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

Question

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 — fGDu94, Apr 20 '19 at 17:34
Yes the union, sorry for the typo – Apocalypse Apr 21 '19 at 05:40 — Apocalypse, Apr 21 '19 at 05:40

score 1 · Accepted Answer · answered Apr 21 '19 at 08:58

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

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

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