proof of the d\'ebut theorem
ProofOfTheDebutTheorem
2008-12-27 13:45:30
2009-02-03 21:45:54
Proof
hitting times are stopping times
0
d\'ebut
stopping time
progressively measurable
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\PMlinkescapeword{right limit}
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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 $t\in\mathbb{T}$.
If $A\subseteq\mathbb{T}\times\Omega$ 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.
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*}
\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 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 right 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.