I am trying to figure out the proof of Theorem 6 in Protter's Stochastic Integration and Differential Equations. I have a hunch that what may have been used is the statement in the title. I've seen it used a couple of places already, but I cannot figure out why it is true. For example in this question Measurability of a stopped random variable., both answers use the statement.
As far as I'm aware, cadlag means that ALMOST every path is cadlag. I showed that if a process $X$ has each sample path cadlag and is adapted, then it is progressively measurable by using the the functions $X^n(s,\omega) = X(0,\omega)\mathbf{1}_{\{0\}}(s)+\sum\limits_{k = 1}^{2^n}X(tk/2^n,\omega)\mathbf{1}_{(t(k-1)/2^n, tk/2^n]}(s)$ on $[0,t]\times \Omega$.
$X^n$ is $\mathcal{B}([0,t]) \bigotimes \mathcal{F}_t$ measurable and by right continuity, $X^n$ converges everywhere to $X$ on $[0,t]\times \Omega$, so X is progressively measurable.
However, if we drop the assumption that $X$ has every path cadlag and instead just almost every path is cadlag (ie $X$ is a cadlag process) we can let $N = \{\omega \in \Omega: X(\cdot,\omega) \text{ is not cadlag}\}$. Then N is a null set. Even if $\mathcal{F}_t$ is complete, $N$ is a measurable null set in $\mathcal{F}_t$ and therefore $[0,t]\times N$ is null in $\mathcal{B}([0,t]) \bigotimes \mathcal{F}_t$. Since $X^n$ converges to $X$ pointwise on $([0,t]\times N)^c$ we have that $X^n$ converges to $X$ almost everywhere. Now, if we don't know that the product measure space ($[0,t] \times \Omega, \mathcal{B}([0,t]) \bigotimes \mathcal{F}_t) $ is complete, then we can't say that the limit $X$ is $\mathcal{B}([0,t]) \bigotimes \mathcal{F}_t$ measurable.
So if anyone can clear up my confusion I would be very appreciative. I've been stumped for weeks now. I think that there must be another way to show the result than how I did with the simpler case above using sequences.
*Update: In theorem 6 of Protter, I was able to show that regardless of whether $\mathcal{F}_t$ is complete or not, $\mathcal{G}^* = \sigma(\{X_T: X \text{ is everywhere cadlag and adapted to } \{\mathcal{F}_t\}\}) = \{A \in \mathcal{F}: \text{ for each } t \geq 0, A \cap T^{-1}([0,t]) \in \mathcal{F_t}\} = \mathcal{F}_T$
However, I still cannot show the analogue for $\mathcal{F_t}$ satisfying the usual conditions.