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Complex Riemann-Stieltjes integral with a step function for integrator

  1. Jan 17, 2009 #1
    Hello all. In my book (Mathematical Analysis by Apostol), the author states and proves the following theorem for real (i.e. f and alpha are functions from [a,b] to R) Riemann-Stieltjes integrals:

    Assume [tex]c\in(a,b)[/tex] and [tex]\alpha[/tex] is such that [tex]\alpha(a)=\alpha(x)[/tex] when [tex]a\leq x<c[/tex], and [tex]\alpha(x)=\alpha(b)[/tex] if [tex]b\geq x > c[/tex], with [tex]\alpha(a)[/tex], [tex]\alpha(c)[/tex], and [tex]\alpha(b)[/tex] arbitrary. If either [tex]f[/tex] or [tex]\alpha[/tex] is continuous from the right AND either [tex]f[/tex] or [tex]\alpha[/tex] is continuous from the left, [tex]\int_a^b f\,d\alpha[/tex] exists and is equal to [tex]f(c)[\alpha(b)-\alpha(a)][/tex]. The author later, when discussing complex Riemann-Stieltjes (f and alpha are functions from [a,b] to C), lists the theorems for the real case that are also true in the complex case, but leaves this one out. However, I have gone over the proof, and it seems to work perfectly for the complex case as well. Am I missing some subtle detail in the proof, or did the author simply not decide to mention it? Does anyone know if this is true for complex Riemann-Stieltjes integrals?
  2. jcsd
  3. Jan 17, 2009 #2


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    If you integrate from a to b on the real line, then it makes sense to talk about "[itex]a\le x< c" or "c< x\le b[/itex]. Integrating in the complex plane, which is not an ordered field, you cannot do that.
  4. Jan 17, 2009 #3
    Hmm? I'm talking about functions from the real interval [tex][a,b][/tex] to the set of complex numbers. [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] are all real numbers here.
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