proof of the début theorem

Let $(\mathcal{F})_{t\in\mathbb{T}}$ be a right-continuous filtration (http://planetmath.org/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 début

D(A)=\inf\left\{t\in\mathbb{T}:(t,\omega)\in A\right\}

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

\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\}

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

\left\{D(A)\leq t\right\}=\bigcap_{n}\left\{D(A)<t_{n}\right\}\in\bigcap_{n}% \mathcal{F}_{t_{n}}=\mathcal{F}_{t+}=\mathcal{F}_{t}.

On the other hand, if $t$ is not a right limit point of $\mathbb{T}$ then

\{D(A)\leq t\}=\{D(A)<t\}\cup\{\omega\in\Omega:(t,\omega)\in A\}\in\mathcal{F}% _{t}.

In either case, $\{D(A)\leq t\}$ is in $\mathcal{F}_{t}$ , so $D(A)$ is a stopping time.

Title	proof of the début theorem
Canonical name	ProofOfTheDebutTheorem
Date of creation	2013-03-22 18:39:15
Last modified on	2013-03-22 18:39:15
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	6
Author	gel (22282)
Entry type	Proof
Classification	msc 60G40
Classification	msc 60G05