# The Unapologetic Mathematician

## The Riemann-Stieltjes Integral I

Today I want to give a modification of the Riemann integral which helps give insight into the change of variables formula.

So, we defined the Riemann integral

$\displaystyle\int\limits_a^bf(x)dx$

to be the limit as we refined the tagged partition $x=((x_0,...,x_n),(t_1,...,t_n))$ of the Riemann sum

$\displaystyle f_x=\sum\limits_{i=1}^nf(t_i)(x_i-x_{i-1})$

But why did we multiply by $(x_i-x_{i-1})$? Well, that was the width of a rectangular strip we were using to approximate part of the area under the graph of $f$. But why should we automatically use that difference as the “width”?

Let’s imagine we’re walking past a fence. Sometimes we walk faster, and sometimes we walk slower, but at time $t$ we can measure the height of the fence right next to us: $f(t)$. So what’s the area of the fence? If we just integrated $f(t)$ we’d get the wrong answer. The samples we made when walking fast made fat rectangles, while the samples we made when we were walking slowly got paired with skinny rectangles, but we gave them the same weight if they took the same time to get through that segment of the partition. We need to reweight our sums to compensate for how fast we’re walking!

Okay, so how wide should we make the rectangles? Let’s say that at time $t$ we’re at position $\alpha(t)$ along the fence. Then in the segment of the partition between times $x_{i-1}$ and $x_i$ we move from position $\alpha(x_{i-1})$ to position $\alpha(x_i)$, so we should make the width come out to $(\alpha(x_i)-\alpha(x_{i-1}))$. We’ll put this into our formalism from before and get the “Riemann-Stieltjes sum”:

$\displaystyle f_{\alpha,x}=\sum\limits_{i=1}^nf(t_i)(\alpha(x_i)-\alpha(x_{i-1}))$

And now we can take the limit over tagged partitions as before to get the “Riemann-Stieltjes integral”:

$\displaystyle\int\limits_{\left[a,b\right]}f(x)d\alpha(x)=\int\limits_a^bf(x)d\alpha(x)$

if this limit exists.

Here we call the function $f$ the “integrand”, and the function $\alpha$ the “integrator”. Clearly, the old Riemann integral is the special case when $\alpha(x)=x$.

Immediately from the definition we can see the same “additivity” (using signed intervals) in the region of integration that the Riemann integral had:

$\displaystyle\int\limits_{\left[x_1,x_3\right]+\left[x_3,x_2\right]}f(x)d\alpha(x)=\int\limits_{\left[x_1,x_3\right]}f(x)d\alpha(x)+\int\limits_{\left[x_3,x_2\right]}f(x)d\alpha(x)$

and the same linearity in the integrand:

$\displaystyle\int\limits_{\left[x_1,x_2\right]}af(x)+bg(x)d\alpha(x)=a\int\limits_{\left[x_1,x_2\right]}f(x)d\alpha(x)+b\int\limits_{\left[x_1,x_2\right]}g(x)d\alpha(x)$

and also a new linearity in the integrator:

$\displaystyle\int\limits_{\left[x_1,x_2\right]}f(x)d(a\alpha(x)+b\beta(x))=a\int\limits_{\left[x_1,x_2\right]}f(x)d\alpha(x)+b\int\limits_{\left[x_1,x_2\right]}f(x)d\beta(x)$

Neat!

February 28, 2008 - Posted by | Analysis, Calculus

1. […] follow on yesterday’s discussion of the Riemann-Stieltjes integral by looking at a restricted sort of integrator. We’ll assume here that is continuously […]

Pingback by The Riemann-Stieltjes Integral II « The Unapologetic Mathematician | February 29, 2008 | Reply

2. […] Riemann-Stieltjes Integral IV Let’s do one more easy application of the Riemann-Stieltjes integral. We know from last Friday that when our integrator is continuously differentiable, we can reduce to […]

Pingback by The Riemann-Stieltjes Integral IV « The Unapologetic Mathematician | March 4, 2008 | Reply

3. […] of Bounded Variation I In our coverage of the Riemann-Stieltjes integral, we have to talk about Riemann-Stieltjes sums, which are of the […]

