3
$\begingroup$

So, the problem, as in the title, is:

If $f$ is $g$-Riemann-Stieltjes integrable on $[a,b]$, $g \in BV[a,b]$, prove that it's $g$-RS-integrable on every subinterval $[a,c] \subset [a,b]$, where $a < c< b$, as well as on $[c,b]$, and that $$\int_{a}^{b} fdg = \int_{a}^{c} fdg + \int_{c}^{b}fdg.$$

Some of the things I've tried: I know that if I have a partition $P$ on $[a,c]$, I can extend the partition to be a partition $\tilde{P}$ on $[a,b]$ so that $\Delta(P)=\Delta(\tilde{P})$, by retaining the $[a,c]$ part and adding points $c+\Delta(P)/2$, $c+\Delta(P)$, ... , $c+k\Delta(P)/2$, as long as these numbers are in $[a,b]$. However, I'm not sure how to use that to derive $$\textrm {diam}(\{\sigma(P,\xi, f, g): P \textrm{ is a partition on } [a,c] \textrm{ and }\Delta(P)<\delta \}) \leq \textrm{diam}(\{\sigma(P_{1},\xi_{1}, f, g): P_{1} \textrm{ is a partition on } [a,b] \textrm{ and }\Delta(P_{1})<\delta \}),$$

which I have a hint might be true. However, even if I were to derive this, I don't know how that would effectively imply the existence of the RS-integral $$ \int_{a}^{c} fdg. $$

I've also tried comparing this to the proof on Riemann-integrability, but the ones I found and could think of use Darboux sums, which aren't as powerful in the Riemann-Stieltjes case.

In case some people use different notation, here are the definitions of the notation I used:

If $P:a=x_{0},...,x_{n-1},x_{n}=b$, then: $\Delta(P) = \max_{1 \leq i \leq n} (x_{i}-x_{i-1})$, $\sigma(P, \xi, f, g) = \sum_{i=1}^{n} g(\xi_{i})(f(x_{i})-f(x_{i-1}))$, $\xi = (\xi_{1},...,\xi_{n}),$ where $\xi_{i} \in [x_{i-1},x_{i}]$.

EDIT: After taking a look at Rudin's "Principlies of mathematical analysis", I found the proof of this claim. However, Rudin uses a different definition from what I consider the Riemann-Stieltjes integral, namely through the analogue of Darboux sums; the definition I'm using can be found on Wikipedia. Also, I'd like to note that there's an exercise in my textbook that states an alternative definition of the Riemann-Stieltjes integral, the same one that can be found in Rudin, and asks the reader to prove that these definitions are not equivalent.

$\endgroup$
2
$\begingroup$

If the integrator $g$ were monotonically increasing then this could be proved in the same way it is done for the Riemann integral using Darboux sums or other another approach.

Here we have $g \in BV([a,b])$. As a function of bounded variation, $g = \beta_1 - \beta_2$, a difference of two increasing functions. There are many such decompositions and it is not always the case that $f$ which is Riemann-Stieltjes integrable with respect to $g$ is also Riemann-Stieltjes integrable with respect to $\beta_1$ and $\beta_2$. However, there is one decomposition where the integrability always extends, and that is where $\beta_1(x) = V_a^x(\beta)$, the total variation on $[a,x]$, and $\beta_2 = \beta_1 - g$. A proof is given here.

With $\beta_1$ and $\beta_2$ defined in this way, it is straightforward to show that

$$\tag{*}\int_a^b f \, dg = \int_a^b f \, d\beta_1 - \int_a^b f \, d\beta_2 .$$

Since $\beta_1$ and $\beta_2$ are increasing it follows from the statement in the first paragraph that $f$ is $\beta_1-$ and $\beta_2-$RS integrable on every subinterval $[a,c] \subset [a,b]$, and for $j = 1,2$,

$$\tag{**}\int_a^b f \, d\beta_j = \int_a^c f \, d\beta_j + \int_c^b f \, d\beta_j.$$

The argument leading to (*) can be reversed to show that the RS-integrability of $f$ with respect to $\beta_1$ and $\beta_2$ on $[a,c]$ implies the RS-integrability of $f$ with respect to $g$ on $[a,c]$, and, finally,

$$\int_a^b f \, dg = \int_a^c f \, dg + \int_c^b f \, dg.$$

$\endgroup$
  • $\begingroup$ So if $g$ is increasing, are the two definitions of the Riemann-Stieltjes integral (one using Darboux sums, one using partitions and choosing arbitrary points within the subsegments) equivalent? Is that why the Riemann-integral proof would work in that case? Or is because of a different reason? $\endgroup$ – Matija Sreckovic May 3 '18 at 18:04
  • $\begingroup$ Because the equality of the lower and upper Stieltjes-Darboux sums doesn't imply the existence of the Riemann-Stieltjes integral (the "standard" definition version), even when the integrator is only increasing. $\endgroup$ – Matija Sreckovic May 3 '18 at 18:09
  • $\begingroup$ I understand that if the statement is proven in the case where $g$ is increasing, we're "done". However, I don't understand why the case where $g$ is increasing is analogous to the Riemann-integral case. My question is, could you elaborate on why the statement holds when $g$ is increasing, and not from $BV[a,b]$, in light of my previous two comments? $\endgroup$ – Matija Sreckovic May 3 '18 at 18:13
  • $\begingroup$ @MatijaSreckovic: I think you are raising several issues, so let me be sure I understand. First for Riemann integrals there are different definitions using (1) Darboux upper and lower integrals, (2) tagged Riemann sums which converge to the integral as the partition is refined, (3) tagged Riemann sums which converge to the integral as the partition norm goes to zero. For simple Riemann integration they are all equivalent, and you can prove this theorem $\int_a^bf(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$ for any definition. Would you agree? $\endgroup$ – RRL May 3 '18 at 18:21
  • 1
    $\begingroup$ What is your textbook? There are two books that deal with these deeper issues about the RS integral and the multiple definitions. Mathematical Analysis by Apostol and Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger.. You need both because they don't overlap completely. $\endgroup$ – RRL May 3 '18 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.