Let functions $f$ and $h$ be Riemann integrable and $H(x) = \int_a^x h(t)dt$. Is it always true that the Riemann-Stieltjes integral $\int_a^bf(x)dH(x)$ exists and $\int_a^b f(x) dH(x) = \int_a^bf(x)h(x) dx$?

I remember seeing this used in a reference without a proof. How is it proved?

The closest I could find was the more restrictive Theorem 6.17 in Principles of Mathematical Analysis by Rudin. He proves $\int_a^b fd\alpha = \int_a^bf(x) \alpha'(x) dx$ when $\alpha$ is differentiable in $[a,b]$ and the derivative $\alpha'$ is Riemann integrable.


In general, $H$ as defined is continuous and of bounded variation. Hence, the integral $\int_a^b f \, dH$ exists. Since $H$ has bounded variation, the derivative $h = H'$ exists almost everywhere and the result follows.

More work is needed to prove this under weaker assumptions than what is given in Rudin's theorem.

See here for an elementary proof using Riemann-Stieltjes sums.

  • $\begingroup$ Thanks your response and link were very helpful. I was looking for the proof. $\endgroup$ – AlRacoon Jan 11 '18 at 22:43

Theorem: Let $h$ be a real Riemann integrable function on $[a,b]$, and let $f$ be a bounded real function on $[a,b]$. Then $f$ is Riemann-Stieltjes integrable with respect to $H(x)=\int_a^x h(t)dt$ iff $fh$ is Riemann integrable on $[a,b]$. And, in that case, $$ \int_{a}^{b}fdH = \int_{a}^{b}fhdx. $$ Proof: If you have an augmented partition $\mathcal{P}$ of $[a,b]$, then $$ \sum_{\mathcal{P}}f(t_j^*)\Delta_j H-\sum_{\mathcal{P}}f(t_j^*)h(t_j^*)\Delta_j x = \sum_{\mathcal{P}}f(t_j^*)\int_{t_{j-1}}^{t_j}\{h(t)-h(t_j^*)\}dt. \;\;(*) $$ Let $M$ be a bounded for $f$. Then the right side is bounded absolutely by $$ M\sum_{\mathcal{P}}\omega(h,I_j)\Delta_j x \le M( \overline{S}_{\mathcal{P}}(h)-\underline{S}_{\mathcal{P}}(h)). $$ Here $\omega(h,I_j)$ is the oscillation of $h$ over the partition interval $I_j$, and the terms on the right are upper and lower sums. So that means the right side of $(*)$ tends to $0$ as the norm of the partition tends to $0$. Therefore, the limit of one of the sums on the left of $(*)$ exists iff the limit of the other sum on the left exists and, in that case, the two limits are equal. $\;\;\blacksquare$

  • $\begingroup$ Thanks for confirming that this is true. $\endgroup$ – AlRacoon Jan 11 '18 at 22:43
  • $\begingroup$ @AlRacoon : I've supplied the proof for you now. Let me know if you spot any errors in it. $\endgroup$ – DisintegratingByParts Jan 11 '18 at 23:15

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