Pingback by Functions of Bounded Variation I « The Unapologetic Mathematician | March 5, 2008 | Reply

4. Hi,

I just wanted to say thanks for the Riemann-Stieltjes pieces. I am taking second semester analysis which heavily emphasizes this integral – along with doing every proof in Rudin in class (with a take no prisoner’s attitude to the homework – i.e. NO partial credit). I also would like to say that the Professor is Dan Oberlin (FSU math) who not only is the greatest math teacher I have ever had but is a man who has so much respect for mathematics that he refuses to let it be watered down into hand-waving. His is a great teacher and even injects some humor into every class (I know this sound weird in an analysis class) – for example – we wanted to sup over a bunch of terms so we came up with “Zupping” – umlaut over the ‘u’ – which we might submit to Colbert. For all of you out there that think you are ‘scooting’ – get hold of a teacher like Dan – it will blow your socks off in terms of how much math you can learn and do.

Comment by Carlie Saunders | March 8, 2008 | Reply

• I don’t believe Colbert would be able to use a joke that depends on math any higher than arithmetic. It has to do with the audience he gathers by violating logic, honesty, and common sense regularly.

Comment by Joe Bob | September 21, 2009 | Reply

5. […] If we want our Riemann-Stieltjes sums to converge to some value, we’d better have our upper and lower sums converge to that value […]

Pingback by Riemann’s Condition « The Unapologetic Mathematician | March 14, 2008 | Reply

6. […] of Bounded Variation Today I want to start considering Riemann-Stieltjes integrals where the integrator is a function of bounded […]

Pingback by Integrators of Bounded Variation « The Unapologetic Mathematician | March 17, 2008 | Reply

7. […] about to do, I’m going to need a couple results about increasing integrators, and how Riemann-Stieltjes integrals with respect to them play nicely with order properties of the real […]

Pingback by Increasing Integrators and Order « The Unapologetic Mathematician | March 18, 2008 | Reply

8. […] of integrable functions From the linearity of the Riemann-Stieltjes integral in the integrand, we know that the collection of functions that are integrable with respect to a […]

Pingback by Products of integrable functions « The Unapologetic Mathematician | March 20, 2008 | Reply

9. […] Let’s consider some conditions under which we’ll know that a given Riemann-Stieltjes integral will exist. First off, we have a straightforward adaptation of our old result that continuous […]

Pingback by Sufficient Conditions for Integrability « The Unapologetic Mathematician | March 24, 2008 | Reply

10. […] Function Integrators Now that we know how a Riemann-Stieltjes integral behaves where the integrand has a jump, we can put jumps together into more complicated functions. […]

Pingback by Step Function Integrators « The Unapologetic Mathematician | March 27, 2008 | Reply

11. […] Integrals I We’ve dealt with Riemann integrals and their extensions to Riemann-Stieltjes integrals. But these are both defined to integrate a function over a finite interval. What if we want to […]

Pingback by Improper Integrals I « The Unapologetic Mathematician | April 18, 2008 | Reply

12. Dear Professor Armstrong,

I do have found your series of posts covering the Riemann-Stieltjes integral VERY useful to me. By using the “PDF Creator” I have already converted them into pdf files for my private use only and further study, because I see them as a reliable source and in a level appropriate to me.
I dare ask you if you intend to post a list of symbols and notation you use here. Although your notation is the standard one (for american mathematicians) as far as I have seen, it might help us in the topics we are not familiar with. To me, for instance, Abstract Algebra.

Américo Tavares
(retired engineer interested in Mathematics)

Comment by Américo Tavares | April 19, 2008 | Reply

13. I’m not sure what I would put on such a list. I try to explain any symbols or notation as I introduce them. Are there any in particular you’re having difficulty with?

