# proof of Egorov’s theorem

Let $$ Since ${f}_{n}\to f$ almost everywhere, there is a set $S$ with $\mu (S)=0$ such that, given $i\in \mathbb{N}$ and $x\in E-S$, there is $m\in \mathbb{N}$ such that $j>m$ implies $$. This can be expressed by

$$E-S\subset \bigcup _{m\in \mathbb{N}}\bigcap _{j>m}{E}_{i,j},$$ |

or, in other words,

$$\bigcap _{m\in \mathbb{N}}\bigcup _{j>m}(E-{E}_{i,j})\subset S.$$ |

Since ${\{{\bigcup}_{j>m}(E-{E}_{i,j})\}}_{m\in \mathbb{N}}$ is a decreasing nested sequence^{} of sets, each of which has finite measure, and such that its intersection^{} has measure $0$, by continuity from above (http://planetmath.org/PropertiesForMeasure) we know that

$$\mu (\bigcup _{j>m}(E-{E}_{i,j}))\underset{m\to \mathrm{\infty}}{\overset{}{\to}}0.$$ |

Therefore, for each $i\in \mathbb{N}$, we can choose ${m}_{i}$ such that

$$ |

Let

$${E}_{\delta}=\bigcup _{i\in \mathbb{N}}\bigcup _{j>{m}_{i}}(E-{E}_{i,j}).$$ |

Then

$$ |

We claim that ${f}_{n}\to f$ uniformly on $E-{E}_{\delta}$. In fact, given $\epsilon >0$, choose $n$ such that $$. If $x\in E-{E}_{\delta}$, we have

$$x\in \bigcap _{i\in \mathbb{N}}\bigcap _{j>{m}_{i}}{E}_{i,j},$$ |

which in particular implies that, if $j>{m}_{n}$, $x\in {E}_{n,j}$; that is, $$. Hence, for each $\epsilon >0$ there is $N$ (which is given by ${m}_{n}$ above) such that $j>N$ implies $$ for each $x\in E-{E}_{\delta}$, as required. This the proof.

Title | proof of Egorov’s theorem |
---|---|

Canonical name | ProofOfEgorovsTheorem |

Date of creation | 2013-03-22 13:47:59 |

Last modified on | 2013-03-22 13:47:59 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 7 |

Author | Koro (127) |

Entry type | Proof |

Classification | msc 28A20 |