# Cadlag and adapted imply progressively measurable (related to Protter theorem 6)

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.

• I don't know if it helps but usually the stochastic basis respects the usual conditions which are : right continuity of filtration + $\mathcal{F}_0$ is complete. As you mention that you know that the result is true in case of completeness of the filtration, it should be true under usual conditions. Often these conditions are implicit because ubiquitous, you need them for existence of càdlàg version of local martingale. Regards – TheBridge Jun 11 '18 at 13:14
• I mentioned that the result is true in the case of every sample path being right continuous, but that in the case of just convergence almost everywhere, if I'm correct, then we would need to know the the product sigma algebra was complete to say that the almost everywhere limit is jointly measurable. – Ceeerson Jun 11 '18 at 13:19