – WoodWorker Jun 05 '19 at 00:46