18
$\begingroup$

I certainly have a question, but i don't know what the best title should be. Please edit the title if there is a better one :)

And I believe, to get a better answer, it would be good to explain exactly what i feel and know..

========

Riemann-Stieltjes Integral (Darboux Sum) can be defined for bounded integrand $f$ and nondecreasing $\alpha$. ($\alpha$ does not need to be continuous). However, as far as i know, Lebesgue-Stieltjes Integral requires integrator to be right or left continuous. Moreover, even if there is a way to define measure associated to nondecreasing discontinuous function, integral w.r.t to this function cannot be identical to that of Riemann-Stieltjes. (I'll explain it below)

Below is the result i have proved:

Theorem

Let $\alpha:\mathbb{R}$ be a monotonically increasing right-continuous function.

Let $f:[a,b]\rightarrow \mathbb{R}$ be a RS-integrable function w.r.t $\alpha$ on $[a,b]$.

Let $\mu_\alpha$ be the LS-measure w.r.t $\alpha$.

Then, there exists $F\in L^1(\mu_\alpha)$ such that $\forall x\in(a,b], F(x)=f(x)$ and $\int_a^b f d\alpha = \int_{(a,b]} F d\mu_\alpha$

Since when one wants to use RS-integration, one's interest is exactly in $[a,b]$. So rather than using full domain of $\alpha$ as illustrated in the above theorem, we can replace it by $\alpha(x)=\alpha(a)$ for $x<a$. Then, this new $\alpha$ is continuous at $a$. Thus $\int_a^b f d\alpha = \int_{[a,b]} F d\mu_\alpha$. Hence, in the case $\alpha$ is right continuous, it can be understood that LS-integration is an extension of RS-integration.

However, in general, if $\alpha$ is not right or left continuous, LS-integration cannot be a generalization of RS-integration.

Let $\alpha$ be a nondecreasing function. Suppose there exists a measure $\mu$ such that for some type of interval $【a,b】$, $\mu(【a,b】)=\alpha(b)-\alpha(a)$. (【a,b】denotes one of open,closed,half-open and whatsoever that has an interval form.) Then, the continuity of $\mu$ shows that the only choice that $\mu$ could be a measure is that $\mu((a,b))=\alpha(b-)-\alpha(a+)$.

Indeed, we can construct such measure. Define $F(x)=\alpha(x+)$. Then $F$ is right-continuous nondecreasing function and the LS-measure $\mu$ associated with $F$ is the measure such that $\mu((a,b))=\alpha(b-) - \alpha(a+)$.

In this case, RS-integral and LS-integral of a bounded function do not coincide in general.

When $\alpha$ is continuous, LS seems more natural than RS to me. However, when $\alpha$ is not continuous, LS does not seem natural at all to me..

I'm an undergraduate, so i don't know many cases where integration is useful. I don't have any clue why one should know LS integration. Is there any natural-science example of a function that is only LS integrable but not RS integrable? Or is there any mathematical example that LS integration is useful?

I understand how Lebesgue Integration flows better than Riemann Integration, but I don't see any reason why LS-integration is in need.

$\endgroup$
  • $\begingroup$ @user117818 Yes, i know. But that seems to give a motivation to "Lebesgue integeation", not LS integration. Why one needs to integrate Dirichlet function w.r.t a nondecreasing function? As far as i know, motivation of Stieltjes integration begins from physics and i haven't seen a discontinuous function that arises from physics. And it seems to me when an integrator is discontinuous, RS integration seems more flexible than LS integration. $\endgroup$ – John. p Feb 3 '14 at 4:24
  • $\begingroup$ Oh sorry I didn't read carefuly. $\endgroup$ – IAmNoOne Feb 3 '14 at 6:13
5
$\begingroup$

What I have read (I believe in "A Modern Approach to Probability Theory" by Fristedt and Gray) is that one of the primary reasons for using Riemann-Stieltjes integration is the convenient integration by parts formula that comes with it.

However, the Riemann-Stieltjes integral suffers several of the same problems as the Riemann integral which the Lebesgue integral lacks. Namely, compared to the Lebesgue integral, the Riemann-Stieltjes Integral can integrate fewer functions.

So what do we do if we want to use that convenient integration by parts for formula for functions which are Lebesgue- but not Riemann-integrable? We use the Lebesgue-Stieltjes integral!

https://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration#Integration_by_parts

This turns out to be especially important in stochastic calculus.

The Riemann-Stieltjes integral also gives convenient formulae for the expectation of a functions of a random variable $\mathbb{E}[f(X)]$. Again, however, the "Riemann" part places restrictions on what functions $f$ we can consider which are lifted by using the Lebesgue-Stieltjes integral instead. Admittedly more often than not $f$ will be continuous and it won't matter which of the two we use, but this will not always be the case.

Moreover, it is usually easier to work with the Lebesgue-Stieltjes integral in the context of abstract measure theory or multiple measures on the same space (as happens often in probability theory). Since one generally does not want to focus too much attention on such generalities, it is usually easier to assume that all integrals are Lebesgue-Stieltjes integrals unless stated otherwise, rather than having to stop at several junctures to explain why the Riemann-Stieltjes integral is compatible in this case (and sometimes the reason would just be that it is equivalent to the Lebesgue-Stieltjes integral for sufficiently "nice" functions).

As far as applications in other fields are concerned, besides integration by parts, I do not know.

$\endgroup$

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.