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why can I say that

$$ \int_0^a t^2 dF(t) = \int_0^a t^2 d(F(t) -1) $$

unfortunately my experience with the Riemann -Stieltjes is practically non existent, so for instance I do not understand, why the interval of integration is not affected by the change of the integrator.

Appart from the lack of understanding mentioned above, my second question is, whether I can basically make sense of the change of the integrator in the same way I do in the Riemann context for a change of variable via the chain rule ?

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Do you know the definition of the Riemann-Stieltjes integral? –  Fabian Feb 4 '12 at 15:05
    
@Fabian: what a dum question in retrospect :) on the upside I won t forget this ever. I thought there would be more behind it, thanks! –  Beltrame Feb 4 '12 at 15:45

2 Answers 2

up vote 3 down vote accepted

Write down a Riemann sum for both sides. Compare; you will see they are the same. If all the Riemann sums are the same, the integrals are the same.

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I'm not sure, if this helped. Could you work out his example? –  draks ... Feb 4 '12 at 15:49
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The idea is that $dF(t) = d(F(t)-1) = d(F(t))-d(1)= d(F(t))-0 = d(F(t))$ –  Pedro Tamaroff Feb 4 '12 at 15:55

Consider the integrator $\alpha(t)-1$ vs $\alpha(t)$. If we make

$$\Delta \alpha_i = \alpha(t_i)- \alpha(t_{i-1})$$

$$\Delta \alpha_i = (\alpha(t_i)-1)- (\alpha(t_{i-1})-1)$$

$$\Delta \alpha_i = \alpha(t_i)-1- \alpha(t_{i-1})+1$$

$$\Delta \alpha_i = \alpha(t_i)- \alpha(t_{i-1})$$

and both sums will end up being identical.

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