| Version current |
Version 2 |
| \PMlinkescapeword{right limit} |
\PMlinkescapeword{right limit} |
| \PMlinkescapeword{point} |
\PMlinkescapeword{point} |
| \PMlinkescapeword{sequence} |
\PMlinkescapeword{sequence} |
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| Let $(\mathcal{F})_{t\in\mathbb{T}}$ be a right-continuous \PMlinkname{filtration}{FiltrationOfSigmaAlgebras} on the measurable space $(\Omega,\mathcal{F})$, It is assumed that $\mathbb{T}$ is a closed subset of $\mathbb{R}$ and that $\mathcal{F}_t$ is universally complete for each $t\in\mathbb{T}$. |
Let $(\mathcal{F})_{t\in\mathbb{T}}$ be a right-continuous \PMlinkname{filtration}{FiltrationOfSigmaAlgebras} on the measurable space $(\Omega,\mathcal{F})$, It is assumed that $\mathbb{T}$ is a closed subset of $\mathbb{R}$ and that $\mathcal{F}_t$ is universally complete for each $t\in\mathbb{T}$. |
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If $A\subseteq\mathbb{T}\times\Omega$ is a progressively measurable set, then we show that its d\'ebut
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If $A\subseteq\mathcal{B}(\mathbb{T})\times\Omega$ is a progressively measurable set, then we show that its d\'ebut
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| \begin{equation*} |
\begin{equation*} |
| D(A)=\inf\left\{t\in\mathbb{T}:(t,\omega)\in A\right\} |
D(A)=\inf\left\{t\in\mathbb{T}:(t,\omega)\in A\right\} |
| \end{equation*} |
\end{equation*} |
| is a stopping time. |
is a stopping time. |
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| As $A$ is progressively measurable, the set $A\cap((-\infty,t)\times\Omega)$ is $\mathcal{B}(\mathbb{T})\times\mathcal{F}_t$-measurable. By the measurable projection theorem it follows that |
As $A$ is progressively measurable, the set $A\cap((-\infty,t)\times\Omega)$ is $\mathcal{B}(\mathbb{T})\times\mathcal{F}_t$-measurable. By the measurable projection theorem it follows that |
| \begin{equation*} |
\begin{equation*} |
| \left\{D(A)<t\right\}=\left\{\omega\in\Omega:(s,\omega)\in A\cap((-\infty,t)\times\Omega)\textrm{ for some }s\in\mathbb{T}\right\} |
\left\{D(A)<t\right\}=\left\{\omega\in\Omega:(s,\omega)\in A\cap((-\infty,t)\times\Omega)\textrm{ for some }s\in\mathbb{T}\right\} |
| \end{equation*} |
\end{equation*} |
| is in $\mathcal{F}_t$. If there exists a sequence $t_n\in\mathbb{T}$ with $t_n>t$ and $t_n\rightarrow t$, then |
is in $\mathcal{F}_t$. If there exists a sequence $t_n\in\mathbb{T}$ with $t_n>t$ and $t_n\rightarrow t$, then |
| \begin{equation*} |
\begin{equation*} |
| \left\{D(A)\le t\right\}=\bigcap_n\left\{D(A)<t_n\right\}\in\bigcap_n\mathcal{F}_{t_n}=\mathcal{F}_{t+}=\mathcal{F}_t. |
\left\{D(A)\le t\right\}=\bigcap_n\left\{D(A)<t_n\right\}\in\bigcap_n\mathcal{F}_{t_n}=\mathcal{F}_{t+}=\mathcal{F}_t. |
| \end{equation*} |
\end{equation*} |
| On the other hand, if $t$ is not a right limit point of $\mathbb{T}$ then |
On the other hand, if $t$ is not a right limit point of $\mathbb{T}$ then |
| \begin{equation*} |
\begin{equation*} |
| \{D(A)\le t\}=\{D(A)<t\}\cup\{\omega\in\Omega:(t,\omega)\in A\}\in\mathcal{F}_t. |
\{D(A)\le t\}=\{D(A)<t\}\cup\{\omega\in\Omega:(t,\omega)\in A\}\in\mathcal{F}_t. |
| \end{equation*} |
\end{equation*} |
| In either case, $\{D(A)\le t\}$ is in $\mathcal{F}_t$, so $D(A)$ is a stopping time. |
In either case, $\{D(A)\le t\}$ is in $\mathcal{F}_t$, so $D(A)$ is a stopping time. |
