# If $f$ is $\mathcal F_{\tau-}$-measurable, is $1_{\{\:s\:<\:\tau\:\}}f$ is $\mathcal F_s$-measurable for all $s\ge0$?

Let

• $$(\Omega,\mathcal A)$$ 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 $$A\in\mathcal G_{\tau-}$$.

• 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. Commented Jun 17, 2022 at 19:51
• @user6247850 It is the definition you can find in the book of Kallenberg, for example. Commented Jun 17, 2022 at 20:50

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