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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 $f\not\in\mathcal{R}(\alpha)$?

Thanks!

Prism
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  • 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
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    @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 04:50

1 Answers1

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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.

ˈjuː.zɚ79365
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  • Thanks for your answer :) I will try to read this more carefully, and then I will accept. – Prism May 29 '13 at 08:46
  • I couldn't understand how ∫α(x)dα(x) does not exist ... can someone explain it to me please? – Arsh Gh Apr 29 '16 at 16:29
  • @Arsh Gh $\int_0^1 \alpha(x) \alpha^\prime(x) dx$ does exist as an improper integral. However, the Riemann sums for $\int_0^1 \alpha d\alpha$ can be made as large positively or negatively as desired. – Richard Hevener May 24 '21 at 00:06