[proofplan]
Part (i) verifies the martingale property of the stopped process $X^T$ by decomposing $X_{\min\{T,t\}}$ according to the values of $T$ and using the martingale property together with measurability of $\{T = s\}$ and $\{T > t-1\}$ with respect to $\mathcal{F}_{t-1}$. Part (ii) establishes the strong optional stopping identity $\mathbb{E}[X_T \mid \mathcal{F}_S] = X_S$ by telescoping $X_T - X_S$ and using the martingale property to annihilate each increment. Parts (iii)-(v) follow by taking expectations or applying the [Dominated Convergence Theorem](/theorems/4) with appropriate dominating [functions](/page/Function).
[/proofplan]
[step:Verify the stopped process $X^T$ is a martingale]
**(i)** We show $\mathbb{E}[X_{\min\{T, t\}} \mid \mathcal{F}_{t-1}] = X_{\min\{T, t-1\}}$ for each $t \geq 1$. Decompose:
\begin{align*}
X_{\min\{T, t\}} = \sum_{s=0}^{t-1} X_s \, \mathbb{1}_{\{T = s\}} + X_t \, \mathbb{1}_{\{T \geq t\}}.
\end{align*}
For $s \leq t-1$, the event $\{T = s\} \in \mathcal{F}_s \subset \mathcal{F}_{t-1}$, and $X_s$ is $\mathcal{F}_s$-measurable, hence $\mathcal{F}_{t-1}$-measurable. By [Taking Out What is Known](/theorems/1151), each term $X_s \, \mathbb{1}_{\{T = s\}}$ passes through the conditional expectation unchanged. The event $\{T \geq t\} = \{T > t-1\} = \{T \leq t-1\}^c \in \mathcal{F}_{t-1}$ (since $T$ is a stopping time), so $\mathbb{1}_{\{T \geq t\}}$ is $\mathcal{F}_{t-1}$-measurable and factors out:
\begin{align*}
\mathbb{E}[X_{\min\{T, t\}} \mid \mathcal{F}_{t-1}] &= \sum_{s=0}^{t-1} X_s \, \mathbb{1}_{\{T = s\}} + \mathbb{1}_{\{T \geq t\}} \, \mathbb{E}[X_t \mid \mathcal{F}_{t-1}].
\end{align*}
The martingale property gives $\mathbb{E}[X_t \mid \mathcal{F}_{t-1}] = X_{t-1}$. Substituting:
\begin{align*}
\mathbb{E}[X_{\min\{T, t\}} \mid \mathcal{F}_{t-1}] = \sum_{s=0}^{t-1} X_s \, \mathbb{1}_{\{T = s\}} + X_{t-1} \, \mathbb{1}_{\{T \geq t\}} = X_{\min\{T, t-1\}},
\end{align*}
where the last equality recombines the sum by noting $X_{t-1} \, \mathbb{1}_{\{T = t-1\}} + X_{t-1} \, \mathbb{1}_{\{T \geq t\}} = X_{t-1} \, \mathbb{1}_{\{T \geq t-1\}}$ and regrouping.
The identity $\mathbb{E}[X_{\min\{T,t\}}] = \mathbb{E}[X_0]$ follows by induction: the averaging property of conditional expectation gives $\mathbb{E}[X_{\min\{T,t\}}] = \mathbb{E}[X_{\min\{T,t-1\}}]$, and iterating down to $t = 0$ yields $\mathbb{E}[X_0]$.
[guided]
The essential ingredient is the measurability of stopping time events. The stopping time condition states $\{T \leq t\} \in \mathcal{F}_t$ for all $t$, which implies $\{T = s\} = \{T \leq s\} \setminus \{T \leq s-1\} \in \mathcal{F}_s$ and $\{T \geq t\} = \{T \leq t-1\}^c \in \mathcal{F}_{t-1}$.
The decomposition of $X_{\min\{T,t\}}$ separates the "already stopped" part (the sum over $s \leq t-1$) from the "not yet stopped" part ($X_t \, \mathbb{1}_{\{T \geq t\}}$). The already-stopped terms are $\mathcal{F}_{t-1}$-measurable and pass through the conditional expectation. The not-yet-stopped term contributes $\mathbb{1}_{\{T \geq t\}} \, \mathbb{E}[X_t \mid \mathcal{F}_{t-1}]$, which uses the martingale property to replace $\mathbb{E}[X_t \mid \mathcal{F}_{t-1}]$ by $X_{t-1}$.
The recombination at the end uses:
\begin{align*}
\sum_{s=0}^{t-1} X_s \, \mathbb{1}_{\{T = s\}} + X_{t-1} \, \mathbb{1}_{\{T \geq t\}} &= \sum_{s=0}^{t-2} X_s \, \mathbb{1}_{\{T = s\}} + X_{t-1} \bigl(\mathbb{1}_{\{T = t-1\}} + \mathbb{1}_{\{T \geq t\}}\bigr) \\
&= \sum_{s=0}^{t-2} X_s \, \mathbb{1}_{\{T = s\}} + X_{t-1} \, \mathbb{1}_{\{T \geq t-1\}} = X_{\min\{T, t-1\}}.
\end{align*}
[/guided]
[/step]
[step:Prove the strong optional stopping identity $\mathbb{E}[X_T \mid \mathcal{F}_S] = X_S$ for bounded stopping times]
**(ii)** Assume $T \leq N$ a.s. for some $N \in \mathbb{N}$. Telescope:
\begin{align*}
X_T - X_S = \sum_{k=0}^{N-1} (X_{k+1} - X_k) \, \mathbb{1}_{\{S \leq k < T\}}.
