0

Let $\sigma:[a,b] \to \mathbb{R}$ be a strictly increasing function. We seek a function $f:[a,b] \to \mathbb{R}$ such that:

  • $f$ is unbounded on $[a,b]$,
  • $f$ is Riemann-Stieltjes integrable with respect to $\sigma$, i.e., $\int_a^b f\,d\sigma$ exists.

Attempts & Thoughts:

If $\sigma(x) = x$, the problem reduces to ordinary Riemann integration, where unbounded functions are not integrable. Thus, $\sigma$ must be more carefully chosen.

A plausible idea is to choose $\sigma$ such that it "neutralizes" the singularities of $f$. For instance, if $f$ has a singularity at some point, say $c \in [a,b]$, and $\sigma$ is constant in a neighborhood of $c$ (or grows very slowly), then the contribution of $f$ near $c$ to the Riemann-Stieltjes integral may still be finite.

Thus, the intuition is to balance the blow-up of $f$ with the flatness (or slowness) of $\sigma$.

Thomas
  • 23,023
Raffi
  • 1
  • 1
  • 1
    An unbounded function is not Riemann-Stieltjes integrable, see https://math.stackexchange.com/questions/2311135/riemann-stieltjes-integral-of-unbounded-function – Thomas May 30 '25 at 08:45
  • if you take the g(x) = 0 (or any constant) on [a,b], then any function with respect to g(x) is Riemann-Stieltjes integrable – Raffi May 30 '25 at 10:39
  • A simple example would be $\sigma(x) = x^3$ and $f(x) = \frac 1x$ on $[-1,1]$, which is easily seen to have integral $0$ – Paul Sinclair Jun 04 '25 at 03:12
  • first your function f is not defined on point 0, also you cannot just compute the integral by taking σ'(x)dx from dσ(x), to do this you should have f continuous on [a,b] and derivative of σ Rieman integrable – Raffi Jun 04 '25 at 12:20

0 Answers0