Let $\mathbb{F} = \{ \mathcal{F}_t \}_{t \geq 0}$ be a continuous time filteration. $\tau : \Omega \to [0, \infty]$ is called an $\mathbb{F}$-stopping time if $\{ \tau \leq t \} \in \mathcal{F}_t$ for all $t\geq 0$.

Let $\tau_n$, $n \geq 1$ be $\mathbb{F}$-stopping times. We know that $sup_{n \geq 1} \tau_n$ is also a $\mathbb{F}$-stopping time. I want to find a counterexample such that $\tau_i$, $i \in I$ (where $I$ is uncountable) be $\mathbb{F}$-stopping times, but $sup_{i \in I} \tau_i$ is not a $\mathbb{F}$-stopping time. Any idea?


1 Answer 1


Let the sample space be $[0,1]$ and let each element of the filtration be the borel $\sigma-algebra$. Let $I$ be some non-measurable subset and $\tau_x=\mathbb{1}(x), x\in [0,1]$. Then $\{\tau_x<t\}$ is the complement of a singleton for $x\le 1$ and $[0,1]$ otherwise so it belongs to each $F_t$, but $\{sup_{x\in I}\tau_x<t\}=\cap_{x\in I}\{\tau_x <t\}=\cap_{x\in I}\bar{x}=\bar{I}$ isn't measurable for $t< 1$ (but is measurable for times greater than 1).

  • $\begingroup$ Yes, I also knew about this example. But I was hoping that whether I can find an example whether the sup function is measurable yet not a stopping time. $\endgroup$
    – m0_as
    Feb 7, 2016 at 0:57
  • $\begingroup$ measurable wrt what $\sigma$-algebra, the limiting filtration? $\endgroup$
    – snarfblaat
    Feb 7, 2016 at 1:24
  • $\begingroup$ I think your example is correct, because showing $\tau$ is a stopping time in fact means showing that it is measurable. So, to find a counterexample, we need to violate measurability. $\endgroup$
    – m0_as
    Feb 7, 2016 at 1:28

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