5

Are there any extensions of Riemann-Stieltjes integration that are able to handle the following integral?

$\int_0^1 \alpha \space d\alpha$

where

$ \alpha(x) = \left\{ \begin{array}{lr} 0 & x < \frac{1}{2} \\ \frac{1}{2} & x = \frac{1}{2} \\ 1 & x > \frac{1}{2} \end{array} \right. $

Everything I've found so far have strong requirements about the funcions' continuity, and are undefined when there are shared discontinuities, even for such a simple function.

My interest: Where integration by parts hold.

What consequence/uses would there be to it?

Emilio Martinez
  • 209
  • 1
    Maybe something in one of these papers will be of use: paper 1 and paper 2 and paper 3 and paper 4 and paper 5. – Dave L. Renfro Feb 23 '19 at 19:50
  • @DaveL.Renfro Thanks. I'll check them later, as I'm interested in knowing many forms of integration. But at first glance it doesn't look like any of those live up to the task. Appreciated nonetheless. – Emilio Martinez Feb 23 '19 at 20:24
  • The Riemann-Stieltjes integral can exist when integrand and integrator have shared discontinuity points. For example, see here. The sufficient condition for non-existence is that they both cannot be discontinuous from the right or from the left. – RRL Feb 23 '19 at 20:25
  • Of course your example meets that sufficient condition. – RRL Feb 23 '19 at 20:26
  • 1
    Regarding wanting to know "many forms of integration", see my answer to Dissertation on Integrals, especially the papers by Peter Bullen and Ralph Henstock (for reasons I give there). – Dave L. Renfro Feb 24 '19 at 07:09
  • @RRL Thanks. That's what I meant by "strong requirements". So I take there is no known way to integrate any function against itself? – Emilio Martinez Feb 24 '19 at 12:18
  • "there is no known way to integrate any function against itself" --- I don't know, but I would be VERY cautious about making such a general statement. For example, I wonder whether a Riemann-Stieltjes version of what is described in A generalization of the Riemann integral by Casper Goffman (1952) might be able to handle what you're dealing with (by overlooking shared discontinuities, at least when they are at most countably many or otherwise confined to a "small set"). – Dave L. Renfro Feb 24 '19 at 18:12
  • @DaveL.Renfro I'll check it out, but I'm hopeless. Avoiding discontinuities works for integrals over dx because values at a single point move along the y axis and don't change the function's mass. But shared discontinuities should pop up as mass, not overlooked. In my highly speculative opinion. – Emilio Martinez Feb 25 '19 at 12:23

0 Answers0