Let $X_{n}$ be an $\mathcal F_{n}$-martingale and let $B\in \mathcal B$.
Show that $T=\min\{n:X_{n}\in B\}$ is an $\mathcal F_{n}$-stopping time.
$\mathcal B$ is Borel $\sigma$-algebra and filtration is $\mathcal F=\sigma(X_{1},\dots,X_{n})$. Thanks for help.

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    $\begingroup$ Bad practice to post identical questions on several sites: math.stackexchange.com/questions/548802/… $\endgroup$ – Drew75 Nov 2 '13 at 8:56
  • $\begingroup$ @Drew75.i need the answer of this question so I put in two sites.is there a problem?. but for you i am delete question in math.st.. .I hope you're not angry about my work.thanks $\endgroup$ – jack Nov 2 '13 at 9:16
  • $\begingroup$ You should post on one site. If there is no response, delete your question then post on another. Read "Is it acceptable to post on multiple sites? NO" meta.stats.stackexchange.com/questions/896/… $\endgroup$ – Drew75 Nov 3 '13 at 8:03
  • $\begingroup$ @Drew75.Thank you for the advise. Gee you'll do $\endgroup$ – jack Nov 4 '13 at 7:08

Note that for $\def\N{\mathbb N}n \in \N$ we have $$\{T\le n\} = \bigcup_{i=1}^n \{X_i \in B\} = \bigcup_{i=1}^n X_i^{-1}(B) $$ As $X_i$ is $\mathcal F$-measurable for $i \le n$ by definition, we have $X_i^{-1}(B)\in \mathcal F_n$ for $i \le n$ and hence $\{T \le n\}\in \mathcal F_n$. That is, $T$ is an $(\mathcal F_n)$-stopping time.


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