The proof for a stopped discrete-time martingale is shown as follows.

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> 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:
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> 
> 1. 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 $\mathcal F\_n-measurable$;
> 2. ${\Bbb E}|M\_{n\land T}| \le \underbrace{{\Bbb E}|M\_1|+…+{\Bbb E}|M\_n|}\_{{\Bbb E}|M\_i| \text{ is integrable, }\forall i\ge0}<\infty$, i.e., $M\_{n\land T}$ is integrable $\forall n \ge 0$;
> 3. 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.