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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.

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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.

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  • $\begingroup$ Great! Thank you! $\endgroup$ – quallenjäger Apr 13 '18 at 12:10

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