Why $1_{T=\infty}X_{\infty}$ is $\mathcal{F}_T$-measurable for a martingale? For a filtration $\mathcal{F}_0\subset\dots\subset \mathcal{F}_t\subset\dots\subset \mathcal{F}_{\infty}$. We know that if $X$ is $\mathcal{F}_0$-measurable, then $X$ is also $\mathcal{F}_t$-measurable for $t\ge 0$.
I have a question about the martingale $X_t$ (which $\mathcal{F}_t$ measurable) is  that given a stopping time $T$,
why $1_{T=\infty}X_{\infty}$ is $\mathcal{F}_T$-measurable?
It is clear that $1_{T=\infty}$ is $\mathcal{F}_T$-measurable. But because $X_{\infty}$ is $\mathcal{F}_{\infty}$-measurable. How can we say $X_{\infty}$ is also $\mathcal{F}_{T}$-measurable?
 A: $X_\infty$ does not need to be $\mathcal F_T$ measurable,  but $1_{T = \infty} X_\infty$ is.  We can show this directly from the definition:
We want to show $\{1_{T=\infty}X_\infty \in B\} \in \mathcal F_T$ for any Borel set $B$.  By the definition of $\mathcal F_T$, this means we must show $\{1_{T=\infty}X_\infty \in B\} \cap \{T \le t\} \in \mathcal F_t$ for all $t > 0$.  On the event $\{T \le t\}$, $1_{T=\infty}X_\infty = 0$, so $$\{1_{T=\infty}X_\infty \in B\} \cap \{T \le t\} = \{T \le t \}$$ if $0 \in B$ and $$\{1_{T=\infty}X_\infty \in B\} \cap \{T \le t\} = \emptyset$$ if $0 \not \in B$.  In either case, $\{1_{T=\infty}X_\infty \in B\} \cap \{T \le t\} \in \mathcal F_t$, so $1_{T = \infty}X_\infty$ is $\mathcal F_T$ measurable.
A: Let $Y_\infty\equiv1_{T=\infty}X_\infty$, and let $S_a \equiv Y_\infty^{-1}(]-\infty,a]) = \left\{\omega : Y_\infty(\omega) \le a\right\}$ for $a\in\mathbb R$. To show that $Y_\infty$ is $\mathcal F_T$-measurable, it is enough to show that every set $S_a$ is in $\mathcal F_T$, as $\sigma(\left\{S_a,a\in\mathbb R\right\}) = \mathscr B(\mathbb R)$.
By definition of $\mathcal F_T$, for any $a\in\mathbb R$, $S_a\in\mathcal F_T$ if and only if $$S_a\cap\left\{T\le t \right\}\in\mathcal F_t \text{ for all } t\in[0,\infty].$$
And this is always the case. Indeed, fix $a\in\mathbb R$, you have that for all $t\in[0,\infty[$,
$$S_a\cap\left\{T\le t \right\} = \left\{\omega : T(\omega)\le t \text{ and } Y_\infty(\omega) \le a \right\} = \left\{T\le t\right\}\cap\left\{a\ge 0\right\} $$
If $a\in \mathbb R^+$ then $S_a\cap\left\{T\le t \right\} = \left\{T\le t \right\} \in \mathcal F_t$ otherwise $S_a\cap\left\{T\le t \right\} =\emptyset \in \mathcal F_t$.
Lastly, for $t=\infty$ you directly have by $\mathcal F_\infty$ measurability of $Y_\infty$ that $S_a\cap\left\{T\le \infty \right\} = S_a$ is in $\mathcal F_\infty$. You can thus conclude that $Y_\infty=1_{T=\infty}X_\infty$ is $\mathcal F_T$ measurable.
