# 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

 $\lim_{n\to\infty}\int f_{n}(x)\,d\mu(x)=\int f(x)\,d\mu(x)$
Title proof of dominated convergence theorem  ProofOfDominatedConvergenceTheorem1 2013-03-22 14:33:58 2013-03-22 14:33:58 rspuzio (6075) rspuzio (6075) 5 rspuzio (6075) Proof msc 28A20 ProofOfDominatedConvergenceTheorem