# proof of dominated convergence theorem

Define the functions ${h}_{n}^{+}$ and ${h}_{n}^{-}$ as follows:

$${h}_{n}^{+}(x)=sup\{{f}_{m}(x):m\ge n\}$$ |

$${h}_{n}^{-}(x)=inf\{{f}_{m}(x):m\ge n\}$$ |

These suprema and infima exist because, for every $x$, $|{f}_{n}(x)|\le g(x)$. These functions enjoy the following properties:

For every $n$, $|{h}_{n}^{\pm}|\le g$

The sequence^{} ${h}_{n}^{+}$ is decreasing and the sequence ${h}_{n}^{-}$ is increasing.

For every $x$, ${lim}_{n\to \mathrm{\infty}}{h}_{n}^{\pm}(x)=f(x)$

Each ${h}_{n}^{\pm}$ is measurable.

The first property follows from immediately from the definition of
supremum. The second property follows from the fact that the
supremum or infimum^{} is being taken over a larger set to define
${h}_{n}^{\pm}(x)$ than to define ${h}_{m}^{\pm}(x)$ when $n>m$. The third
property is a simple consequence of the fact that, for any sequence
of real numbers, if the sequence converges^{}, then the sequence has an
upper limit^{} and a lower limit which equal each other and equal the
limit. As for the fourth statement, it means that, for every real
number $y$ and every integer $n$, the sets

$$\{x\mid {h}_{n}^{-}(x)\ge y\}\text{and}\{x\mid {h}_{n}^{+}(x)\le y\}$$ |

are measurable. However, by the definition of ${h}_{n}^{\pm}$, these sets can be expressed as

$$\bigcup _{m\le n}\{x\mid {f}_{n}(x)\le y\}\text{and}\bigcup _{m\ge n}\{x\mid {f}_{n}(x)\le y\}$$ |

respectively. Since each ${f}_{n}$ is assumed to be measurable, each set
in either union is measurable. Since the union of a countable^{}
number of measurable sets^{} is itself measurable, these unions are
measurable, and hence the functions ${h}_{n}^{\pm}$ are measurable.

Because of properties 1 and 4 above and the assumption^{} that $g$ is
integrable, it follows that each ${h}_{n}^{\pm}$ is integrable. This
conclusion^{} and property 2 mean that the monotone convergence theorem^{}
is applicable so one can conclude that $f$ is integrable and that

$$\underset{n\to \mathrm{\infty}}{lim}\int {h}_{n}^{\pm}(x)\mathit{d}\mu (x)=\int \underset{n\to \mathrm{\infty}}{lim}{h}_{n}^{\pm}(x)d\mu (x)$$ |

By property 3, the right hand side equals $\int f(x)\mathit{d}\mu (x)$.

By construction, ${h}_{n}^{-}\le {f}_{n}\le {h}_{n}^{+}$ and hence

$$\int {h}_{n}^{-}\le \int {f}_{n}\le \int {h}_{n}^{+}$$ |

Because the outer two terms in the above inequality^{} tend towards the
same limit as $n\to \mathrm{\infty}$, the middle term is squeezed into
converging to the same limit. Hence

$$\underset{n\to \mathrm{\infty}}{lim}\int {f}_{n}(x)\mathit{d}\mu (x)=\int f(x)\mathit{d}\mu (x)$$ |

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

Canonical name | ProofOfDominatedConvergenceTheorem1 |

Date of creation | 2013-03-22 14:33:58 |

Last modified on | 2013-03-22 14:33:58 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 5 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 28A20 |

Related topic | ProofOfDominatedConvergenceTheorem |