In Karatzas&Shreve Brownian Motion and Stochastic Calculus P.23 Remarks 4.6: If $A$ is an increasing process and $X$ is progressively measurable, define the Lebesgue-Stieltjes integrals $I_{t}^{\pm}\left(\omega\right)=\int_{\left(0,t\right]}X_{s}^{\pm}\left(\omega\right)dA_{s}\left(\omega\right)$, then $I_{t}=I_{t}^{+}-I_{t}^{-}$ is well defined and finite for all $t\geq0$, then $I$ is right-continuous and progressively measurable.
Here is the definition of an increasing process： An adapted process A is called increasing if for $P$-a.e. $\omega\in\Omega$ we have
• (a) $A_{0}\left(\omega\right)=0 $
• (b) $t\mapsto A_{t}\left(\omega\right)$ is a nondecreasing, right-continuous function,
• (c) $E\left(A_{t}\right)<\infty$ holds for every $t\in\left[0,\infty\right)$.
Now I want to show that $I$ is right-continuous and progressively measurable. It is well known that every right-continuous adapted process is progressively measurable. So I just need to show that $I$ is right-continuous and adapted. But how can I show that $I$ is right-continuous and adapted?
I assume you also need the property that $X$ has bounded variation paths. Then right-continuity follows from $$ \bigg| \int_{(0;t+\varepsilon]} X^+_s(\omega) \mathrm dA_s(\omega) - \int_{(0;t]} X^+_s(\omega) \mathrm dA_s(\omega) \bigg| \\= \bigg| \int_{(t;t+\varepsilon]} X^+_s(\omega) \mathrm dA_s(\omega) \bigg| \le \int_{(t;t+\varepsilon]} X^+_s(\omega) \mathrm dA_s(\omega) \le \int_{(t;t+\varepsilon]} X^+_{t+\varepsilon}(\omega) \mathrm dA_s(\omega) = X^+_{t+\varepsilon}(\omega) (A_{t+\varepsilon}(\omega) - A_t(\omega) ) $$ This converges to $0$ because $A$ is right-continuous and $X^+_{t+\varepsilon}(\omega) < \infty$. Adaptedness is a little bit trickier from scratch.