[proofplan]
Fix a deterministic time $n\in\mathbb Z_{\ge 0}$. The event $\{\tau+\sigma\le n\}$ can occur only when both stopping times take finite values, and it decomposes as a finite union over pairs $(i,j)$ with $i+j\le n$. We first verify that each exact-time event $\{\tau=i\}$ and $\{\sigma=j\}$ is measurable at its own time, then use filtration monotonicity to promote each finite intersection to $\mathcal F_n$. Closure of $\mathcal F_n$ under finite unions gives $\{\tau+\sigma\le n\}\in\mathcal F_n$, which is precisely the stopping-time condition.
[/proofplan]
[step:Show that finite exact-time events are measurable at their own times]
Fix $i\in\mathbb Z_{\ge 0}$. If $i=0$, then
\begin{align*}
\{\tau=0\}=\{\tau\le 0\}.
\end{align*}
Since $\tau$ is a [stopping time](/page/Stopping%20Time), $\{\tau\le 0\}\in\mathcal F_0$, hence $\{\tau=0\}\in\mathcal F_0$.
If $i\ge 1$, then
\begin{align*}
\{\tau=i\}=\{\tau\le i\}\setminus\{\tau\le i-1\}.
\end{align*}
Because $\tau$ is a stopping time, $\{\tau\le i\}\in\mathcal F_i$ and $\{\tau\le i-1\}\in\mathcal F_{i-1}$. Since $(\mathcal F_n)_{n\in\mathbb Z_{\ge 0}}$ is a filtration, $\mathcal F_{i-1}\subset\mathcal F_i$, so $\{\tau\le i-1\}\in\mathcal F_i$. Therefore $\{\tau=i\}\in\mathcal F_i$.
Now fix $j\in\mathbb Z_{\ge 0}$. If $j=0$, then
\begin{align*}
\{\sigma=0\}=\{\sigma\le 0\}.
\end{align*}
Since $\sigma$ is a stopping time, $\{\sigma\le 0\}\in\mathcal F_0$, hence $\{\sigma=0\}\in\mathcal F_0$.
If $j\ge 1$, then
\begin{align*}
\{\sigma=j\}=\{\sigma\le j\}\setminus\{\sigma\le j-1\}.
\end{align*}
Because $\sigma$ is a stopping time, $\{\sigma\le j\}\in\mathcal F_j$ and $\{\sigma\le j-1\}\in\mathcal F_{j-1}$. Since $(\mathcal F_n)_{n\in\mathbb Z_{\ge 0}}$ is a filtration, $\mathcal F_{j-1}\subset\mathcal F_j$, so $\{\sigma\le j-1\}\in\mathcal F_j$. Therefore $\{\sigma=j\}\in\mathcal F_j$.
[guided]
We need exact-time events because the sum condition will be described by saying that $\tau$ equals some finite value $i$ and $\sigma$ equals some finite value $j$. The stopping-time definition gives events of the form $\{\tau\le m\}$, so we first convert exact equality into inequalities.
Fix $i\in\mathbb Z_{\ge 0}$. For $i=0$, no previous time exists, and the exact event is simply
\begin{align*}
\{\tau=0\}=\{\tau\le 0\}.
\end{align*}
Since $\tau$ is a stopping time, the event on the right belongs to $\mathcal F_0$. Hence $\{\tau=0\}\in\mathcal F_0$.
Now suppose $i\ge 1$. Then $\tau=i$ holds exactly when $\tau\le i$ holds but $\tau\le i-1$ does not hold, so
\begin{align*}
\{\tau=i\}=\{\tau\le i\}\setminus\{\tau\le i-1\}.
\end{align*}
The stopping-time property gives $\{\tau\le i\}\in\mathcal F_i$ and $\{\tau\le i-1\}\in\mathcal F_{i-1}$. The filtration property says that information only increases with time, so $\mathcal F_{i-1}\subset\mathcal F_i$. Thus both sets in the difference belong to $\mathcal F_i$. Since $\mathcal F_i$ is a $\sigma$-algebra, it is closed under set difference, and therefore $\{\tau=i\}\in\mathcal F_i$.
Now we perform the corresponding verification for $\sigma$. Fix $j\in\mathbb Z_{\ge 0}$. If $j=0$, then
\begin{align*}
\{\sigma=0\}=\{\sigma\le 0\}.
