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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.

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    $\begingroup$ 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 $\int_{a}^{b-}fd\Phi$. $\endgroup$ Nov 5, 2014 at 15:14
  • $\begingroup$ 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. $\endgroup$ Nov 5, 2014 at 18:00

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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.)

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  • $\begingroup$ 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!!! $\endgroup$ Nov 5, 2014 at 18:18
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    $\begingroup$ 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/… $\endgroup$ Nov 5, 2014 at 18:24
  • $\begingroup$ 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!!! $\endgroup$ Nov 5, 2014 at 19:44
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    $\begingroup$ @DavideZena : The reason, of course, that the classical works so well is that, after a refinement of partitions at the evaluation points, you get a summation by parts that is exact. And it is these types of arguments that are ideally suited for functions of bounded variation. The continuous is just a little easier to use than the discrete, but you could argue all of this using summation by parts instead if you wanted to, but the continuous is just so much eaiser to use, and it connects with the Riemann integral as well. These arguments seem to get to the heart of the issue, more than tricks. $\endgroup$ Nov 5, 2014 at 19:57
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    $\begingroup$ @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. $\endgroup$ Nov 6, 2014 at 0:09

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