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If we want to check whether $\tau$ is a stopping time with respect to some filtration $\lbrace \mathcal{F}_{n}\rbrace$, where $n\in\mathbb{N}$, we can verify that $\lbrace\tau= n\rbrace\in\mathcal{F}_{n}$ for all $n$ rather than $\lbrace\tau\leq n\rbrace\in\mathcal{F}_{n}$. Is it true because $\lbrace\tau\leq n\rbrace=\bigcup_{k=0}^{n}\lbrace\tau=k\rbrace\in\mathcal{F}_{n}$? (if we assume the above, the $k$-th set in the sum of the right-hand side of the equation lies in $\mathcal{F}_{k}$, so in $\mathcal{F}_{n}$ as well, since $k\leq n$, and the sum is also there as the sum of elements of a $\sigma$-algebra). Did I get it right?

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Yes looks right to me, we can say that $\{\tau\le n\}$ is in $\mathcal{F}_n$ since $\{\tau = k\}\in\mathcal{F}_n$ for $k=0,\ldots,n$ and by using the properties of $\mathcal{F}_n$ (closure under countable unions). –  Erik Miehling Feb 1 '13 at 20:35
    
Also use: $\mathcal F_k \subseteq \mathcal F_n$ if $k \le n$. –  GEdgar Feb 1 '13 at 21:02
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