[proofplan]
We prove the equivalence by converting between cumulative events and exact-time events. In the forward direction, the event $\{\tau=n\}$ is the difference between the two cumulative events $\{\tau\le n\}$ and $\{\tau\le n-1\}$, with the case $n=0$ handled separately. In the reverse direction, the event $\{\tau\le n\}$ is the finite union of the exact-time events $\{\tau=k\}$ for $0\le k\le n$. The only structural input is that a filtration is increasing and each $\mathcal F_n$ is a $\sigma$-algebra.
[/proofplan]
[step:Derive exact-time measurability from stopping-time measurability]
Assume that $\tau$ is a [stopping time](/page/Stopping%20Time) with respect to $(\mathcal F_n)_{n\in\mathbb Z_{\ge 0}}$. Fix $n\in\mathbb Z_{\ge 0}$.
If $n=0$, then, since $\tau$ takes values in $\mathbb Z_{\ge 0}\cup\{\infty\}$,
\begin{align*}
\{\omega\in\Omega:\tau(\omega)=0\}=\{\omega\in\Omega:\tau(\omega)\le 0\}.
\end{align*}
The right-hand side belongs to $\mathcal F_0$ by the stopping-time assumption.
Now suppose $n\ge 1$. Define the events
\begin{align*}
A_n:=\{\omega\in\Omega:\tau(\omega)\le n\}
\end{align*}
and
\begin{align*}
A_{n-1}:=\{\omega\in\Omega:\tau(\omega)\le n-1\}.
\end{align*}
By the stopping-time assumption, $A_n\in\mathcal F_n$ and $A_{n-1}\in\mathcal F_{n-1}$. Since $(\mathcal F_m)_{m\in\mathbb Z_{\ge 0}}$ is a filtration and $n-1\le n$, we have $\mathcal F_{n-1}\subset\mathcal F_n$, so $A_{n-1}\in\mathcal F_n$. Because $\mathcal F_n$ is a $\sigma$-algebra, it is closed under complements in $\Omega$ and finite intersections. Hence
\begin{align*}
A_n\cap A_{n-1}^c\in\mathcal F_n.
\end{align*}
For every $\omega\in\Omega$, the condition $\tau(\omega)\le n$ and not $\tau(\omega)\le n-1$ is equivalent to $\tau(\omega)=n$, since $\tau(\omega)$ lies in $\mathbb Z_{\ge 0}\cup\{\infty\}$. Therefore
\begin{align*}
\{\omega\in\Omega:\tau(\omega)=n\}=A_n\cap A_{n-1}^c\in\mathcal F_n.
\end{align*}
Thus $\{\tau=n\}\in\mathcal F_n$ for every $n\in\mathbb Z_{\ge 0}$.
[guided]
Assume that $\tau$ is a stopping time. This means precisely that for every $m\in\mathbb Z_{\ge 0}$, the cumulative event
\begin{align*}
\{\omega\in\Omega:\tau(\omega)\le m\}
\end{align*}
belongs to $\mathcal F_m$.
We must prove that the exact-time event $\{\omega\in\Omega:\tau(\omega)=n\}$ belongs to $\mathcal F_n$ for each fixed $n\in\mathbb Z_{\ge 0}$. The first time $n=0$ has to be separated because there is no $\mathcal F_{-1}$. Since $\tau$ takes only the values $0,1,2,\dots$ and $\infty$, the condition $\tau(\omega)=0$ is the same as $\tau(\omega)\le 0$. Hence
\begin{align*}
\{\omega\in\Omega:\tau(\omega)=0\}=\{\omega\in\Omega:\tau(\omega)\le 0\}\in\mathcal F_0,
\end{align*}
where the membership follows from the stopping-time assumption with $m=0$.
Now fix $n\in\mathbb Z_{\ge 0}$ with $n\ge 1$. Define
\begin{align*}
A_n:=\{\omega\in\Omega:\tau(\omega)\le n\}
\end{align*}
and
\begin{align*}
A_{n-1}:=\{\omega\in\Omega:\tau(\omega)\le n-1\}.
\end{align*}
The stopping-time assumption gives $A_n\in\mathcal F_n$ and $A_{n-1}\in\mathcal F_{n-1}$. To compare both events inside the same $\sigma$-algebra, we use the filtration property: since $n-1\le n$, one has $\mathcal F_{n-1}\subset\mathcal F_n$. Thus $A_{n-1}\in\mathcal F_n$ as well.
The event that $\tau$ occurs exactly at time $n$ is obtained by keeping the outcomes where $\tau\le n$ and removing the outcomes where $\tau\le n-1$. Formally,
\begin{align*}
\{\omega\in\Omega:\tau(\omega)=n\}=A_n\cap A_{n-1}^c.
\end{align*}
This identity is valid because the only values of $\tau$ below or equal to $n$ are $0,1,\dots,n$, and excluding the values $0,1,\dots,n-1$ leaves exactly the value $n$. Since $\mathcal F_n$ is a $\sigma$-algebra, it is closed under complements in $\Omega$ and finite intersections. Therefore $A_n\cap A_{n-1}^c\in\mathcal F_n$, which proves
\begin{align*}
\{\omega\in\Omega:\tau(\omega)=n\}\in\mathcal F_n.
\end{align*}
This establishes the exact-time measurability condition for every $n\in\mathbb Z_{\ge 0}$.
[/guided]
[/step]
[step:Recover stopping-time measurability from exact-time measurability]
Assume conversely that
\begin{align*}
\{\omega\in\Omega:\tau(\omega)=k\}\in\mathcal F_k
\end{align*}
for every $k\in\mathbb Z_{\ge 0}$. Fix $n\in\mathbb Z_{\ge 0}$. For each $k\in\{0,1,\dots,n\}$, the filtration property gives $\mathcal F_k\subset\mathcal F_n$, and therefore
\begin{align*}
\{\omega\in\Omega:\tau(\omega)=k\}\in\mathcal F_n.
\end{align*}
Because $\mathcal F_n$ is a $\sigma$-algebra, it is closed under finite unions. Since $\tau$ takes values in $\mathbb Z_{\ge 0}\cup\{\infty\}$,
\begin{align*}
\{\omega\in\Omega:\tau(\omega)\le n\}=\bigcup_{k=0}^{n}\{\omega\in\Omega:\tau(\omega)=k\}.
\end{align*}
The right-hand side is a finite union of events in $\mathcal F_n$, so it belongs to $\mathcal F_n$. Since $n\in\mathbb Z_{\ge 0}$ was arbitrary, $\tau$ is a stopping time with respect to $(\mathcal F_n)_{n\in\mathbb Z_{\ge 0}}$.
[/step]
