If $X$ is a continuous adapted stochastic process, and $Y$ is a square integrable martingale, then how do we know that $I\left(t\right)\dot{=}\int_{0}^{t}X_{s}dY_{s}$ is also adapted?
I know that by definition $$\int_{0}^{t}X_{s}dY_{s}\dot{=}\lim_{n\rightarrow\infty}^{p}I_{n},$$ where $I_{n}\left(t\right)\dot{=}\sum_{k}X_{t_{k}^{\left(n\right)}\wedge t}\left(Y_{t_{k+1}^{\left(n\right)}\wedge t}-Y_{t_{k}^{\left(n\right)}\wedge t}\right)$. I understand that the $I_{n}\left(t\right)$ is $\mathcal{F}_{t}$-measurable for each $n$, since it is the linear combination of $\mathcal{F}_{t}$-measurable variables, but I have problems with the limit.
As far as I know, the space of random variables is complete with the $p$-convergence, i.e. all Cauchy series of it is convergent. Basically this is the reason why the $p$-limit of $I_{n}$ also exists. But does it also mean that the $p$-limit of $I_{n}$ is also $\mathcal{F}_{t}$-measurable? I mean, otherwise it wouldn't make sense to “calculate” $\mathbf{P}\left(\omega:\left|I_{n}-I\right|>\varepsilon\right)$ in the definition of the $p$-limit: “calculating” the measure of the set $\left\{ \omega:\left|I_{n}-I\right|>\varepsilon\right\}$ implicitly assumes that the set is measurable, i.e. the terms in it, such as $I$ is also measurable. But this reasoning just doesn't feel good, I feel like I mixing the definition of "measurable" and $\mathcal{F}_{t}$-measurable.
So in short: what guarantees that the process $I$ is adapted, i.e. $\mathcal{F}_{t}$-measurable for all $t$?
(There is a similar question with a similar name, but there I haven't found my answer that I've been looking for: Why are stochastic integrals adapted?)
Convergence in probability of $(I_n)$ implies the existence of a subsequence $(I_{n(k)})$ that converges a.s. to $I$. As each r.v. $I_{n(k)}$ is $\mathcal F_t$ measurable, so is $J:=\liminf_k I_{n(k)}$. Because $I=J$ a.s., so $I$ is equal a.s. to an $\mathcal F_t$ measurable r.v. Under the "usual conditions", $I$ is even $\mathcal F_t$ measurable.