# Complex Riemann-Stieltjes integral with a step function for integrator

1. Jan 17, 2009

### uman

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 $$c\in(a,b)$$ and $$\alpha$$ is such that $$\alpha(a)=\alpha(x)$$ when $$a\leq x<c$$, and $$\alpha(x)=\alpha(b)$$ if $$b\geq x > c$$, with $$\alpha(a)$$, $$\alpha(c)$$, and $$\alpha(b)$$ arbitrary. If either $$f$$ or $$\alpha$$ is continuous from the right AND either $$f$$ or $$\alpha$$ is continuous from the left, $$\int_a^b f\,d\alpha$$ exists and is equal to $$f(c)[\alpha(b)-\alpha(a)]$$. 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. Jan 17, 2009

### HallsofIvy

If you integrate from a to b on the real line, then it makes sense to talk about "$a\le x< c" or "c< x\le b$. Integrating in the complex plane, which is not an ordered field, you cannot do that.

3. Jan 17, 2009

### uman

Hmm? I'm talking about functions from the real interval $$[a,b]$$ to the set of complex numbers. $$a$$, $$b$$, and $$c$$ are all real numbers here.