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Let $F,G$ be non-decreasing real functions of bounded variation on $\mathbb{R}$ and $\mu_F$ the Lebesgue-Stieltjes measure defined by it. Kolmogorov-Fomin's says (p. 452 here) that we can commute the order of the functions in their Lebesgue-Stieltjes convolution, i.e. we can write $$G\star F:=\int_{-\infty}^\infty G(x-\xi)dF(\xi)=\int_{-\infty}^\infty F(x-\xi)dG(\xi)=F\star G$$where $\int_{-\infty}^\infty G(x-\xi)dF(\xi):=\int_{\mathbb{R}}G(x-\xi)d\mu_{F(\xi)}$.

How can such an equality be proved? There is even something that I am not even able to convince myself of: if $\forall x\in\mathbb{R}\quad F(x)= 1$, then I would say that $G\star F=0$, but $F\star G=\mu_G(\mathbb{R})$: what am I missing? I thank you all very much!

EDIT: I originally asked whether $\int_a^b G(x)dF(x)=\int_a^bF(x)dG(x)$ because I thought that was the key to prove the desired result, but I have now rearranged the question because the last equality equality does not hold, as John, whom I thank very much, has pointed out in his comment.

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    $\begingroup$ No. This does not hold for example when $G(x) =x^2$, $F(x)=x$. $\endgroup$ – user99914 Nov 30 '14 at 12:48
  • $\begingroup$ @John I thank you so much! Therefore, I see that we cannot take that road to prove the commutativity of Lebesgue-Stieltjes convolution and I've edited the o.p. $\endgroup$ – Self-teaching worker Nov 30 '14 at 13:15

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