[proofplan] We fix a deterministic time $n$ and rewrite the event that the entrance time has occurred by time $n$ as a finite union of the events that the process is in $A$ at one of the times $0,\dots,n$. Adaptedness makes each event $\{X_k\in A\}$ measurable with respect to $\mathcal F_k$. The filtration property then upgrades this measurability to $\mathcal F_n$ for every $k\le n$, and closure of $\mathcal F_n$ under finite unions gives the stopping-time condition. [/proofplan] [step:Rewrite the event of entrance by time $n$ as a finite union] Fix $n\in\mathbb Z_{\ge 0}$. For each $k\in\{0,\dots,n\}$, define the event \begin{align*} B_k:=\{\omega\in\Omega:X_k(\omega)\in A\}=X_k^{-1}(A). \end{align*} We claim that \begin{align*} \{\omega\in\Omega:\tau_A(\omega)\le n\}=\bigcup_{k=0}^{n}B_k. \end{align*} Indeed, if $\omega\in\bigcup_{k=0}^{n}B_k$, then $X_k(\omega)\in A$ for some $k\le n$, so the infimum defining $\tau_A(\omega)$ is at most $k$ and hence at most $n$. Conversely, if $\tau_A(\omega)\le n$, then the set $\{m\in\mathbb Z_{\ge 0}:X_m(\omega)\in A\}$ is nonempty and has least element $\tau_A(\omega)$, so $X_{\tau_A(\omega)}(\omega)\in A$ and $\tau_A(\omega)\in\{0,\dots,n\}$; hence $\omega\in B_{\tau_A(\omega)}\subseteq\bigcup_{k=0}^{n}B_k$. [guided] Fix $n\in\mathbb Z_{\ge 0}$. The stopping-time condition asks for measurability of the event that $\tau_A$ has occurred by time $n$, so we first express that event using only the process values up to time $n$. For each $k\in\{0,\dots,n\}$, define \begin{align*} B_k:=\{\omega\in\Omega:X_k(\omega)\in A\}=X_k^{-1}(A). \end{align*} This is the event that the process enters the measurable set $A$ exactly at the observation time $k$, without requiring that this be the first such time. We prove the set identity \begin{align*} \{\omega\in\Omega:\tau_A(\omega)\le n\}=\bigcup_{k=0}^{n}B_k. \end{align*} First suppose $\omega\in\bigcup_{k=0}^{n}B_k$. Then there exists $k\in\{0,\dots,n\}$ such that $\omega\in B_k$, meaning $X_k(\omega)\in A$. Therefore $k$ belongs to the set \begin{align*} \{m\in\mathbb Z_{\ge 0}:X_m(\omega)\in A\}. \end{align*} The infimum of this set is at most $k$, so $\tau_A(\omega)\le k\le n$. Hence $\omega\in\{\tau_A\le n\}$. Conversely suppose $\omega\in\Omega$ satisfies $\tau_A(\omega)\le n$. Since $n$ is finite, $\tau_A(\omega)\ne\infty$, so the set \begin{align*} \{m\in\mathbb Z_{\ge 0}:X_m(\omega)\in A\} \end{align*} is nonempty. Because it is a nonempty subset of $\mathbb Z_{\ge 0}$, it has a least element, and this least element is exactly $\tau_A(\omega)$. Thus $X_{\tau_A(\omega)}(\omega)\in A$. Also $\tau_A(\omega)\le n$, so $\tau_A(\omega)\in\{0,\dots,n\}$. Therefore $\omega\in B_{\tau_A(\omega)}\subseteq\bigcup_{k=0}^{n}B_k$. This proves the identity. [/guided] [/step] [step:Use adaptedness and the filtration property to prove measurability] For each $k\in\{0,\dots,n\}$, the set $A$ belongs to $\mathcal E$ and the map $X_k:(\Omega,\mathcal F_k)\to(E,\mathcal E)$ is measurable by adaptedness. Hence \begin{align*} B_k=X_k^{-1}(A)\in\mathcal F_k. \end{align*} Since $(\mathcal F_m)_{m\in\mathbb Z_{\ge 0}}$ is a filtration and $k\le n$, we have $\mathcal F_k\subseteq\mathcal F_n$, so $B_k\in\mathcal F_n$ for every $k\in\{0,\dots,n\}$. Because $\mathcal F_n$ is a $\sigma$-algebra, it is closed under finite unions. Therefore \begin{align*} \{\omega\in\Omega:\tau_A(\omega)\le n\}=\bigcup_{k=0}^{n}B_k\in\mathcal F_n. \end{align*} Since $n\in\mathbb Z_{\ge 0}$ was arbitrary, $\tau_A$ is a [stopping time](/page/Stopping%20Time) with respect to $(\mathcal F_n)_{n\in\mathbb Z_{\ge 0}}$. [/step]