I'm looking for an example of a stochastic process $X$ that is $\mathbb{F}$-adapted, but not progressively measurable. One example I found is the following:
$(\Omega, \mathfrak{A}) = (\mathbb{R^+}, \mathfrak{B}(\mathbb{R^+}))$, $\mathcal{F}_t = \sigma({x}:x \in \mathbb{R}^+)$ for all $t \geq 0$;
$X(t,\omega) := \cases{0, \mbox{if}~ \omega \neq t,\\
1, \mbox{if}~ \omega = t}$.
I don't see why this process is adapted, but not progressively measurable.
For $X_t$ we have
$X_t (\omega) = \cases{0, \mbox{if}~ \omega \neq t,\\
1, \mbox{if}~ \omega = t}.$
Then
$X_t^{-1}(\{0\}) = \emptyset$
$X_t^{-1}(\{1\}) = \{t\}$
and since $\emptyset,\{t\} \in \mathcal{F}_t$ for all $t \geq 0$, $X_t$ is $\mathcal{F}_t$-measurable and therefore adapted. (correct so far?)
But can someone please explain, how I can see, that this process is not progressively measurable?
Or does someone have another easy understandable example?
Best Answer
I find the following example easier to understand:
Consider any probability space $(\Omega,\mathcal{A},\mathbb{P})$. Set $\mathcal{F}_t := \mathcal{A}$ for all $t \geq 0$ and let $A \subseteq [0,T]$ be a set which is not Borel-measurable. The process
$$X_t(\omega) := 1_A(t)$$
takes only values in $\{0,1\}$ and for each fixed $t \geq 0$, we have either $X_t = 1$ or $X_t = 0$. This implies in particular that $\omega \mapsto X_t(\omega)$ is measurable (it is a constant!). On the other hand, we have
$$\{(t,\omega) \in [0,T] \times \Omega; X_t(\omega) = 1\} = A \times \Omega \notin \mathcal{B}(\mathbb{R}) \otimes \mathcal{A}$$
and this means that the mapping
$$X: [0,T] \times \Omega \to \mathbb{R}$$
is not measurable, i.e. $(X_t)_t$ is not progressively measurable.