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So far I've already shown that the sum and the maximum of two stopping times is a stopping time, but the minimum is giving me some problems which I just can't get around. This is what I've tried:

Let $\mathbb{F}$ be a filtration, and $S, T$ be stopping times. Then: \begin{align*} \{S \wedge T\} &= \{\min(S,T) = n\} \\ &= \left(\bigcup_{i=n}^\infty \{S = i\} \cap \{T=n\}\right) \bigcup \left(\bigcup_{i=n}^\infty \{T = i\} \cap \{S=n\}\right) \end{align*} I thought we might be able to use the identity $\bigcup_{i=n}^\infty \{S=i\} = \left(\bigcup_{i=0}^{n-1}\{S=i\} \cup \{S=\infty\}\right)^c$ to convert this to the union/intersection/complement of $\mathcal{F}_n$ measurable sets, but the problem is that $\{TS= \infty\}$ is only $\mathcal{F}_\infty$ measurable, so it seems to be a dead end.

Is it a stopping time? It seems since we are dealing with a union from $n$ to $\infty$ it shouldn't be.

For clarification, these are stopping times for discrete martingales, so $T:\Omega \rightarrow \{0,1,\dots\}\cup \{\infty\}$ is a stopping time iff $\{T = n\} \in \mathcal{F}_n$ for all $n \in \mathbb{N}$.

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look where $S \wedge T > n$ which is an intersection – mike Oct 23 '12 at 13:14
By definition $T$ is a stopping if $\{T\leqslant t\}\in \mathcal{F}_t$ for all $t$. $\{S\wedge T\leqslant t\} =\{S\leqslant t\} \cup \{T\leqslant t\}$ both of which $\in\mathcal{F}_t$ – Igor Oct 23 '12 at 14:37
I'm a retard, major brainfart about this. Thanks, for some reason I was confusing minimum in maximum halfway my argument. – BallzofFury Oct 23 '12 at 16:03

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