Adaptedness of random variables

Question

Suppose we have an RCLL adapted process $(X_t)$. Moreover we are given a stopping time $T$. We define $\mathcal{F}_T=\{A\in\mathcal{F}\mid A\cap\{T\le t\}\in \mathcal{F}_t, \text{ for all }t\ge0\}$. Now I have several question concerning measurability of products of r.v.:

1. Why is $X_T\mathbf1_{\{T<t\}}\in\mathcal{F}_{T\wedge t}$?
2. For $H\in\mathcal{F}_t$, why is $H\mathbf1_{\{T\ge t\}}\in \mathcal{F}_T$?
3. Why does 2. imply: $E[X_T\mathbf1_{\{T\ge t\}}|\mathcal{F}_{t\wedge T}]=E[X_T|\mathcal{F}_t]\mathbf1_{T\ge t}$?

where $\mathbf1_A$ denotes the indicator function of a set $A$. Many thanks in advance!

Answer 1

I'll take the following results for granted. If you're not familiar with them, you should prove them.

Result 1: Let $X:\Omega\to\mathbb{R}$ be a mapping. Then $X$ is $\mathcal{F}$-$\mathcal{B}(\mathbb{R})$-measurable if and only if $X^{-1}(B)\in\mathcal{F}$ for all $B\in\mathcal{B}(\mathbb{R})$ with $0\notin B$.

Result 2: If $\sigma$ and $\tau$ are two stopping times and $A\in\mathcal{F}_\sigma$, then $A\cap \{\sigma\leq \tau\}\in\mathcal{F}_{\tau}$.

1) Since $\mathcal{F}_{T\wedge t}=\mathcal{F}_T\cap \mathcal{F}_t$, we need to show that $X_T\mathbf{1}_{T<t}$ is both $\mathcal{F}_T$- and $\mathcal{F}_t$-measurable.

The $\mathcal{F}_T$-measurability: Since $(X_t)$ is RCLL it is also jointly measurable, and hence $X_T$ is $\mathcal{F}_T$-measurable. Since also $\{T<t\}\in\mathcal{F}_T$ we have that $X_T\mathbf{1}_{T<t}$ is $\mathcal{F}_T$-measurable.

The $\mathcal{F}_t$-measurability: For $B\in\mathcal{B}(\mathbb{R})$ with $0\notin B$ we have $$ (X_T\mathbf{1}_{T<t})^{-1}(B)=X_T^{-1}(B)\cap \{T<t\}=\bigcup_n \left(X_T^{-1}(B)\cap\left\{T\leq t-\tfrac1n\right\}\right)\in\mathcal{F}_t, $$ by definition of $\mathcal{F}_T$.

2) $(H\mathbf{1}_{T\geq t})^{-1}(B)=H^{-1}(B)\cap \{T\geq t\}\in\mathcal{F}_{T}$ for all $B\in\mathcal{B}(\mathbb{R})$ with $0\notin B$ by Result 2.

3) Using Result 2 with $\sigma=t$ and $\tau=T$, we have $A\cap\{T\geq t\}\in\mathcal{F}_{T}$ for all $A\in\mathcal{F}_t$.

Now, let $Y={\rm E}[X_T\mid\mathcal{F}_t]$. Then we have to argue that $Y\mathbf{1}_{T\geq t}$ is $\mathcal{F}_{t\wedge T}$-measurable and that ${\rm E}[X_T\mathbf{1}_{T\geq t}\mathbf{1}_A]={\rm E}[Y\mathbf{1}_{T\geq t}\mathbf{1}_A]$ for all $A\in\mathcal{F}_{t\wedge T}$.

The measurability follows from the fact that for $B\in\mathcal{B}(\mathbb{R})$ with $0\notin B$ we have $$ \left(Y\mathbf{1}_{T\geq t}\right)^{-1}(B)=Y^{-1}(B)\cap\{T\geq t\}\in\mathcal{F}_T\cap\mathcal{F}_t $$ since $Y^{-1}(B)\in\mathcal{F}_t$ and $\{T\geq t\}\in\mathcal{F}_t$.

For the last part, let $A\in\mathcal{F}_{T\wedge t}$, then $$ {\rm E}[X_T\mathbf{1}_{T\geq t}\mathbf{1}_A]={\rm E}[X_T\mathbf{1}_{\{T\geq t\}\cap A}]={\rm E}[Y\mathbf{1}_{\{T\geq t\}\cap A}]={\rm E}[Y\mathbf{1}_{T\geq t}\mathbf{1}_A], $$ where the middle equality follows by the definition of $Y$ and the fact that $\{T\geq t\}\cap A\in\mathcal{F}_t$.

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Proof of Result 1: Suppose $X^{-1}(B)\in\mathcal{F}$ for all $B\in\mathcal{B}(\mathbb{R})$ with $0\notin B$. Now suppose that $B\in\mathcal{B}(\mathbb{R})$ and $0\in B$. Then $0\notin B^c$, and hence $X^{-1}(B^c)\in\mathcal{F}$ and therefore also $$ X^{-1}(B)=(X^{-1}(B^c))^c\in\mathcal{F}. $$