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Non Stieltjes integrable function

  1. Dec 9, 2011 #1
    I know that this function [itex] f : [-2,2] \to \mathbb{R} [/itex]
    [itex] f(x) = \begin{cases} 1 & \textrm{ if } x \geq 0 \\ 0 & \textrm{ if } x < 0 \end{cases} [/itex]
    is not Riemann-Stieltjes integrable with itself (that is, taking [itex] g = f [/itex] then [itex] f \not\in R(g) [/itex])
    That is because both share a point of discontinuity, namely 0.

    What I don't know is how do I justify this? for wich partition does the limit of the superior sums and the inferior sums don't equal? Shouldn't this integral be equal to 1? (all the terms in the sum cancel except at zero, where [itex] \triangle g = 1-0 = 1 [/itex], and [itex] f(0) = 1[/itex]

    I think I'm not seeing it correctly.
     
  2. jcsd
  3. Dec 9, 2011 #2

    Stephen Tashi

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    Science Advisor

    In computing the upper and lower sums, the contribution of an interval like [-0.1, 0] involves more than f(0). You have to use the infimum of f on the interval when computing the lower sum.

    I found a PDF of class notes that supposedly proves the theorm that if the two functions share a common point of discontinuity then they are not Riemann-Stieljes integrable. (page 3 of 5) It looks like it's hard to prove!

    http://www.google.com/url?sa=t&rct=...sg=AFQjCNEL7sm8dOrwngWsUyk-yVKajXGqQA&cad=rja
     
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