• $(\Omega,\mathcal A)$ be a measurable space;
  • $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$;
  • $\tau:\Omega\to[0,\infty]$ be an $(\mathcal F_t)_{t\ge0}$-stopping time on $(\Omega,\mathcal A)$, $$\mathcal G_{\tau-}:=\left\{A\cap\{t<\tau\}:A\in\mathcal F_t\text{ and }t\ge0\right\}$$ and $$\mathcal F_{\tau-}:=\sigma(\mathcal F_0\cup\mathcal G_{\tau-}).$$

Let $f:\Omega\to\mathbb R$ be $\mathcal F_{\tau-}$-measurable. Can we show that $1_{\{\:s\:<\:\tau\:\}}f$ is $\mathcal F_s$-measurable for all $s\ge0$?

Intuiteively, $\mathcal F_{\tau-}$ should contain all the information immediately before $\tau$. So, on the event $\{s<\tau\}$, we should have the information in $\mathcal F_s$.

But how can we prove this? Most probably (if at all) by a monotone class argument, but I'm failing to see this even for $f=1_A$ with $A\in\mathcal G_{\tau-}$.

  • 2
    $\begingroup$ Are you sure that's the definition of $\mathcal G_{\tau-}$ you want? The definition I'm familiar with is that $\mathcal G_{\tau-} = \{A \in \mathcal F_{\infty}: A \cap \{t < \tau\} \in \mathcal F_t \text{ for all } t \ge 0\}$. It isn't clear to me that the definition you wrote is even a $\sigma$-algebra. $\endgroup$ Jun 17, 2022 at 19:51
  • $\begingroup$ @user6247850 It is the definition you can find in the book of Kallenberg, for example. $\endgroup$
    – 0xbadf00d
    Jun 17, 2022 at 20:50

1 Answer 1


Please consider the following example. Let $\{\xi_n,n\ge 1\}$ be a sequence of independent non-degenerate random variables on probability space $(\Omega,\mathscr{F},\mathsf{P})$. For $t\ge 0$, set \begin{equation*} \mathscr{F}_t=\mathscr{F}_{[t]}=\sigma(\xi_i,1\le i\le t)\vee \mathscr{N}. \tag{1} \end{equation*} Taking stopping time $\tau=3$ then \begin{equation*} f=\xi_2\in \mathscr{F}_{\tau-}(=\mathscr{F}_{2}). \tag{2} \end{equation*}
Now, for $s=1$, \begin{equation*} 1_{s<\tau}f=1_{1<3}\xi_2=\xi_2\notin \mathscr{F}_1(=\mathscr{F}_s). \tag{3} \end{equation*} (3) means that, in this case, $1_{s<\tau}f$ is not $\mathscr{F}_s$-meausrable.

  • $\begingroup$ The index set in OP is clearly $[0,\infty)$. $\endgroup$
    – zhoraster
    Jun 28, 2022 at 12:56

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