# $\tau > \sigma$ is a stoppinig time when $\sigma$ is a stopping time and $\tau$ measurable regarding $\mathcal{F}_\sigma$

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.

• You want the union of those two sets – fGDu94 Apr 20 '19 at 17:34
• Yes the union, sorry for the typo – Apocalypse Apr 21 '19 at 5:40

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