# proof of dominated convergence theorem

It is not difficult to prove that $f$ is measurable. In fact we can write

$$f(x)=\underset{n}{sup}\underset{k\ge n}{inf}{f}_{k}(x)$$ |

and we know that measurable functions^{} are closed under the $sup$ and $inf$ operation.

Consider the sequence ${g}_{n}(x)=2\mathrm{\Phi}(x)-|f(x)-{f}_{n}(x)|$. Clearly ${g}_{n}$ are nonnegative functions since $f-{f}_{n}\le 2\mathrm{\Phi}$. So, applying Fatou’s Lemma, we obtain

$\underset{n\to \mathrm{\infty}}{lim}{\displaystyle {\int}_{X}}|f-{f}_{n}|\mathit{d}\mu \le \underset{n\to \mathrm{\infty}}{lim\; sup}{\displaystyle {\int}_{X}}|f-{f}_{n}|\mathit{d}\mu $ | ||||

$=$ | $-\underset{n\to \mathrm{\infty}}{lim\; inf}{\displaystyle {\int}_{X}}-|f-{f}_{n}|d\mu $ | |||

$=$ | ${\int}_{X}}2\mathrm{\Phi}\mathit{d}\mu -\underset{n\to \mathrm{\infty}}{lim\; inf}{\displaystyle {\int}_{X}}2\mathrm{\Phi}-|f-{f}_{n}|d\mu $ | |||

$\le $ | ${\int}_{X}}2\mathrm{\Phi}\mathit{d}\mu -{\displaystyle {\int}_{X}}2\mathrm{\Phi}-\underset{n\to \mathrm{\infty}}{lim\; sup}|f-{f}_{n}|d\mu $ | |||

$=$ | ${\int}_{X}}2\mathrm{\Phi}\mathit{d}\mu -{\displaystyle {\int}_{X}}2\mathrm{\Phi}\mathit{d}\mu =0.$ |

Title | proof of dominated convergence theorem^{} |
---|---|

Canonical name | ProofOfDominatedConvergenceTheorem |

Date of creation | 2013-03-22 13:30:02 |

Last modified on | 2013-03-22 13:30:02 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 4 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 28A20 |

Related topic | SecondProofOfDominatedConvergenceTheorem |

Related topic | SecondProofOfDominatedConvergenceTheorem2 |