# Coincidence of functions defining Riemann-Stieltjes integral

I read in Kolmogorov-Fomin's Элементы теории функций и функционального анализа that (p. 372 here), if the Riemann-Stieltjes integrals $$\int_a^b f(x) d\Phi_1(x)\quad\text{ and }\quad\int_a^b f(x) d\Phi_2(x),$$ which are the same as Lebesgue-Stieltjes integrals for $f\in C[a,b]$, with respect to two functions of bounded variation $\Phi_1:[a,b]\to\mathbb{R}$ and $\Phi_2:[a,b]\to\mathbb{R}$, are equal for all $f\in C[a,b]$, then $\Phi_1$ and $\Phi_2$ coincide in every point where $\Phi_1-\Phi_2$ is continuous up to a constant (and I know that a function of bounded variation on $[a,b]$ is continuous except for countably many point).

That does not seem so trivial to me, although the book does not prove it. I know that if $\Phi_1-\Phi_2$ is constant except for a countable number of points contained in $(a,b)$ then $\forall f\in C[a,b]$ $\int_a^b f(x) d\Phi_1(x)=\int_a^b f(x) d\Phi_2(x)$, but I cannot prove the converse to myself. Could anybody prove this interesting fact? $\infty$ thanks!!!

EDIT: Corrected imprecision thanks to T.A.E.

• A couple of technicalities. You used $\Phi_{1}$ twice in your first sentence. The second should be $\Phi_{2}$. Also, $\int_{a}^{b}fd\Phi$ is not defined unless $\Phi$ is defined on $[a,b]$, unless you are taking a limit $\int_{a}^{b-}fd\Phi$. Nov 5, 2014 at 15:14
• Thank you so much! Sorry, I wrongly wrote the index in the second $\Phi_k$. As to $\Phi$'s domain, Kolmogorov-Fomin's original text, which I notice now to be slightly different from the English one, which I hadn't carefully read, uses a $\Phi$ defined and continuous from the left on $[a,b)$ and uses $\Phi(x_n):=\Phi(b^-)$ (where $x_n=b$) in the Riemann-Stieltjes sum. I don't know why it does so, but it does, really. Nov 5, 2014 at 18:00

I'm going to use the Riemann-Stieltjes integral to prove this. Assume that $\int_{a}^{b}fd\Phi=0$ for all $f\in C[a,b]$ and some $\Phi$ of bounded variation on $[a,b]$. Suppose $t \in [a,b)$ is a point of continuity of $\Phi$. For any $\epsilon \in [0,b-t)$, define $$f_{t,\epsilon}(s)=\left\{\begin{array}{ll} 1, & a \le s \le t,\\ 1-\frac{1}{\epsilon}(s-t), & t < s \le t+\epsilon,\\ 0, & t+\epsilon < t \le b. \end{array}\right.$$ This function $f_{t,\epsilon}$ is continuous on $[a,b]$. The nice part about Riemann-Stieltjes is that one may always integrate by parts. Hence, $$0= \int_{a}^{b}f_{t,\epsilon}d\Phi = \int_{a}^{t}d\Phi +\int_{t}^{t+\epsilon}f_{t,\epsilon}d\Phi \\ = \Phi(t)-\Phi(a)+f_{t,\epsilon}\Phi|_{t}^{t+\epsilon}-\int_{t}^{t+\epsilon}\Phi df_{t,\epsilon}\\ = \Phi(t)-\Phi(a)-\Phi(t)+\frac{1}{\epsilon}\int_{t}^{t+\epsilon}\Phi(t)dt \\ = \frac{1}{\epsilon}\int_{t}^{t+\epsilon}\Phi(t)\,dt-\Phi(a).$$ Therefore, if $\Phi$ is continuous at $t$, $$0=\lim_{\epsilon\downarrow 0}\int_{a}^{b}f_{\epsilon,t}d\Phi = \lim_{\epsilon\downarrow 0}\frac{1}{\epsilon}\int_{t}^{t+\epsilon}\Phi(s)\,ds -\Phi(a) = \Phi(t)-\Phi(a).$$ It follows that $\Phi(t)=\Phi(a)$ for all points of continuity $t \in [a,b)$. Similarly, you can argue that $\Phi(t)=\Phi(b)$ for all points of continuity $t\in (a,b]$. (In fact, if you look carefully at the expressions, you see $\Phi(a)=\Phi(t+0)$ for all $t\in [a,b)$ and $\Phi(t-0)=\Phi(b)$ for all $t \in (a,b]$, but these identities can be deduced from the weaker statement, too.)
• Thank you very much! You use a wonderfully interesting properties my book doesn't talk about: I think you mean $\Phi\cdot f|_a^b=\int_a^bfd\Phi+\int_a^b\Phi df$ for two left-continuous functions of bounded variation. I find nothing about it: could you give a link to a proof (or write one)? $\infty$ thanks in any case!!! Nov 5, 2014 at 18:18
• Here's the theorem: Let $f$, $g$ be functions on $[a,b]$ (no assumptions.) The Riemann-Stieljtes integral $\int_{a}^{b}f dg$ exists iff $\int_{a}^{b}g df$ exists and, in that case, $\int_{a}^{b}f dg = (fg)|_{a}^{b}-\int_{a}^{b}gdf$. It doesn't get any better than that in a classical setting. Of course, if $g$ is of bounded variation and $f$ is continuous, then both integrals exist, and 'integration-by-parts' holds. No normalization required. en.wikipedia.org/wiki/… Nov 5, 2014 at 18:24
• Very interesting fact. I'll search the Web for a proof and, if I don't find one, I'll post a specific question since it's a topic that I think worth a separate thread. Anyhow I suspect that the same result can be inferred from the fact that $|\int_t^{t+\epsilon} fd\Phi|\leq\max_{s\in[t,t+\epsilon]}|f(s)|\cdot V_t^{t+\epsilon}(\Phi)$ and $V_t^{x}(\Phi)$ is continuous where $\Phi(x)$ is. Thank you so much again!!! Nov 5, 2014 at 19:44
• @DavideZena : The only identity I needed to prove the result was $f(r)[g(s)-g(r)]=fg|_{r}^{s}-g(s)[f(s)-f(r)]$. When considering $\sum f(t_{n}^{\star})\Delta_{n}g$, you can refine $[t_{j-1},t_{j}]$ to $[t_{j-1},t_{j}^{\star}]\cup[t_{j}^{\star},t_{j+1}]$ and use $t_{j}^{\star}$ on both new intervals. That doesn't change the sum. Then you rewrite with identity as $fg|_{a}^{b}-\sum_{j}g(t_{n}^{\star\star})\Delta_{n}f$ and, if $\int_{a}^{b}gdf$ exists, the right side is within $\epsilon$ of $fg|_{a}^{b}-\int_{a}^{b}gdf$ if $\|\mathcal{P}\| < \delta$. So you get $\int_{a}^{b}fdg$ and the equality. Nov 6, 2014 at 0:09