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The problem is depicted in the title. I want to know to complete my proof.

For any $t\ge0$ we want to show $$\{\omega;\tau(\omega)\ge t\}\in\mathcal{F}_t$$ Since we have $\tau\ge\sigma$. The left hand side could be split into two disjoint part depending on $\sigma$ that $$\{\omega;\tau(\omega)\ge t\}=\{\omega;\sigma(\omega)\ge t\}\cup\{\omega;\tau(\omega)\ge t>\sigma(\omega)\ge0\}$$ Now $\{\omega;\sigma(\omega)\ge t\}\in \mathcal{F}_t$ since $\sigma$ is a stopping time. I think the second part is also in $\mathcal{F}_t $ from the fact that $\tau$ is $\mathcal{F}_\sigma$ measurable, but I can't get my mind clear on this.

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  • $\begingroup$ You want the union of those two sets $\endgroup$ – fGDu94 Apr 20 '19 at 17:34
  • $\begingroup$ Yes the union, sorry for the typo $\endgroup$ – Apocalypse Apr 21 '19 at 5:40
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I answer the question myself. I think this is the correct way of thinking:

By the definition of $\mathcal{F}_\sigma$ that $$\mathcal{F}_\sigma=\{F\in\mathcal{F}_\infty;\forall t\in\mathbb{R}_+,F\cap[\sigma\le t]\in\mathcal{F}_t\}$$ If $\tau$ is mesurable regarding $\mathcal{F}_\sigma$, we have $[\tau\ge t]\in\mathcal{F}_\sigma$, so $[\tau\ge t]\cap[\sigma\le t]\in\mathcal{F}_t$, which is what we want

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