# Dominated convergence for Lebesgue-Stieltjes integral

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 ?

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 $$\int f_n dg = \int f_n d\mu_g^+ - \int f_n d\mu_g^- \to \int \alpha d\mu_g^+ - \int \alpha d\mu_g^- = \int \alpha dg$$
• 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 ? – W. Volante May 7 '18 at 17:18