\end{align*}
Since $\sigma$ is a stopping time, the event on the right belongs to $\mathcal F_0$. Hence $\{\sigma=0\}\in\mathcal F_0$.
Suppose instead that $j\ge 1$. Then $\sigma=j$ holds exactly when $\sigma\le j$ holds and $\sigma\le j-1$ fails, so
\begin{align*}
\{\sigma=j\}=\{\sigma\le j\}\setminus\{\sigma\le j-1\}.
\end{align*}
The stopping-time property gives $\{\sigma\le j\}\in\mathcal F_j$ and $\{\sigma\le j-1\}\in\mathcal F_{j-1}$. Since $(\mathcal F_n)_{n\in\mathbb Z_{\ge 0}}$ is a filtration, $\mathcal F_{j-1}\subset\mathcal F_j$. Therefore both sets in the difference belong to $\mathcal F_j$. Since $\mathcal F_j$ is a $\sigma$-algebra, it is closed under set difference, and hence $\{\sigma=j\}\in\mathcal F_j$.
[/guided]
[/step]
[step:Decompose the event that the sum has occurred by time $n$]
Fix $n\in\mathbb Z_{\ge 0}$, and define the finite index set
\begin{align*}
I_n:=\{(i,j)\in\mathbb Z_{\ge 0}\times\mathbb Z_{\ge 0}: i+j\le n\}.
\end{align*}
We claim that
\begin{align*}
\{\tau+\sigma\le n\}=\bigcup_{(i,j)\in I_n}\bigl(\{\tau=i\}\cap\{\sigma=j\}\bigr).
\end{align*}
Indeed, if $\omega\in\{\tau+\sigma\le n\}$, then extended addition implies $\tau(\omega)\ne\infty$ and $\sigma(\omega)\ne\infty$. Set $i:=\tau(\omega)$ and $j:=\sigma(\omega)$. Then $i,j\in\mathbb Z_{\ge 0}$, $i+j\le n$, and $\omega\in\{\tau=i\}\cap\{\sigma=j\}$.
Conversely, if $\omega\in\{\tau=i\}\cap\{\sigma=j\}$ for some $(i,j)\in I_n$, then
\begin{align*}
(\tau+\sigma)(\omega)=i+j\le n.
\end{align*}
Thus $\omega\in\{\tau+\sigma\le n\}$, proving the claimed equality.
[/step]
[step:Use the common filtration to make every summand measurable at time $n$]
Let $(i,j)\in I_n$. Then $i\le n$ and $j\le n$. From the previous exact-time measurability step, $\{\tau=i\}\in\mathcal F_i$ and $\{\sigma=j\}\in\mathcal F_j$. Since $(\mathcal F_m)_{m\in\mathbb Z_{\ge 0}}$ is a filtration,
\begin{align*}
\mathcal F_i\subset\mathcal F_n
\end{align*}
and
\begin{align*}
\mathcal F_j\subset\mathcal F_n.
\end{align*}
Hence $\{\tau=i\}\in\mathcal F_n$ and $\{\sigma=j\}\in\mathcal F_n$. Because $\mathcal F_n$ is a $\sigma$-algebra, it is closed under finite intersections, so
\begin{align*}
\{\tau=i\}\cap\{\sigma=j\}\in\mathcal F_n.
\end{align*}
[/step]
[step:Take the finite union and conclude the stopping-time property]
The set $I_n$ is finite because $0\le i\le n$ and $0\le j\le n$ for every $(i,j)\in I_n$. By the decomposition above,
\begin{align*}
\{\tau+\sigma\le n\}=\bigcup_{(i,j)\in I_n}\bigl(\{\tau=i\}\cap\{\sigma=j\}\bigr).
\end{align*}
Each event in this finite union belongs to $\mathcal F_n$, and $\mathcal F_n$ is closed under finite unions. Therefore
\begin{align*}
\{\tau+\sigma\le n\}\in\mathcal F_n.
\end{align*}
Since $n\in\mathbb Z_{\ge 0}$ was arbitrary, $\tau+\sigma$ is a stopping time with respect to $(\mathcal F_n)_{n\in\mathbb Z_{\ge 0}}$.
[/step]
