I'm trying to convince myself on the existence of a measurable version of a process, $X:[0,T]\times \Omega\to E$ which is assumed to be stochastically continuous with $E$ a separable Banach space. Recall that measurability in this context means that it is measurable jointly in $(t,\omega)$.
The text I'm following suggests first identifying a sequence of partitions of $[0,T]$ and then using these to construct a sequence of piecewise constant (in $t$) left continuous approximations, $X_m(t,\omega)$. This makes sense and these are obviously measurable.
The next step is to look at the set on which these are convergent, $$ A = \{(t,\omega): \{X_m(t,\omega)\}\;\text{is Cauchy}\}, $$ and define $$ Y(t,\omega) = 1_{A}(t,\omega)(\lim_{m\to \infty} X_m(t,\omega)). $$ By Borel-Cantelli, you can then argue that $Y$ is a version of $X$.
However, there is one point that sticks out for me, that I do not know how to address, and the text I'm looking at is very vague about - the measurability of $A$, which is needed to conclude $Y$ is measurable. I'm hoping someone can either clarify why $A$ is obviously measurable, or point me to a reference that clarifies this issue.
You can easily check that all $X_m$ are measurable according to the product-$\sigma$-algebra $\mathbb{B}([0,T]) \otimes \mathcal{F}.$ The set $A$ can be written as $$A= \bigcap_{k=1}^\infty \bigcup_{N=1}^\infty \bigcap_{n>m>N }\{ \|X_n -X_m\| <1/k\}$$ and is therefore measurable.