# proof of monotone convergence theorem

It is enough to prove the following

###### Theorem 1

Let $(X,\mu)$ be a measurable space   and let $f_{k}\colon X\to{\mathbb{R}}\cup\{+\infty\}$ be a monotone increasing sequence of positive measurable functions  (i.e. $0\leq f_{1}\leq f_{2}\leq\ldots$). Then $f(x)=\lim_{k\to\infty}f_{k}(x)$ is measurable and

 $\lim_{n\to\infty}\int_{X}f_{k}\,d\mu=\int_{X}f(x)\,d\mu.$

First of all by the monotonicity of the sequence we have

 $f(x)=\sup_{k}f_{k}(x)$

hence we know that $f$ is measurable. Moreover being $f_{k}\leq f$ for all $k$, by the monotonicity of the integral, we immediately get

 $\sup_{k}\int_{X}f_{k}\,d\mu\leq\int_{X}f(x)\,d\mu.$

So take any simple measurable function $s$ such that $0\leq s\leq f$. Given also $\alpha<1$ define

 $E_{k}=\{x\in X\colon f_{k}(x)\geq\alpha s(x)\}.$

The sequence $E_{k}$ is an increasing sequence of measurable sets. Moreover the union of all $E_{k}$ is the whole space $X$ since $\lim_{k\to\infty}f_{k}(x)=f(x)\geq s(x)>\alpha s(x)$. Moreover it holds

 $\int_{X}f_{k}\,d\mu\geq\int_{E_{k}}f_{k}\,d\mu\geq\alpha\int_{E_{k}}s\,d\mu.$

Since $s$ is a simple measurable function it is easy to check that $E\mapsto\int_{E}s\,d\mu$ is a measure  and hence

 $\sup_{k}\int_{X}f_{k}\,d\mu\geq\alpha\int_{X}s\,d\mu.$

But this last inequality  holds for every $\alpha<1$ and for all simple measurable functions $s$ with $s\leq f$. Hence by the definition of Lebesgue integral

 $\sup_{k}\int_{X}f_{k}\,d\mu\geq\int_{X}f\,d\mu$
Title proof of monotone convergence theorem ProofOfMonotoneConvergenceTheorem 2013-03-22 13:29:56 2013-03-22 13:29:56 paolini (1187) paolini (1187) 7 paolini (1187) Proof msc 28A20 msc 26A42