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Riemann-Stieltjies Integrals

  1. Nov 26, 2012 #1
    I'm having trouble visualizing the riemann-stieltjies integral...

    Our textbook states:

    We assume throughout this section that F is an increasing function on a closed interval [a,b]. To avoid trivialities we assume F(a)<F(b). All left-hand and right-hand limits exist...We use the notation

    [tex]F(t^-)= lim_{x \rightarrow t^-} F(x)[/tex] and [tex]F(t^+)= lim_{x \rightarrow t^+} F(x)[/tex]

    For a bounded function f on [a,b] and a partition [tex]P={a=t_0 < t_1 < ... < t_n = b}[/tex] of [a,b], we write

    [tex]J_F(f,P) = \sum_{k=0}^n f(t_k) [F(t_k^+) - F(t_k^-)][/tex]

    The upper Darboux-Stieltjes sum is

    [tex]U_F(f,P) = J_F(f,P) + \sum_{k=1}^n max(f, (t_{k-1}, t_k) [F(t_k^+) - F(t_{k-1}^-)][/tex]

    I'm having trouble visualizing this...also, by F(x), do they mean the integral of f(x)?

    Thanks in advance
  2. jcsd
  3. Nov 26, 2012 #2


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    Homework Helper

    No F (a confusing choice of variable) is a function that determines the size of intervals. In a Riemann integral
    ∫f dx
    the interval [a,b] has size b-a
    In a Riemann-Stieltjies integral
    ∫f dF
    the interval [a,b] has size F(b+)- F(a-)

    and of course when F(x)=x the Riemann-Stieltjies integral reduces to the Riemann integral

    This is helpful in many ways.
    -We can take sums as a type of integral and unify sums and integrals
    -We can have impulse function like the Dirac delta function which concentrate a change to a single point.
    Last edited: Nov 26, 2012
  4. Nov 26, 2012 #3
    Thanks a lot...is it ok if you give an example or something? So I can understand the difference better (between usual Reimann integrals and Reimann-Stieltjes Integrals)?
  5. Nov 26, 2012 #4

    Stephen Tashi

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    Science Advisor

    Something short of a complete example:

    I see from your other posts that you know something about statistics. Suppose we have a random variable X whose distribution is defined by the statement:

    There is a 0.3 probability that X = 0.5 and if X is not equal to 0.5 then the other possibilities for X are uniformly distributed on the intervals [0,0.5) and (0.5, 1].

    How would you compute the expected value a function f(X) ? ( e.g. the case f(X) = X would be the expected value of X). I think the common sense way is;

    [itex] \bar{f(x)} = (0.3) f(0.5) + (1.0 - 0.3) \ ( \ (0.5) \int_0^{0.5} f(x) u_1(x) dx + (0.5) \int_{0.5}^{1} f(x) u_2(x) dx\ ) [/itex]

    Where [itex] u_1(x) [/itex] is the uniform distribution on [0,0.5) and [itex] u_2(x) [/itex] is the uniform distribution on (0.5, 1] and the integrals are Riemann integrals.

    It would be convenient to define a single distribution function for X and write [itex] \bar{f(x)} [/itex] as a single integral (even if the practical computation of that integral amounted to the work above). However, a Riemann integral can't handle the "point mass" probability at X = 0.5 because, in a manner of speaking, it sits on a rectangle whose base has zero length.

    From the viewpoint of probability theory, a Riemann-Stieljes integral can be regarded as way of defining a new form of integration that handles such "point masses". ( You can define a nondecresasing function [itex] F(x) [/itex] which has a jump of size 0.3 at x = 0.5 )
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