# Function that is Riemann-Stieltjes integrable but not Riemann integrable?

This is my first question, so please go easy on me :3 - I've searched, and I haven't found any questions that are particularly similar to this one.

I'm reading Rudin's Principles of Mathematical Analysis, and I've reached the chapter on integration. It surprised me that in the text Rudin referred to the RS integral as a "generalization" of the Riemann integral without expounding at all upon what in fact makes it a generalization. I understand that picking an interesting value for the monotone increasing function $\alpha$ might be practical in some applications, but why is it useful to teach the Riemann integral as a special case of the Riemann-Stieltjes integral?

I can only assume that the set of functions in $R(\alpha)$ is a proper superset of the set of functions in $R(x)$, but thus far I have been unable to come up with an example of a function that is integrable with respect to some $\alpha$ but not to the monotone increasing function $x$. Can anyone offer an illuminating example of such a function, or are there no such functions?

• Choose the interval $[0,1]$, and choose $\alpha$ to be a step function at, say, $x=\frac{3}{4}$, and then take $f=1_{\mathbb{Q} \cap [0,\frac{1}{2}]}$. $f$ is not in ${\cal R}(x)$, but is in ${\cal R}(\alpha)$. May 8, 2013 at 18:26
• Thanks for your answer. Pardon my ignorance, but can you give a more verbose definition of $f$? I've never seen that notation before. May 8, 2013 at 18:37
• Indicator function of the set $\mathbb{Q} \cap [0,\frac{1}{2}]$. May 8, 2013 at 18:41
• @copper.hat: Why not leave this as an answer? (By the way, what if $\alpha$ is not constant on any nontrivial subinterval?) May 8, 2013 at 18:53
• @copper.hat: I understood your comment. The reason I asked you to leave it as an answer is that it is in fact a complete answer, and surely an optimally simple one. (Thanks for making it into an answer.) I was just asking a followup question because the counterexample is in a sense "trivial" and was wondering if there are more serious counterexamples. May 8, 2013 at 21:00

## 2 Answers

Imo, the point of the Riemann-Stieltjes integral is not that "We can now integrate more functions, which makes us happy." (btw. I'm doubtful this is actually true). The point is, that we now have a more general way of expressing certain things as an integrals, or a clearer view of what an integral naturally should be.

For example: We may think of the expression $\int_a^b \rho \, dx$ as the total mass of some segment $[a,b]$, where $\rho: [a,b] \to \mathbb R$ is the mass density on $[a,b]$. In this situation, the Riemann integral is sufficient.

But what the Riemann integral allows us to do may not be sufficient in some cases. Sometimes, we really would like to talk about point-like objects, such as point-masses. Intuitively, the density $\rho$ of a point-mass situated at $x=0$, say, should be zero everywhere, except at $0$. At the same time, we would like to have $$\int_{\mathbb R} \rho \, dx = \mathrm{mass}.$$ Now, with a Riemann integral, we can never achieve this. There simply is no function $\rho$, such that $\rho(x)=0$ for $x\ne 0$, and yet $\int_{\mathbb R}\rho \, dx \ne 0$.

Bring in the Riemann-Stieltjes integral: Let $F$ be given by $F(x) = 0$, if $x<0$, and $F(x) = \mathrm{mass}$, if $x\ge 0$. Then $F(x)$ is the total mass on $(-\infty, x]$. Now, $F$ does not have a density of the form $\rho\, dx$, but it does have a density $dF$ in the R-S sense. And this does what we want: We have

$$\int_{-\infty}^x \, dF = \begin{cases} 0 & x< 0\\ \mathrm{mass} & x\ge 0.\end{cases}$$

So, using R-S integrals, we have gained a generalized notion of "density". In fact, we learn from this that when speaking about the density of something, we should not think of it in terms of a function $\rho$, but rather in terms of an expression like $\rho dx$, or $dF$. Said differently, we see that the Riemann integral, or the $dx$ density, is just a very special way of measuring something, which pertains to certian situations but not to others. There are situations, where measurement should be done in a way which is not captured by the Riemann integral.

This realization may have been the first step in the direction of the notion of a measure. A concept which underlies much of modern analysis.

It is in this way, that the Riemann-Stieltjes integral generalizes the Riemann integral.

Choose the interval $[0,1]$, and choose $\alpha$ to be constant on $[0,\frac{1}{2}]$. Then take $f=1_{\mathbb{Q} \cap [0,\frac{1}{2}]}$ (ie, the indicator function of $\mathbb{Q} \cap [0,\frac{1}{2}]$).

Then f is not in ${\cal R}(x)$, but is in ${\cal R}(\alpha)$.