This is a standard setting in stochastic stieltjes integration. I have some questions about the measurability issues in this setting.
Let $A$ be a right-continuous increasing process, i.e. the paths of $A: t \mapsto A_t(\omega)$ are non-decreasing and right-continuous. Then for a fixed $\omega$, this function induces a measure $\mu_A(\omega,ds)$ on $\mathbb{R}_+$. If $f$ is a bounded, Borel function on $\mathbb{R}_+$, then $\int_0^t f(s)\mu_A(\omega,ds)$ is well-defined for each $t>0$. We denote this integral by $\int_0^t f(s)dA_s(\omega)$. If $F_s = F(s,\omega)$ is bounded and jointly measurable, we can define, $\omega$-by-$\omega$, the integral $I(t,\omega) = \int_0^t F(s,\omega) dA_s(\omega)$. $I$ is right continuous in $t$ and jointly measurable.
In the above case, how can we explicitly show that $I$ is right continuous in $t$ and jointly measurable? It seems like the function $\omega \mapsto dA_s(\omega)$ is measurable, but I cannot find the details to these anywhere. I would greatly appreciate some formal explanations to these.
Fot $t > 0$ and $\omega$, we define $I(t,\omega)$ as \begin{align*} I(t,\omega)=\int_{0}^t F(s,\omega )\,dA_s(\omega). \end{align*}
Step1. We first show that for each $t > 0$, the map $\omega \mapsto I(t,\omega)$ is measurable. By a monotone class argument, we may assume that $F(s,\omega)=f(s)g(\omega)$ for some bounded measurable functions $f:\mathbb{R}_{+} \to \mathbb{R}$ and $g:\Omega \to \mathbb{R}$. Notice that for any $a,b \in \mathbb{R}_{+}$, the map $\omega \mapsto \mu_A(\omega ,(a,b])$ is measurable. This together with the fact that $f$ is approximated by simple measurable functions on $\mathbb{R}_{+}$, we see that the map $\omega \mapsto \int_{0}^t f(s)\,dA_s(\omega)$ is measurable. Therefore, the map $\omega \mapsto I(t,\omega)(=g(\omega ) \times \int_{0}^t f(s)\,dA_s(\omega) )$ is also measurable.
Step2. Let $n \in \mathbb{N}$ and $T>0$. For $(\omega,s) \in \Omega \times [0,T]$, we set \begin{align*} I_n(t,\omega)=\mathbf{1}_{\{0\}}(t)I(0,\omega)+\sum_{k=1}^{2^n}\mathbf{1}_{\{( (k-1)T/2^n, kT/2^n]\}}(t )I((k-1)T/2^n,\omega). \end{align*} By the dominated convergence theorem, we see that the map $t \mapsto I(t,\omega)$ is right continuous on $\mathbb{R_{+}}$, and $I(t,\omega)$ is the pointwise limit of $I_{n}(t,\omega)$. Therefore, we arrive at the conclusion from Step1.