[proofplan]
We prove measurability by testing an arbitrary measurable target set $B\in\mathcal E$. The preimage $\{X_\tau\in B\}$ is decomposed according to the disjoint alternatives $\{\tau=k\}$ for finite $k$ and the terminal alternative $\{\tau=\infty\}$. The finite exact-time pieces are measurable because $\tau$ is a [stopping time](/page/Stopping%20Time) and $X_k$ is $\mathcal F_k$-$\mathcal E$ measurable. Intersecting with $\{\tau\le n\}$ removes the infinite-time branch and leaves only a finite union of events in $\mathcal F_n$, which is precisely the defining condition for membership in $\mathcal F_\tau$.
[/proofplan]
[step:Derive exact-time measurability from the stopping-time condition]
For each $k\in\mathbb N_0$, define the event
\begin{align*}
T_k:=\{\omega\in\Omega:\tau(\omega)=k\}.
\end{align*}
We claim that $T_k\in\mathcal F_k$ for every $k\in\mathbb N_0$. If $k=0$, then $T_0=\{\tau\le 0\}$, so $T_0\in\mathcal F_0$ by the stopping-time hypothesis. If $k\ge 1$, then
\begin{align*}
T_k=\{\tau\le k\}\cap\{\tau\le k-1\}^c.
\end{align*}
Here $\{\tau\le k\}\in\mathcal F_k$ by the stopping-time hypothesis, and $\{\tau\le k-1\}\in\mathcal F_{k-1}\subset\mathcal F_k$ by filtration monotonicity. Since $\mathcal F_k$ is a $\sigma$-algebra, it is closed under complements and finite intersections; hence $T_k\in\mathcal F_k$.
Also define the event
\begin{align*}
T_\infty:=\{\omega\in\Omega:\tau(\omega)=\infty\}.
\end{align*}
Because $\tau$ takes values in $\mathbb N_0\cup\{\infty\}$,
\begin{align*}
T_\infty=\Omega\setminus\bigcup_{m=0}^{\infty}\{\tau\le m\}.
\end{align*}
For each $m\in\mathbb N_0$, the event $\{\tau\le m\}$ belongs to $\mathcal F_m\subset\mathcal F$. Since $\mathcal F$ is a $\sigma$-algebra, $T_\infty\in\mathcal F$.
[guided]
The stopping-time definition gives information about the events $\{\tau\le n\}$, while the stopped value $X_\tau$ is naturally analysed on the exact-time events $\{\tau=k\}$. We therefore first convert the inequalities $\tau\le n$ into exact-time events.
For each $k\in\mathbb N_0$, set
\begin{align*}
T_k:=\{\omega\in\Omega:\tau(\omega)=k\}.
\end{align*}
When $k=0$, the equality $\tau=0$ is the same as $\tau\le 0$, because $\tau$ takes values in $\mathbb N_0\cup\{\infty\}$. Thus
\begin{align*}
T_0=\{\tau\le 0\}\in\mathcal F_0
\end{align*}
by the stopping-time hypothesis.
Now let $k\ge 1$. The event that $\tau$ equals $k$ is the event that $\tau$ has occurred by time $k$ but has not occurred by time $k-1$:
\begin{align*}
T_k=\{\tau\le k\}\cap\{\tau\le k-1\}^c.
\end{align*}
The stopping-time hypothesis gives $\{\tau\le k\}\in\mathcal F_k$ and $\{\tau\le k-1\}\in\mathcal F_{k-1}$. Since $(\mathcal F_n)_{n\in\mathbb N_0}$ is a filtration, $\mathcal F_{k-1}\subset\mathcal F_k$, so $\{\tau\le k-1\}\in\mathcal F_k$. The $\sigma$-algebra $\mathcal F_k$ is closed under complements and intersections, hence $T_k\in\mathcal F_k$.
We also need to handle the branch where the stopping time never occurs. Define
\begin{align*}
T_\infty:=\{\omega\in\Omega:\tau(\omega)=\infty\}.
\end{align*}
Since the only finite values of $\tau$ are $0,1,2,\dots$, the complement of $T_\infty$ is the event that $\tau$ is finite, equivalently that $\tau\le m$ for some $m\in\mathbb N_0$. Therefore
\begin{align*}
T_\infty=\Omega\setminus\bigcup_{m=0}^{\infty}\{\tau\le m\}.
\end{align*}
Each event $\{\tau\le m\}$ lies in $\mathcal F_m$ by the stopping-time hypothesis, and each $\mathcal F_m$ is a sub-$\sigma$-algebra of $\mathcal F$. Thus the countable union lies in $\mathcal F$, and closure of $\mathcal F$ under complements gives $T_\infty\in\mathcal F$.
[/guided]
[/step]
[step:Decompose the preimage of a measurable target set]
Fix $B\in\mathcal E$, and define the event
\begin{align*}
A_B:=X_\tau^{-1}(B)=\{\omega\in\Omega:X_\tau(\omega)\in B\}.
\end{align*}
For each $k\in\mathbb N_0$, define
\begin{align*}
C_k(B):=T_k\cap X_k^{-1}(B).
\end{align*}
Because $X_k:(\Omega,\mathcal F_k)\to(E,\mathcal E)$ is measurable, $X_k^{-1}(B)\in\mathcal F_k$. Since $T_k\in\mathcal F_k$ from the previous step, $C_k(B)\in\mathcal F_k\subset\mathcal F$.
Define $D_B:=T_\infty$ if $e_*\in B$, and define $D_B:=\varnothing$ if $e_*\notin B$. In both cases $D_B\in\mathcal F$. By the definition of $X_\tau$,
\begin{align*}
A_B=\left(\bigcup_{k=0}^{\infty} C_k(B)\right)\cup D_B.
\end{align*}
The right-hand side is a countable union of events in $\mathcal F$, so $A_B\in\mathcal F$.
[/step]
[step:Check the stopped sigma-algebra condition at every deterministic time]
Let $n\in\mathbb N_0$ be fixed. Since the event $D_B$ is contained in $T_\infty$ and $T_\infty\cap\{\tau\le n\}=\varnothing$, the infinite-time branch disappears after intersection with $\{\tau\le n\}$. Also, for $k>n$, $T_k\cap\{\tau\le n\}=\varnothing$. Hence
\begin{align*}
A_B\cap\{\tau\le n\}=\bigcup_{k=0}^{n} C_k(B).
\end{align*}
For each $k\in\{0,\dots,n\}$, the previous step gives $C_k(B)\in\mathcal F_k$, and filtration monotonicity gives $\mathcal F_k\subset\mathcal F_n$. Therefore $C_k(B)\in\mathcal F_n$ for every $k\le n$. Since $\mathcal F_n$ is closed under finite unions,
\begin{align*}
A_B\cap\{\tau\le n\}\in\mathcal F_n.
\end{align*}
[/step]
[step:Conclude measurability of the stopped map]
We have shown that, for the arbitrary set $B\in\mathcal E$, the preimage $A_B=X_\tau^{-1}(B)$ satisfies both defining conditions for membership in $\mathcal F_\tau$: first $A_B\in\mathcal F$, and second $A_B\cap\{\tau\le n\}\in\mathcal F_n$ for every $n\in\mathbb N_0$. Thus
\begin{align*}
X_\tau^{-1}(B)\in\mathcal F_\tau
\end{align*}
for every $B\in\mathcal E$. By the definition of measurability between measurable spaces, $X_\tau:(\Omega,\mathcal F_\tau)\to(E,\mathcal E)$ is measurable.
[/step]
