In the Protter's book, they want to show that for an adapted Cadlag process $X$, the stopped process $X_T$ is $\mathcal F_T$ measurable, where $$\mathscr F_T:=\{A\in \mathscr F \mid A\cap\{T<t\}\in\mathscr F_t, \text{all }t>0\}.$$

What they did is to construct $\varphi:\{T\leq t\}\rightarrow [0,\infty)\times\Omega$ by $\varphi(\omega)=(T(\omega),\omega)$. Then since $X$ is adapted and cadlag, we have $X_T=X\circ\varphi$ is a measurable mapping from $({T\leq t},\mathcal F_t\cap\{T\leq t\} )$ into $(\Bbb R, \mathcal B(\Bbb R))$.

What I couldn't follow is the last argumentation:

  1. Why is the map $X_T$ measurable with respect to $\mathcal F_t\cap \{T\leq t\}$? Do I proof it via the composition map or directly? If I prove it as a composition map, I would require the $X_t$ to be progressively measurable since $\varphi$ is a mapping to the product space. However, is it given here?
  2. Why do we need the cadlag property of the process?

I would appreciate for any hints.


1 Answer 1


Using a standard approximation procedure, it is not difficult to see that any adapted càdlàg process is progressively measurable, i.e.

$$([0,t] \times \Omega,\mathcal{B}([0,t]) \otimes \mathcal{F}_t) \ni (s,\omega) \mapsto X(s,\omega) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$

is measurable for any $t \geq 0$. Since

$$(\{T \leq t\},\mathcal{F}_t) \ni \omega \mapsto \varphi(\omega) = (T(\omega),\omega) \in ([0,t] \times \Omega,\mathcal{B}([0,t]) \otimes \mathcal{F}_t)$$

is measurable, it follows that the composition

$$(\{T \leq t\},\mathcal{F}_t) \ni \omega \mapsto X_T(\omega) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$

is measurable, i.e.

$$\{T \leq t\} \cap \{X_T \in B\} \in \mathcal{F}_t$$

for any Borel set $B$. This is equivalent to saying that $X_T$ is $\mathcal{F}_T$-measurable.

  • $\begingroup$ Great! Thank you! $\endgroup$ Apr 13, 2018 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.