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Let $(X_t)$ a progressively measurable process, i.e. $[0,t]\times \Omega \ni (t,\omega )\mapsto X_t(\omega )\in (\mathbb R, \mathcal B(\mathbb R))$ is $\mathcal B([0,t])\otimes \mathcal F$ measurable. Prove that $X_T$ is $\mathcal F_T$ measurable on $\{T<\infty \}$.

Q1) What does it mean that $X_T$ is $\mathcal F_T$ measurable on $\{T<\infty \}$ ? Does it mean that for all Borel set $B$, we have $(X_T\boldsymbol 1_{\{T<\infty \}})^{-1}(B)\in \mathcal F_T$ ?

Suppose my definition is correct. Then that's enough to prove

$\{X_T\boldsymbol 1_{\{T<\infty \}}\leq x\}\in \mathcal F_T$ for all $x$. ($*$)

Q2) Any idea on how to prove ($*$) ?

I don't see how I can prove that

$\{X_T\boldsymbol 1_{\{T<\infty \}} \} \in \mathcal F_\infty $ and $\{X_T\boldsymbol 1_{\{T<\infty \}}\leq x\}\cap \{T\leq t\}\in \mathcal F_t$ for all $t$.

Marine Galantin
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user657324
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3 Answers3

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Hint

Let $B$ a Borel set.

$$\{X_{T}\in B\}\cap \{T\leq t\}=\{X_{T\wedge t}\in B\}\cap \{T\leq t\}.$$

Therefore, if you can prove that the stopped process is progressively measurable, the claim follow.

Surb
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As mentionned Surb, it isn't a complete proof as it is for discrete and not continuous time. I leave it for sake of learning purpose.

After some research by myself, I was finally able to solve that question :

Write :

$$ \begin{align} \small (X^T_n)^{-1}(B)&=\bigcup_{k\geq 0}(X_{k\wedge n}^{-1}(B)\cap \{\omega\mid T(\omega)=k\}) \\&=\bigcup_{k=0}^{n-1}(X_k^{-1}(B)\cap\{\omega\mid T(\omega)=k\}) \cup (X_n^{-1}(B)\cap \{\omega\mid T(\omega)\leq n\}^c) \\& \in\mathcal F_n. \end{align} $$

Notice that the first equality is true iff $ T < \infty $. If you need more detail let me know.

Marine Galantin
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    $T$ is not supposed discrete. In fact, it's not discrete (since $(X_t)$ is a process in continuous time). So your proof doesn't hold. – Surb Feb 18 '20 at 16:39
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Proof: provided that $T$ takes a finite value, for every positive value $t$ it will take, we will be able to decide whether $X_t$ belongs to a Borel set $B$, whatever set we choose.

Formaly, we need to prove for all Borel set $B$ and $t\ge0$ that $$ \left\{\omega\in\Omega,\;X_{T(\omega)}(\omega)\in B\right\}\cap\left\{\omega\in\Omega,\;T(\omega)\le t\right\} $$ is $\mathcal F_t$-measurable. But \begin{alignat*}{2} \left\{X_{T}\in B\right\}\cap\left\{T\le t\right\} & = \left\{\omega\in\Omega:\; T(\omega) = s, s \le t, X_s(\omega) \in B \right\} \end{alignat*} which is $\mathcal F_t$-mesurable.

thomasb
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  • I am sorry it isn't really clear to me. Possibly, could you write it differently please ? I don't get why you say that you want to show that a set is sigma measurable and what you're showing. – Marine Galantin Feb 18 '20 at 15:11
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    This proof doesn't work as well. There is no reason for $\bigcup_{s<t}{T=s, X_s\in B}$ to be measurable. – Surb Feb 18 '20 at 17:12