proof of monotone convergence theorem

It is enough to prove the following

Theorem 1

Let $\mathrm{(}X\mathrm{,}\mu \mathrm{)}$ be a measurable space and let $f_k\mathrm{:}X\mathrm{\to }\mathrm{R}\mathrm{\cup }\mathrm{\{}\mathrm{+}\mathrm{\infty }\mathrm{\}}$be a monotone increasing sequence of positive measurable functions (i.e. $\mathrm{0}\mathrm{\le }{f}_{\mathrm{1}}\mathrm{\le }{f}_{\mathrm{2}}\mathrm{\le }\mathrm{\dots }$). Then $f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}{\lim }_{k\mathrm{\to }\mathrm{\infty }}\mathit{}{f}_k\mathit{}\mathrm{(}x\mathrm{)}$ is measurable and

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_X{f}_k𝑑\mu ={\int }_{X}f(x)𝑑\mu .$$

First of all by the monotonicity of the sequence we have

$$f(x)=\underset{k}{sup}{f}_k(x)$$

hence we know that $f$ is measurable. Moreover being $f_k\le f$ for all $k$, by the monotonicity of the integral, we immediately get

$$\underset{k}{sup}{\int }_X{f}_k𝑑\mu \le {\int }_{X}f(x)𝑑\mu .$$

So take any simple measurable function $s$ such that $0\le s\le f$. Given also define

$$E_k=\{x\in X:f_k(x)\ge \alpha s(x)\}.$$

The sequence $E_k$ is an increasing sequence of measurable sets. Moreover the union of all $E_k$ is the whole space $X$ since${lim}_{k\to \mathrm{\infty }}{f}_k(x)=f(x)\ge s(x)>\alpha s(x)$. Moreover it holds

$${\int }_X{f}_k𝑑\mu \ge {\int }_{E_k}{f}_k𝑑\mu \ge \alpha {\int }_{E_k}s𝑑\mu .$$

Since $s$ is a simple measurable function it is easy to check that$E\mapsto {\int }_{E}s𝑑\mu$ is a measure and hence

$$\underset{k}{sup}{\int }_X{f}_k𝑑\mu \ge \alpha {\int }_{X}s𝑑\mu .$$

But this last inequality holds for every and for all simple measurable functions $s$ with $s\le f$. Hence by the definition of Lebesgue integral

$$\underset{k}{sup}{\int }_X{f}_k𝑑\mu \ge {\int }_{X}f𝑑\mu$$

which completes the proof.