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# proof of dominated convergence theorem

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

$h_{n}^{+}(x)=\sup\{f_{m}(x)\colon m\geq n\}$ |

$h_{n}^{-}(x)=\inf\{f_{m}(x)\colon m\geq n\}$ |

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

For every $n$, $|h_{n}^{\pm}|\leq g$

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

For every $x$, $\lim_{{n\to\infty}}h_{n}^{\pm}(x)=f(x)$

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

$\{x\mid h_{n}^{-}(x)\geq y\}\hbox{ and }\{x\mid h_{n}^{+}(x)\leq y\}$ |

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

$\bigcup_{{m\leq n}}\{x\mid f_{n}(x)\leq y\}\hbox{ and }\bigcup_{{m\geq n}}\{x\mid f_{n}(x)\leq y\}$ |

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

$\lim_{{n\to\infty}}\int h_{n}^{\pm}(x)\,d\mu(x)=\int\lim_{{n\to\infty}}h_{n}^{% \pm}(x)\,d\mu(x)$ |

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

By construction, $h_{n}^{-}\leq f_{n}\leq h_{n}^{+}$ and hence

$\int h_{n}^{-}\leq\int f_{n}\leq\int h_{n}^{+}$ |

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

$\lim_{{n\to\infty}}\int f_{n}(x)\,d\mu(x)=\int f(x)\,d\mu(x)$ |

## Mathematics Subject Classification

28A20*no label found*

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