Let $M$ be a stochastic process, and let $T$ be a stopping time. We call $M$ a local martingale up to $T$ if there exists a sequence $(T_n)$ of stopping times such that $T = \text{sup} T_n$ and $M^{T_n}$ is a martingale for all $n$. Moreover, a martingale is a local martingale if it is a local martingale up to $\infty$.
Now if $M$ is a continuous local martingale up to $T$, how can I show that $M^T$ is a local martingale? Is continuity necessary for this statement? Thanks for any idea :-)
Let $S_n:=\inf\{t\ge 0:|M_t|\ge n\}$. Since $M$ is continuous $S_n\nearrow\infty$ a.s. It suffices to show that $\{S_n\}$ reduces $M^{T}$, i.e. we nts that $(M_{t\wedge T\wedge S_n})$ is a martingale. Let $X\equiv M^T$. Then for each $k\ge 1$, $X^{S_n\wedge T_k}$ is a bounded martingale and $T_k\nearrow T$ a.s. Therefore, for all $s\le t$, $$ \mathsf{E}[X_{t\wedge S_n}\mid \mathcal{F}_s]=\lim_{k\to\infty}\mathsf{E}[X_{t\wedge S_n\wedge T_k}\mid \mathcal{F}_s]=\lim_{k\to\infty}X_{s\wedge S_n\wedge T_k}=X_{s\wedge S_n} \quad\text{a.s.} $$