I have a question about probability theory.
Let $(\Omega,\mathcal{F},P)$ be a probability space. The completion of $\mathcal{F}$ w.r.t $P$ is denoted by $\mathcal{F}^{P}$
Given a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{F}$, we define $\mathcal{G}^{P}$, the completion of $\mathcal{G}$ in $\mathcal{F}^{P}$ by \begin{align*} \mathcal{G}^{P}:=\sigma\{\mathcal{G},\mathcal{N}_{P}\}\quad{\text{where}\quad}\mathcal{N}_{P}:=\{N \in \mathcal{F}^{P}:P(N)=0\} \end{align*}
ADD we can show $\displaystyle \mathcal{G}^{P}=\{A \subset \Omega: A \bigtriangleup B \subset N {\rm \,for\,some\,} B \in \mathcal{G},N \in \mathcal{F},P(N)=0\}$
Let $\mathcal{F_{t}},t \in [0,\infty]$, be sub-$\sigma$-algebras of $\mathcal{F}$. $(\mathcal{F}_{t})_{t \in [0,\infty]}$ is called a filtration on $(\Omega,\mathcal{F})$ if $\mathcal{F}_{s} \subset \mathcal{F}_{t}$ for $s \leq t$ and $\displaystyle \mathcal{F}_{\infty}=\sigma\left(\bigcup_{t \in [0,\infty[} \mathcal{F}_{t}\right)$. we set $\mathcal{F}=\mathcal{F}_{\infty}$
Let $\widetilde{\Omega} \in \mathcal{F},P(\widetilde{\Omega})=1$. We define a filtered probability space $(\widetilde{\Omega},\widetilde{\mathcal{F}},\widetilde{P},(\widetilde{\mathcal{F}_{t}})_{t \in [0,\infty]})$ as follows: \begin{align*} \widetilde{\mathcal{F}}:=\widetilde{\Omega}\cap{\mathcal{F}},\widetilde{P}:=P\restriction_{\widetilde{\mathcal{F}}},\widetilde{\mathcal{F_{t}}}:=\widetilde{\Omega} \cap {\mathcal{F}_{t}} \end{align*}
Question: Let $\widetilde{\sigma}:\widetilde{\Omega}\to [0,\infty]$ be a $(\widetilde{\mathcal{F_{t}}}^{\widetilde{P}})_{t \in[0,\infty]}$-stopping time.
Is there $({\mathcal{F_{t}}}^{{P}})_{t \in [0,\infty]} $-stopping time $\sigma:\Omega \to [0,\infty]$ such that $\sigma=\widetilde{\sigma}$ on $\widetilde{\Omega}$ ?
My attempt:
Define $\sigma:\Omega \to [0,\infty]$ as follows \begin{align*} \sigma(\omega)= \begin{cases} \widetilde{\sigma}(\omega) &; \omega \in \widetilde{\Omega}\\ 0 &; \omega \in \Omega \setminus \widetilde{\Omega} \end{cases} \end{align*}
It is enough to show that $\forall t \geq0, \{\omega \in \Omega: \sigma(\omega)\leq t \} \in \mathcal{F_{t}}^{P}$.
\begin{align*} \{\omega \in \Omega: \sigma(\omega)\leq t \} &= \{\omega \in \tilde{\Omega}: \sigma(\omega)\leq t \}\cup \{\omega \in \Omega \setminus \tilde{\Omega}: \sigma(\omega)\leq t \}\\ &=\{\omega \in \tilde{\Omega}: \widetilde{\sigma}(\omega)\leq t \}\cup (\Omega \setminus \tilde{\Omega})\\ \end{align*}
$P(\Omega \setminus \widetilde{\Omega})=0$ implies $\Omega \setminus \widetilde{\Omega} \in {\mathcal{F_{0}}}^{{P}}$.
Since $\widetilde{\sigma}$ be a $(\widetilde{\mathcal{F_{t}}}^{\widetilde{P}})_{t \in[0,\infty]}$-stopping time, $\{\omega \in \tilde{\Omega}: \widetilde{\sigma}(\omega)\leq t \} \in \widetilde{\mathcal{F_{t}}}^{\widetilde{P}} $.
I think the equation looked like $\widetilde{\mathcal{F_{t}}}^{\widetilde{P}} = \widetilde{\Omega}\cap \mathcal{F}_{t}^{P}$ holds.
Please advise me if you have any better idea/method. Thank you in advance.