[proofplan] We verify the three defining properties of a $\sigma$-algebra directly from the definition of $\mathcal{F}_\tau$. The stopping-time hypothesis gives $\{\tau \leq n\} \in \mathcal{F}_n$ for each time $n$, which puts $\Omega$ in $\mathcal{F}_\tau$. Complements and countable unions are handled after intersecting with the event $\{\tau \leq n\}$, where the corresponding closure properties of the fixed-time sigma-algebra $\mathcal{F}_n$ apply. [/proofplan] [step:Put the whole sample space in the stopped sigma-algebra] For each integer $n \geq 0$, define the event $T_n \subset \Omega$ by \begin{align*} T_n = \{\omega \in \Omega: \tau(\omega) \leq n\}. \end{align*} Since $\tau$ is a [stopping time](/page/Stopping%20Time), $T_n \in \mathcal{F}_n$ for every integer $n \geq 0$. Also $\Omega \in \mathcal{F}$, and \begin{align*} \Omega \cap T_n = T_n. \end{align*} Therefore $\Omega \cap T_n \in \mathcal{F}_n$ for every integer $n \geq 0$, so $\Omega \in \mathcal{F}_\tau$. [/step] [step:Show complements remain measurable at every stopped time level] Let $A \in \mathcal{F}_\tau$. Since $A \in \mathcal{F}$ and $\mathcal{F}$ is a $\sigma$-algebra, $A^c \in \mathcal{F}$. Fix an integer $n \geq 0$. By the definition of $\mathcal{F}_\tau$, \begin{align*} A \cap T_n \in \mathcal{F}_n. \end{align*} Since $T_n \in \mathcal{F}_n$ and $\mathcal{F}_n$ is a $\sigma$-algebra, the set difference \begin{align*} T_n \setminus (A \cap T_n) \end{align*} belongs to $\mathcal{F}_n$. But \begin{align*} A^c \cap T_n = T_n \setminus (A \cap T_n). \end{align*} Thus $A^c \cap T_n \in \mathcal{F}_n$ for every integer $n \geq 0$, and hence $A^c \in \mathcal{F}_\tau$. [guided] Let $A \in \mathcal{F}_\tau$. To prove that $A^c \in \mathcal{F}_\tau$, we must check both parts of the definition of $\mathcal{F}_\tau$. First, because $A \in \mathcal{F}_\tau$, the definition already gives $A \in \mathcal{F}$. Since $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$, it is closed under complements, so $A^c \in \mathcal{F}$. Now fix an integer $n \geq 0$. The defining condition for membership in $\mathcal{F}_\tau$ asks us to prove \begin{align*} A^c \cap T_n \in \mathcal{F}_n, \end{align*} where $T_n=\{\omega \in \Omega:\tau(\omega)\leq n\}$. Since $\tau$ is a stopping time, $T_n \in \mathcal{F}_n$. Since $A \in \mathcal{F}_\tau$, we also have \begin{align*} A \cap T_n \in \mathcal{F}_n. \end{align*} The point is that once we have intersected with $T_n$, everything is happening inside the fixed-time sigma-algebra $\mathcal{F}_n$. Because $\mathcal{F}_n$ is closed under relative complements between its elements, the set \begin{align*} T_n \setminus (A \cap T_n) \end{align*} belongs to $\mathcal{F}_n$. This set is exactly $A^c \cap T_n$, because an outcome $\omega$ lies in it precisely when $\omega \in T_n$ and $\omega \notin A$. Hence \begin{align*} A^c \cap T_n = T_n \setminus (A \cap T_n) \in \mathcal{F}_n. \end{align*} Since this holds for every integer $n \geq 0$, the definition of $\mathcal{F}_\tau$ gives $A^c \in \mathcal{F}_\tau$. [/guided] [/step] [step:Show countable unions remain measurable at every stopped time level] Let $(A_k)_{k \geq 1}$ be a sequence of sets in $\mathcal{F}_\tau$. Since $A_k \in \mathcal{F}$ for every integer $k \geq 1$ and $\mathcal{F}$ is closed under countable unions, \begin{align*} \bigcup_{k=1}^{\infty} A_k \in \mathcal{F}. \end{align*} Fix an integer $n \geq 0$. For each integer $k \geq 1$, the definition of $\mathcal{F}_\tau$ gives \begin{align*} A_k \cap T_n \in \mathcal{F}_n. \end{align*} Since $\mathcal{F}_n$ is closed under countable unions, \begin{align*} \bigcup_{k=1}^{\infty}(A_k \cap T_n) \in \mathcal{F}_n. \end{align*} By distributivity of intersection over unions, \begin{align*} \left(\bigcup_{k=1}^{\infty} A_k\right)\cap T_n = \bigcup_{k=1}^{\infty}(A_k \cap T_n). \end{align*} Therefore \begin{align*} \left(\bigcup_{k=1}^{\infty} A_k\right)\cap T_n \in \mathcal{F}_n \end{align*} for every integer $n \geq 0$. Hence $\bigcup_{k=1}^{\infty} A_k \in \mathcal{F}_\tau$. [/step] [step:Conclude that the stopped collection is a sigma-algebra] We have shown that $\Omega \in \mathcal{F}_\tau$, that $\mathcal{F}_\tau$ is closed under complements in $\Omega$, and that $\mathcal{F}_\tau$ is closed under countable unions. These are exactly the defining closure properties of a $\sigma$-algebra on $\Omega$. Therefore $\mathcal{F}_\tau$ is a $\sigma$-algebra on $\Omega$. [/step]