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The notation$\int_0^t\vert f\vert\,\vert dg\vert$ represents the Lebesgue-Stieltjes integral of the function $\vert f \vert$ with respect to the total variation of the function $ g $.

Suppose that the function $f$ satisfies $\int_0^t\vert f \vert\,\vert dg\vert<\infty $, so it is integrable with respect to $g$ in the Lebesgue-Stieltjes sense. If $\vert f_n\vert\le\vert f \vert$ is a sequence of bounded functions tending to a limit function $\alpha$ then how can we conclude that

$$\int_0^t f_n\,dg\rightarrow\int_0^t\alpha\,dg $$

Is there some sort of dominated convergence for signed measures ?

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If $\mu$ is a signed measure then the Jordan decomposition asserts the existence of (mutually singular) non-negative measures $\mu^+$ and $\mu^-$ such that $\mu = \mu^+ - \mu^-$. If $\mu_g$ is the measure arising from $g$ then we can be explicit here and define the measures $\mu_g^{\pm}$ by $$\mu_g^{\pm}((0,t]) = \frac{V(g)_t \pm g(t)}{2}$$ where $V(g)_t$ is the total variation of $g$ up to time $t$.

In particular, if $\mu_g$ is the measure associated to the function $g$, the measure associated to the total variation of $g$ is $|\mu_g| = \mu_g^+ + \mu_g^-$. Then e.g. $\int |f| d \mu_g^+ \leq \int |f| d (\mu_g^+ + \mu_g^-) = \int |f| |dg| < \infty$ and similarly for $\mu_g^-$. Hence, by the usual dominated convergence theorem, $$\int f_n d \mu_g^{\pm} \to \int \alpha d \mu_g^{\pm}.$$ Then, since the quantities involved are all finite, we have $$\int f_n dg = \int f_n d\mu_g^+ - \int f_n d\mu_g^- \to \int \alpha d\mu_g^+ - \int \alpha d\mu_g^- = \int \alpha dg$$

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  • $\begingroup$ Thank you for your answer. How do you know that the measure associated with the total variation of $g$ is the sum of the two measures of the Jordan decomposition ? $\endgroup$ – W. Volante May 7 '18 at 17:18
  • $\begingroup$ @W.Volante I've added an explicit description of these measures that is available in this case to my first paragraph. Hopefully this makes it clear that the measure associated with the total variation of g is the sum of the two measures described. $\endgroup$ – Rhys Steele May 7 '18 at 19:22

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