Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a given function $\alpha: [0, 1]\to \mathbb{R}$, let's denote $\mathcal{R}(\alpha)$ to be the set of all Riemann-Stieltjes integral functions with respect to $\alpha$. It is a well-known fact that (for example, see Theorem 6.8 in Rudin's Principles of Mathematical Analysis) if $f: [0,1]\to\mathbb{R}$ is a continuous function, and $\alpha: [0,1]\to\mathbb{R}$ is a monotonically increasing, then $f\in\mathcal{R}(\alpha)$.

I am interested in the following:

Question 1: What is an example of function $\alpha:[0,1]\to\mathbb{R}$ and a continuous function $f:[0,1]\to\mathbb{R}$ such that $f\not\in\mathcal{R}(\alpha)$?

Question 2: Does there exist an example of continuous function $\alpha:[0,1]\to\mathbb{R}$ and a continuous function $f:[0,1]\to\mathbb{R}$ such that $f\not\in\mathcal{R}(\alpha)$?


share|cite|improve this question
Well the result you stated still holds if $ \alpha $ is of bounded variation which is the difference of two monotone functions. So my guess will be taking $f$ as identity and $\alpha $ as $ x\sin(1/x) $ with $\alpha(0) = 0 $ might work for both cases. If it doesn't work then maybe you can try changing $f$ to the cantor function. – smiley06 Apr 18 '13 at 14:56
@smiley06 I don't think taking $f$ as identity works. Because identity function is Riemann-Stieltjes integrable with respect to any function $\alpha$, since upper and lower sums are both $\alpha(1)-\alpha(0)$ for every partition $P$ of $[0,1]$. – Prism Apr 19 '13 at 4:50
up vote 2 down vote accepted

The Riemann-Stieltjes integral works well as long as $\alpha$ has bounded variation (equivalently, can be written as the difference of two bounded increasing functions). The continuous function $$\alpha(x)=\begin{cases} \sqrt{x}\cos(\pi/x),\quad &x\in (0,1] \\ 0 & x=0\end{cases} \tag1$$ has infinite variation on $[0,1]$. Indeed, $\alpha(1/n)=(-1)^n/\sqrt{n}$, which implies $\sum_{n=1}^\infty |\alpha(1/n)-\alpha(1/(n+1))|=\infty$. (Aside: all real analysis book I know use $\sin$ instead of $\cos$ in this example, and mess with extra $\pi/2$ in calculations.)

I claim that $\int_0^1 \alpha(x)\,d\alpha(x)$ does not exist. First of all, the definition that is based on decomposing $\alpha$ as the difference of two bounded increasing functions breaks down. On the more basic level, each interval $[1/(n+1),1/n]$ contributes $$\approx\frac{1}{\sqrt{n}} \frac{2}{\sqrt{n}}$$ to the upper sum, and $$\approx -\frac{1}{\sqrt{n}} \frac{2}{\sqrt{n}}$$ to the lower sum.

share|cite|improve this answer
Thanks for your answer :) I will try to read this more carefully, and then I will accept. – Prism May 29 '13 at 8:46
I couldn't understand how ∫α(x)dα(x) does not exist ... can someone explain it to me please? – Arsh Gh Apr 29 at 16:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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