\end{align*}
For any $A \in \mathcal{F}_S$, the [set](/page/Set) $A \cap \{S \leq k\} \cap \{T > k\}$ belongs to $\mathcal{F}_k$: indeed $A \in \mathcal{F}_S \subset \mathcal{F}_k$ for $k \geq S$ (and the indicator vanishes when $k < S$), $\{S \leq k\} \in \mathcal{F}_k$ since $S$ is a stopping time, and $\{T > k\} = \{T \leq k\}^c \in \mathcal{F}_k$ since $T$ is a stopping time. Therefore $\mathbb{1}_A \, \mathbb{1}_{\{S \leq k < T\}}$ is $\mathcal{F}_k$-measurable.
The martingale property gives $\mathbb{E}[X_{k+1} - X_k \mid \mathcal{F}_k] = 0$ a.s. By [Taking Out What is Known](/theorems/1151) applied with the bounded $\mathcal{F}_k$-[measurable function](/page/Measurable%20Functions) $\mathbb{1}_A \, \mathbb{1}_{\{S \leq k < T\}}$:
\begin{align*}
\mathbb{E}\bigl[(X_{k+1} - X_k) \, \mathbb{1}_{\{S \leq k < T\}} \, \mathbb{1}_A\bigr] = \mathbb{E}\bigl[\mathbb{E}[X_{k+1} - X_k \mid \mathcal{F}_k] \cdot \mathbb{1}_{\{S \leq k < T\}} \, \mathbb{1}_A\bigr] = 0.
\end{align*}
Summing over $k = 0, \ldots, N-1$:
\begin{align*}
\mathbb{E}[(X_T - X_S) \, \mathbb{1}_A] = 0 \quad \text{for all } A \in \mathcal{F}_S.
\end{align*}
Since $X_S$ is $\mathcal{F}_S$-measurable and $\mathbb{E}[X_T \mathbb{1}_A] = \mathbb{E}[X_S \mathbb{1}_A]$ for all $A \in \mathcal{F}_S$, the uniqueness of conditional expectation gives $\mathbb{E}[X_T \mid \mathcal{F}_S] = X_S$ a.s.
[guided]
The telescoping identity writes the difference $X_T - X_S$ as a sum of martingale increments $X_{k+1} - X_k$, each "switched on" only during the random interval $[S, T)$. The indicator $\mathbb{1}_{\{S \leq k < T\}}$ equals $1$ when the process has started (past time $S$) but has not yet stopped (before time $T$).
The critical measurability check is that $A \cap \{S \leq k < T\} \in \mathcal{F}_k$ for $A \in \mathcal{F}_S$. This holds because:
- $A \in \mathcal{F}_S$ means $A \cap \{S \leq k\} \in \mathcal{F}_k$ (this is part of the definition of $\mathcal{F}_S$, the stopped $\sigma$-algebra).
- $\{T > k\} \in \mathcal{F}_k$ by the stopping time property of $T$.
The expectation $\mathbb{E}[(X_{k+1} - X_k) \cdot \mathbb{1}_{\{S \leq k < T\}} \cdot \mathbb{1}_A]$ vanishes because we can condition on $\mathcal{F}_k$: the martingale increment $X_{k+1} - X_k$ has zero conditional expectation, and the remaining factor is $\mathcal{F}_k$-measurable. This is exactly the [Taking Out What is Known](/theorems/1151) property.
[/guided]
[/step]
[step:Deduce $\mathbb{E}[X_T] = \mathbb{E}[X_S]$ for bounded stopping times]
**(iii)** Taking unconditional expectations in part (ii) and applying the averaging property ([Basic Properties of Conditional Expectation](/theorems/1148), part (i)):
\begin{align*}
\mathbb{E}[X_S] = \mathbb{E}[\mathbb{E}[X_T \mid \mathcal{F}_S]] = \mathbb{E}[X_T].
\end{align*}
[/step]
[step:Prove $\mathbb{E}[X_T] = \mathbb{E}[X_0]$ under uniform domination]
**(iv)** Assume $|X_n| \leq Y$ a.s. for all $n$ with $Y \in L^1$, and $T < \infty$ a.s. By part (i), $\mathbb{E}[X_{\min\{T,t\}}] = \mathbb{E}[X_0]$ for all $t$. As $t \to \infty$, $X_{\min\{T,t\}} \to X_T$ a.s. (since $T < \infty$ a.s.). Since $|X_{\min\{T,t\}}| \leq Y$ for all $t$, the [Dominated Convergence Theorem](/theorems/4) gives:
\begin{align*}
\mathbb{E}[X_T] = \lim_{t \to \infty} \mathbb{E}[X_{\min\{T,t\}}] = \mathbb{E}[X_0].
\end{align*}
[/step]
[step:Prove $\mathbb{E}[X_T] = \mathbb{E}[X_0]$ under bounded increments and finite expected stopping time]
**(v)** Assume $|X_{n+1} - X_n| \leq M$ a.s. for all $n$ and $\mathbb{E}[T] < \infty$. For each $t$, the triangle inequality gives:
\begin{align*}
|X_{\min\{T,t\}}| \leq |X_0| + \sum_{k=0}^{\min\{T,t\}-1} |X_{k+1} - X_k| \leq |X_0| + M \cdot T.
\end{align*}
Since $\mathbb{E}[|X_0| + M \cdot T] = \mathbb{E}[|X_0|] + M \, \mathbb{E}[T] < \infty$, the dominating function $|X_0| + M \cdot T$ is integrable. As $t \to \infty$, $X_{\min\{T,t\}} \to X_T$ a.s. The [Dominated Convergence Theorem](/theorems/4) gives $\mathbb{E}[X_T] = \lim_{t} \mathbb{E}[X_{\min\{T,t\}}] = \mathbb{E}[X_0]$.
[/step]
