[proofplan] Fix a time $n$ and prove that the stopped [random variable](/page/Random%20Variable) $Y_n = X_{n \wedge \tau}$ is $\mathcal F_n$-measurable. For an arbitrary measurable set $B \in \mathcal E$, decompose the event $\{Y_n \in B\}$ according to whether the stopping time has exceeded $n$ or has taken one of the values $0,\dots,n$. The stopping-time property gives measurability of the time events, adaptedness gives measurability of the process events, and filtration monotonicity places every piece in $\mathcal F_n$. Since this works for every $B \in \mathcal E$, $Y_n$ is $\mathcal F_n$-measurable for every $n$, which is exactly adaptedness. [/proofplan] [step:Fix a time and decompose the stopped inverse image] Fix $n \in \mathbb N_0$. Define the map $Y_n: \Omega \to E$ by \begin{align*} Y_n(\omega) := X_{n \wedge \tau(\omega)}(\omega). \end{align*} This is well-defined because $n \wedge \tau(\omega) \in \{0,\dots,n\}$ for every $\omega \in \Omega$, using the convention $n \wedge \infty = n$. Let $B \in \mathcal E$. The event $Y_n^{-1}(B)=\{\omega \in \Omega : Y_n(\omega) \in B\}$ admits the decomposition \begin{align*} Y_n^{-1}(B)=\left(\{\tau>n\}\cap X_n^{-1}(B)\right)\cup\bigcup_{k=0}^{n}\left(\{\tau=k\}\cap X_k^{-1}(B)\right). \end{align*} Indeed, for each $\omega \in \Omega$, either $\tau(\omega)>n$, in which case $n \wedge \tau(\omega)=n$, or $\tau(\omega)=k$ for a unique $k \in \{0,\dots,n\}$, in which case $n \wedge \tau(\omega)=k$. [guided] Fix $n \in \mathbb N_0$. Our goal is to prove that the $E$-valued random variable stopped at time $n$ is measurable with respect to $\mathcal F_n$. Define the map $Y_n: \Omega \to E$ by \begin{align*} Y_n(\omega) := X_{n \wedge \tau(\omega)}(\omega). \end{align*} This definition is meaningful because $\tau(\omega)$ takes values in $\mathbb N_0 \cup \{\infty\}$ and the convention $n \wedge \infty = n$ ensures that $n \wedge \tau(\omega)$ is always one of the finitely many indices $0,\dots,n$. To prove that $Y_n$ is $\mathcal F_n$-$\mathcal E$ measurable, it is enough to prove that $Y_n^{-1}(B) \in \mathcal F_n$ for every $B \in \mathcal E$. Fix such a set $B$. We split $\Omega$ according to the value of $\tau$ relative to $n$. If $\tau(\omega)>n$, then the stopped index is $n$, so $Y_n(\omega)=X_n(\omega)$. If $\tau(\omega)=k$ for some $k \in \{0,\dots,n\}$, then the stopped index is $k$, so $Y_n(\omega)=X_k(\omega)$. Therefore \begin{align*} Y_n^{-1}(B)=\left(\{\tau>n\}\cap X_n^{-1}(B)\right)\cup\bigcup_{k=0}^{n}\left(\{\tau=k\}\cap X_k^{-1}(B)\right). \end{align*} The union is finite because only the stopping values $0,\dots,n$ can occur before or at time $n$. This finite decomposition is the central point of the proof: it reduces measurability of the stopped variable to measurability of finitely many ordinary process variables and finitely many stopping-time events. [/guided] [/step] [step:Place the stopping-time events in the correct sigma-algebras] The event $\{\tau>n\}$ belongs to $\mathcal F_n$ because \begin{align*} \{\tau>n\}=\Omega\setminus\{\tau\le n\}, \end{align*} and $\{\tau\le n\}\in\mathcal F_n$ by the stopping-time property. For $k=0$, the event $\{\tau=0\}$ satisfies \begin{align*} \{\tau=0\}=\{\tau\le 0\}, \end{align*} so $\{\tau=0\}\in\mathcal F_0$. For $1 \le k \le n$, \begin{align*} \{\tau=k\}=\{\tau\le k\}\setminus\{\tau\le k-1\}. \end{align*} The stopping-time property gives $\{\tau\le k\}\in\mathcal F_k$ and $\{\tau\le k-1\}\in\mathcal F_{k-1}$. Since $(\mathcal F_m)_{m \in \mathbb N_0}$ is a filtration, $\mathcal F_{k-1}\subset \mathcal F_k$, hence $\{\tau=k\}\in\mathcal F_k$. [/step] [step:Use adaptedness and filtration monotonicity to prove measurability of every piece] Since $(X_m)_{m \in \mathbb N_0}$ is adapted, each map $X_m: \Omega \to E$ is $\mathcal F_m$-$\mathcal E$ measurable. Therefore $X_k^{-1}(B)\in\mathcal F_k$ for every $0 \le k \le n$, and $X_n^{-1}(B)\in\mathcal F_n$. For each $0 \le k \le n$, filtration monotonicity gives $\mathcal F_k\subset \mathcal F_n$. Hence \begin{align*} \{\tau=k\}\cap X_k^{-1}(B)\in\mathcal F_n. \end{align*} Also $\{\tau>n\}\cap X_n^{-1}(B)\in\mathcal F_n$. Since $\mathcal F_n$ is closed under finite unions, the decomposition from the first step gives \begin{align*} Y_n^{-1}(B)\in\mathcal F_n. \end{align*} [/step] [step:Conclude that the stopped process is adapted] We have shown that for the fixed $n \in \mathbb N_0$ and every $B \in \mathcal E$, the inverse image $Y_n^{-1}(B)$ belongs to $\mathcal F_n$. Thus $Y_n$ is $\mathcal F_n$-$\mathcal E$ measurable. Since $n \in \mathbb N_0$ was arbitrary, $(Y_n)_{n \in \mathbb N_0}=(X_{n\wedge\tau})_{n \in \mathbb N_0}$ is adapted to $(\mathcal F_n)_{n \in \mathbb N_0}$. [/step]