# proof of the dΓ©but theorem

Let ${(\mathrm{\beta \x84\pm})}_{t\beta \x88\x88\mathrm{\pi \x9d\x95\x8b}}$ be a right-continuous filtration (http://planetmath.org/FiltrationOfSigmaAlgebras) on the measurable space^{} $(\mathrm{\Xi \copyright},\mathrm{\beta \x84\pm})$, It is assumed that $\mathrm{\pi \x9d\x95\x8b}$ is a closed subset of $\mathrm{\beta \x84\x9d}$ and that ${\mathrm{\beta \x84\pm}}_{t}$ is universally complete for each $t\beta \x88\x88\mathrm{\pi \x9d\x95\x8b}$.

If $A\beta \x8a\x86\mathrm{\pi \x9d\x95\x8b}\Gamma \x97\mathrm{\Xi \copyright}$ is a progressively measurable set, then we show that its dΓ©but

$$D\beta \x81\u2019(A)=inf\beta \x81\u2018\{t\beta \x88\x88\mathrm{\pi \x9d\x95\x8b}:(t,\mathrm{{\rm O}\x89})\beta \x88\x88A\}$$ |

is a stopping time.

As $A$ is progressively measurable, the set $A\beta \x88\copyright ((-\mathrm{\beta \x88\x9e},t)\Gamma \x97\mathrm{\Xi \copyright})$ is $\mathrm{\beta \x84\neg}\beta \x81\u2019(\mathrm{\pi \x9d\x95\x8b})\Gamma \x97{\mathrm{\beta \x84\pm}}_{t}$-measurable. By the measurable projection theorem it follows that

$$ |

is in ${\mathrm{\beta \x84\pm}}_{t}$. If there exists a sequence ${t}_{n}\beta \x88\x88\mathrm{\pi \x9d\x95\x8b}$ with ${t}_{n}>t$ and ${t}_{n}\beta \x86\x92t$, then

$$ |

On the other hand, if $t$ is not a right limit point of $\mathrm{\pi \x9d\x95\x8b}$ then

$$ |

In either case, $\{D\beta \x81\u2019(A)\beta \x89\u20act\}$ is in ${\mathrm{\beta \x84\pm}}_{t}$, so $D\beta \x81\u2019(A)$ is a stopping time.

Title | proof of the dΓ©but 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 |