Commutativity and associativity in Stieltjes convolution algebra

I've been trying to prove the commutativity and associativity within Stieltjes convolution algebra but haven't succeeded. In the literature it says that the convolution should be both commutative and associative but they show no proof and when I try for some simple example I can't get it to work.

As I understand it, the Stieltjes convolution for non-decreasing and differentiable functions $F$ and $G$, is

$F \star G = \int_0^t F(t-x)dG(x) = \int_0^t F(t-x)G'(x)dx$

my question is if it's the same as

$G \star F = \int_0^t G(t-x)dF(x) = \int_0^t G(t-x)F'(x)dx$

from being commutative and how to prove this?

The simple example I've tried to solve is for $F(x) = 1-e^{-ax}$, $G(x) = 1-e^{-bx}$, $F'(x) = ae^{-ax}$ and $G'(x) = be^{-bx}$, for which I get that it's not commutative (at least not according to the equations above), but my feeling is that I've done something wrong. The first link down below has proofs of commutativity and associativity, but I don't understand the commutativity proof exactly.

So, if anyone is familiar with Stieljes convolution algebra and knows how to prove the commutativity and associativity, I would be very grateful.

The literature:

The Stieltjes Convolution and a Functional Calculus for Non-negative Operators

Stochastic Processes Model and its Application in Operations Research

Dependability analysis of semi-Markov systems

• Are you familiar with product measures? The only proof I know writes both $(F \star G)(t)$ and $(G \star F)(t)$ both as the product measure of $\{(x,y):x+y \leq t\}$. Incidentally, you cannot write $dG(x)$ as $G'(x)dx$ simply because $G$ is differentiable. – Kavi Rama Murthy May 2 '18 at 8:25
• I'm not familiar with product measures, but I will look it up. Alright, can I interpret $dG(x)$ in any other way? Thank you for the answer! – C.bo May 2 '18 at 10:13