I know that if $F:[a,b]\to\mathbb{R}$ is a non-decreasing absolutely continuous function then$$\int_a^b f(x)dF(x)=\int_a^b f(x)F'(x)d\mu$$where the first integral is the Lebesgue-Stieltjes integral $\int_{[a,b]} f(x)d\mu_F$, i.e. a Lebesgue integral with respect to the Lebesgue-Stieltjes measure, $\mu_F$, defined by $F$, and the second one is a Lebesgue integral where $\mu$ is the "canonical" linear Lebesgue measure.

Although the proof given in Kolmogorov-Fomin's seems to have some possibly misleading simplification, it appears clear to me from the proof that the second integral $\int_a^b f(x)F'(x)d\mu$ exists if the first one does.

I wonder whether $\int_a^b f(x)dF(x)$ exists if $\int_a^b f(x)F'(x)d\mu$ does, since Kolmogorov-Fomin's is silent about that. Does anybody know more about that? I thank everybody for any answer!

  • 1
    $\begingroup$ The important idea to take here is that when the non-decreasing function $f$ is absolutely continuous, we have $$dF = F'(x)d\mu$$ or to say $$F'(x) = \frac{dF}{d\mu}.$$ Then the existence of the integrals are automatically implied. $\endgroup$ – Xiao Nov 4 '14 at 2:41

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