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Let $\big( X_t , t\geq 0 \big)$ be a measurable process, that is, $$\big( t, \omega \big) \in \mathbb{R}_+\times\Omega\longmapsto X_t(\omega)\in\mathbb{R} \quad\text{is $\mathscr{B}\big( \mathbb{R}_+ \big)\otimes\mathscr{F} \big/\mathscr{B}\big(\mathbb{R} \big)$-measurable. } $$ $T : \Omega\to \big[ 0, +\infty \big]$ be a random time.

Show that the collection $\mathscr{G}$ of all sets of the form $\big( X_T\in A \big)$ and $\big( X_T\in A \big)\cup \big( T = \infty \big)$, $A\in\mathscr{B}(\mathbb{R} )$ is a $\sigma$-algebra on $\Omega$.

Problem 1.17. Karatzas et Shreve.

My confusion is how to understand this question given that $X_T$ is undefined on $\big( T = +\infty \big)$.


EDIT: there would be no problem if we make a change as follows:

Show that the collection $\mathscr{G}$ of all sets of the form $\big( X_T\in A , T < +\infty \big)$ and $\big( X_T\in A , T < +\infty\big)\cup \big( T = \infty \big)$, $A\in\mathscr{B}(\mathbb{R} )$ is a $\sigma$-algebra on $\Omega$.

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  • $\begingroup$ $\big( X_T\in A , T < +\infty \big)=\big( X_T\in A \big)$, so there is no undefined question in this problem. $\endgroup$
    – Danielsen
    May 12, 2015 at 13:06

1 Answer 1

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One way of defining $X_T$ on the whole of $\Omega$ is to let $$ X_T(\omega):= \begin{cases} \lim\limits_{n\to\infty} X_{T(\omega)\wedge n}(\omega)\quad &\text{if the limits exists and is finite},\\ 0 &\text{otherwise}. \end{cases} $$ With this definition we have a random variable $X_T$ defined on $\Omega$ with $X_T(\omega)=X_{T(\omega)}(\omega)$ for $\omega\in\{T<\infty\}$.

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