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'proof of the d\'ebut theorem'
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| Title of object: |
proof of the d\'ebut theorem |
| Canonical Name: |
ProofOfTheDebutTheorem |
| Type: |
Proof |
| Created on: |
2008-12-27 13:45:30 |
| Modified on: |
2008-12-27 13:45:30 |
| Classification: |
msc:60G40, msc:60G05 |
| Keywords: |
d\'ebut, stopping time, progressively measurable |
Revision comment (for changes between this and next version):
| use the measurable projection theorem |
Preamble:
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Content:
\PMlinkescapeword{properties}
\PMlinkescapeword{analytic sets}
\PMlinkescapeword{analytic}
\PMlinkescapeword{analytic set}
\PMlinkescapeword{projection}
\PMlinkescapeword{universal}
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 $t\in\mathbb{T}$.
If $A\subseteq\mathcal{B}(\mathbb{T})\times\Omega$ is a progressively measurable set, then we show that its d\'ebut
\begin{equation*}
D(A)=\inf\left\{t\in\mathbb{T}:(t,\omega)\in A\right\}
\end{equation*}
is a stopping time.
The proof requires the notion of analytic sets on a measurable space $(X,\mathcal{A})$ and makes use of the following properties.
\begin{enumerate}
\item every measurable set is analytic.
\item every analytic set is universally measurable.
\item Let $(X,\mathcal{A})$ be a measurable space, $Y$ a Polish space and $\pi\colon X\times Y\rightarrow X$ be the projection $\pi(x,y)=x$.
Then, for any analytic set $A$ on the measurable space $(X\times Y,\mathcal{A}\otimes\mathcal{B}(Y))$, the projection $\pi(A)$ is an analytic subset of $X$.
\end{enumerate}
As $A$ is progressively measurable, the set $A\cap((-\infty,t)\times\Omega)$ is $\mathcal{B}(\mathbb{T})\otimes\mathcal{F}_t$-measurable and, therefore, analytic. By projection onto $\Omega$ it follows that
\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\}
\end{equation*}
is $\mathcal{F}_t$-analytic and, by universal completeness, 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*}
\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*}
On the other hand, if $t$ is not a left limit point of $\mathbb{T}$ then
\begin{equation*}
\{D(A)\le t\}=\{D(A)<t\}\cup\{\omega\in\Omega:(t,\omega)\in A\}\in\mathcal{F}_t.
\end{equation*}
In either case, $\{D(A)\le t\}$ is in $\mathcal{F}_t$, so $D(A)$ is a stopping time.
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