The proof for a stopped discrete-time martingale is shown as follows.
Let $M=(M_n)_{n\ge0}$ be a discrete-time martinglae w.r.t. the filtration $(\mathcal F_n)_{n\ge0}$, and let $M^T=(M_{n\land T})_{n\ge0}$ be the stopped martingale, where $T$ is a stopping time w.r.t. $(\mathcal F_n)_{n\ge0}$. Since:
We have $$\begin{align} M_{n\land T}&=1_{T\ge n+1}M_n+1_{T\le n}M_T\\ &=1_{T\ge n+1}M_n+ \sum_{k=1}^n 1_{T=k}M_k\\ \end{align}$$ where $1_{T\ge n+1}, M_n, 1_{T=k}$, and $M_k$ are all $\mathcal F_n-measurable$, hence $M_{n\land T}$ is $\mathcal F_n-measurable$;
${\Bbb E}|M_{n\land T}| \le \underbrace{{\Bbb E}|M_1|+…+{\Bbb E}|M_n|}_{{\Bbb E}|M_i| \text{ is integrable, }\forall i\ge0}<\infty$, i.e., $M_{n\land T}$ is integrable $\forall n \ge 0$;
We have $$\begin{align} {\Bbb E}(M_{n+1 \land T}|{\mathcal F_n}) &= {\Bbb E}(M_{n \land T}+1_{T \ge n+1}(M_{n+1}-M_n)|{\mathcal F_n})\\ &=M_{n \land T}+1_{T \ge n+1}{\Bbb E}(M_{n+1}-M_n)\\ &=M_{n \land T}\\ \end{align}$$
Therefore, the stopped martingale satisfies the definition of a discrete martingale. Proof complete.
However, I am not able to extend these three parts of proof to a continuous version, because I cannot devide the time into separate spots with one next to another as the discrete version did. So I really wonder how to give a proof for a continuous-time martingale.
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$\begingroup$We are going to prove the following proposition:
If $M$ is a continuous martingale and $T$ a stopping time, the stopped process $M^{T}$, i.e., $\{M(t \wedge T), t \geq 0\}$ is a martingale with respect to $\left(\mathcal{F}_{t}\right)$.
Here we go.
The process $M^{T}$ is obviously continuous and adapted. Firstly we use a weak form of optional stopping theorem, saying that a martingale has equal expectation at any bounded stopping time. If $S$ is a bounded stopping time, so is $S \wedge T$; hence $$ E\left[M_{S}^{T}\right]=E\left[M_{S \wedge T}\right]=E\left[M_{0}\right]=E\left[M_{0}^{T}\right]. $$ Then we use this conclusion twice to get our desired equality. If $s<t$ and $A \in \mathcal{F}_{s}$ the r.v. $T=t \mathbf{1}_{A^{c}}+s \mathbf{1}_{A}$ is a stopping time and consequently $$ E\left[X_{0}\right]=E\left[X_{T}\right]=E\left[X_{t} \mathbf{1}_{A^{c}}\right]+E\left[X_{s} \mathbf{1}_{A}\right] .$$ On the other hand, $t$ itself is a stopping time, and $$ E\left[X_{0}\right]=E\left[X_{t}\right]=E\left[X_{t} \mathbf{1}_{A^{c}}\right]+E\left[X_{t} \mathbf{1}_{A}\right]. $$
Comparing the two equalities yields $X_{s}=E\left[X_{t} | \mathcal{F}_{s}\right].$
$\endgroup$ $\begingroup$With dyadics, page 2 : . The idea is to discretize the time, build discrete stopping times on these time slots, and conclude using backward martingales.
Optional sampling theorem : If $X = (X_t,\mathcal{F}_t)$ is a supermartingale and $T$ is an arbitrary stopping time, then the stopped process $X^T=(X_{T\wedge t},\mathcal{F}_t)$ is also a supermartingale.
Proof : Here we give a sketch of the proof. Assume $X$ is a supermartingale. First, for a fixed $n\ge 1$, let$$ D_n = \left\{\frac{k}{2^n}, k= 0,1,2,\ldots\right\}\subset D_{n+1}\subset \cdots $$be the set of non-negative dyadic rationals of order no greater than $n$. It follows that$$ X = (X_t,\mathcal{F}_t; t\in D_n) $$is a super martingale (discrete time).
Second, we construct a stopping time $T_n$ such that $T_n\ge T$ and $T_n$ only take values in $D_n$. Indeed, let$$ T_n(\omega) = \inf\left\{t\in D_n; t\ge T(\omega)\right\}. $$Then $T_n\ge T_{n+1} \ge \cdots $ and $T_n$ is a stopping time. Fix $0\le s\le t$, we wish to show that$$ \mathbb{E}\left[X_{t\wedge T}|\mathcal{F_s}\right]\le X_s $$almost surely. Similarly define$$ t_n = \inf\left\{u\in D_n; u\ge t\right\} \ge t_{n+1}\ge\cdots $$and$$ s_n = \inf\left\{u\in D_n; u\ge s\right\} \ge s_{n+1}\ge\cdots $$It follows from the discrete time optional sampling theorem that$$ \mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s_m}\right] \le X_{s_m \wedge T_n} $$for any integers $m\ge n$. Letting $m\to\infty$, we have $s_m \to s$ decreasing and $\mathcal{F}_{s_m}\to\mathcal{F}_s$. By Lévy's Downward theorem and the continuity of process $X$, we have$$ \mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s}\right] \le X_{s_m \wedge T_n} $$Observe that $(X_{t_n\wedge T_n},\mathcal{F}_{t_n\wedge T_n})$ is a backward martingale, whence it is uniformly integrable. Letting $n\to\infty$, we arrive at$$ \mathbb{E}\left[X_{t\wedge T}|\mathcal{F}_{s}\right] \le X_{s\wedge T}. $$This completes the proof.
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