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### martingale, stopping times

It is immediate that $$\{T\leqslant t\}\in \mathcal F_t$$ for all $t$. Now, since $\sigma+\tau$ and $T$ are stopping times, we have  \{(\sigma+\tau)\wedge T\leqslant t\} = \{\sigma+\tau\leqslant t\...
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