Comment by John Armstrong | April 19, 2008 | Reply

I think of it as a list that would start with relative few symbols and that would be extended to show new ones, when you find you have posted anything not yet there.
In principle that would require a separate page or post regularly updated.
Behind the symbols what really is important are the definitions. But that would help, anyway, I think.
For the purpose of giving you a particular example I went to one of your posts (https://unapologetic.wordpress.com/2007/02/15/cosets-and-quotients/) categorized as “Subgroups and Quotients Groups” by clicking this sub-category. There I found, for instance, among several algebraic symbols (e.g. the subgroup {e,(12)}) the “coset” concept that I still do not know what is it. This means only that I should studied the subject.
I searched for this word and found the posts where it appears.
I do realize that one cannot learn the new symbols without the associated definitions.
After being constructed this list would be very similar with a formal list of symbols in a textbook, the main difference would be that it would cover the symbols of your blog instead of the textbook.

Comment by Américo Tavares | April 19, 2008 | Reply

15. […] Subtracting off the integral of we get our result. (Technically to do this, we need to extend the linearity properties of Riemann-Stieltjes integrals to improper integrals, but this is […]

Pingback by Absolute Convergence « The Unapologetic Mathematician | April 22, 2008 | Reply

16. […] say we take an integrator of bounded variation on an interval and a function that’s Riemann-Stieltjes integrable with respect to over that interval. Then we know that is also integrable with respect to over […]

Pingback by The Integral as a Function of the Interval « The Unapologetic Mathematician | March 14, 2009 | Reply

17. […] Across a Jump In the discussion of necessary conditions for Riemann-Stieltjes integrability we saw that when the integrand and integrator are discontinuous from the same side of the same […]

Pingback by Integrating Across a Jump « The Unapologetic Mathematician | March 14, 2009 | Reply

18. Aloha, What about a definition for a higher S-I? You use the first differences: $\nabla\,x=x_i-x_{i-1}$ in the definition above, but might to comment on the higher differences, such as $\nabla^2\,x=x_i-2x_{i-1}+x_{i-2}$. Do these work? Mahalo

Comment by dan | October 14, 2009 | Reply

19. I’m not really sure, since I haven’t considered it. It’s not just iterating integration, since the Riemann-Stieltjes integral gives back a number and not another function. What are you thinking you might gain from using these second-differences?

Comment by John Armstrong | October 14, 2009 | Reply

20. […] of Partial Integrals There are some remaining topics to clean up in the theory of the Riemann-Stieltjes integral. First up is a question that seems natural from the perspective of iterated integrals: […]

Pingback by Continuity of Partial Integrals « The Unapologetic Mathematician | January 12, 2010 | Reply

21. I’ld really appreciate any references to Riemann-Stieltjes integration in n-dimensions. I’ve seen probability books evidently making use of Riemann-Stieltjes integration in n-dimensions, but nowhere — as far as I have found — are such integrals defined. I realize that one could make sense of Riemann-Stieltjes integration in n-dimensions in terms of Lebesgue-Stieltjes integration w.r.t an induced measure, but that is *NOT* what I’m looking for…

I’ld like to know of any results relating to computing the Lebesgue-Stieltjes integral (of, say, a continuous integrand) w.r.t. the measure induced by a joint distribution function dF(x_1, …, x_n) in terms of iterated Riemann-Stieltjes integrals involving (something like) various marginals… For example — in the 2 dimensional case — perhaps something like

\int \int f(x_1,x_2) dF(x_1,x_2)
= \int ( \int f(x_1,x_2) d G_{x_1}(x_2) ) dF(x_1)

where G_{t}(x) has argument x but is parametrized by t, and has value G_{x_1}(x_2) = F(x_1,x_2).

Any pointers would be appreciated!

Comment by michael | March 18, 2011 | Reply

22. Unfortunately, I’m not really much of an analyst or a statistician. My best guess for the definition would parallel this one-dimensional case, but using n-dimensional “intervals” — basically rectangular parallelepipeds with edges that line up with the coordinate axes.

I think that when I defined the n-dimensional integral I used these “intervals” to define tagged partitions, and so on as usual for the Riemann integral. The difference between the Riemann and Riemann-Stieltjes integrals is using $\alpha(x)$ on the tag point rather than $x$ itself when setting up the Riemann sums. That should generalize pretty directly.

Comment by John Armstrong | March 18, 2011 | Reply