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 ofsupremum. The second property follows from the fact that thesupremum 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 thirdproperty is a simple consequence of the fact that, for any sequenceof real numbers, if the sequence converges, then the sequence has anupper limit and a lower limit which equal each other and equal thelimit. As for the fourth statement, it means that, for every realnumber $y$ 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 setscan 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 setin either union is measurable. Since the union of a countablenumber of measurable sets is itself measurable, these unions aremeasurable, and hence the functions $h_n^{\pm }$ are measurable.

Because of properties 1 and 4 above and the assumption that $g$ isintegrable, it follows that each $h_n^{\pm }$ is integrable. Thisconclusion and property 2 mean that the monotone convergence theoremis applicable so one can conclude that $f$ is integrable and that

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

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

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 thesame limit as $n\to \mathrm{\infty }$, the middle term is squeezed intoconverging to the same limit. Hence

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