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Let's assume we are on a filtered probability space $(\Omega, \mathcal{F}, P, \mathbb{F})$ where $\mathbb{F} := \{\mathcal{F}_t\}_{t \ge 0}$ is a filtration satisfying the usual conditions (right continuous + complete). It is obvious that for a sequence of stopping times $\tau_n$, if they converge a.s. to some function $\tau$, that $\tau$ is a stopping time. I am curious to know if the same thing holds for subsequences of $\{\tau_n\}_{n\in \mathbb{N}}$, where the subsequence may depend on $\omega \in \Omega$. Specifically, suppose for almost every $\omega \in \Omega$, there exists a subsequence $n_i(\omega)$ such that $$\lim_{i \rightarrow \infty} \tau_{n_i(\omega)}(\omega) := \tau(\omega) \qquad (1)$$ exists. Are there conditions under which $\tau(\omega)$ is a stopping time? The standard proof that works for the typical setup does not work, since $$\{\omega : \tau(\omega) \leq t\} = \bigcap_{k=1}^\infty\left\{\omega : \exists i\in \mathbb{N} : \forall j \ge i, \tau_{n_j(\omega)}(\omega) \leq t + \frac{1}{k} \right\},$$ and one cannot write this set in terms of a union/intersection of sets of the form $\{\tau_n \leq t+1/k\}$ since the choice of $i$ and the sequence $n_j$ both depend on $\omega$.

My ideas: Suppose we are working on canonical space (i.e. $\Omega = C[0,\infty)$, the set of continuous functions on $[0, \infty)$ and we work with the filtration generated by the functions $X_t(\omega) := \omega(t)$). Then, it simply suffices to show that the function $\tau_{n_i(\omega)}(\omega)$ is measurable, due to Galmarino's test (see Proving Galmarino's Test for details).

To show measurability in such a setup, this is the idea I have. Suppose each $\tau_n$ depends on a real number, $x$, and, for almost every $\omega \in \Omega$, the family $\{\tau_n(x)\}_{n \in \mathbb{N}}$ is uniformly bounded and, say, Lipschitz continuous in the random variable $x$ with common Lipschitz coefficient $M$. The Arzela Ascoli Theorem implies the condition $(1)$. Is it possible to make the selections of the subsequences in a $\mathcal{F}$ measurable way?

RobPratt
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qp212223
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2 Answers2

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Example: To be concrete, let the setting be that of a standard $1$-dimensional Brownian motion $(B_t)_{t\ge 0}$ with its complete natural filtration. The random time $$ \tau:=\sup\{t\in[0,1]: B_t=0\} $$ is not a stopping time, as is well known. Let $\{q_n\}$ be an enumeration of the rational numbers in $[0,1]$, and let $\tau_n(\omega)=q_n$ for each $n$ and each sample point $\omega$. Each $\tau_n$, being constant, is a stopping time. To define $n_i(\omega)$, observe that the set $$ \{n\in\Bbb N: |q_n-\tau(\omega)|<i^{-1}\} $$ is non-empty. Let $n_i(\omega)$ be its least element. Then $\tau_{n_i(\omega)}(\omega)=q_{n_i(\omega)}$ differs from $\tau(\omega)$ by at most $i^{-1}$, so $$ \lim_i \tau_{n_i(\omega)}(\omega)=\tau(\omega). $$

(There's nothing really special about the $\tau$ used here except that it's not a stopping time.)

John Dawkins
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I have a potential answer. Suppose that $(x,\omega) \mapsto\tau_n(x,\omega)$ is uniformly equicontinuous. Then you can still use Arzela-Ascoli (since $C[0, \infty)$ is separable) to state that function $\tau(x,\omega)$ is jointly continuous in $(x,\omega)$ and hence measurable. In fact, this argument avoids the entire problem of the subsequences being different for each $\omega$, since there is a single subsequence along with the convergence is uniform on $\mathbb{R} \times \Omega$. Hence, one does not even need Galmarino anymore.

I believe this is valid, but please let me know if it isn't. Also, I would be happy to know, in general, if one can choose the sequences measurably (as this would of course be a more general result).

qp212223
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