proof of monotone convergence theorem


It is enough to prove the following

Theorem 1

Let (X,μ) be a measurable spaceMathworldPlanetmathPlanetmath and let fk:XR{+} be a monotone increasing sequence of positive measurable functionsMathworldPlanetmath (i.e. 0f1f2). Then f(x)=limkfk(x) is measurable and

limnXfk𝑑μ=Xf(x)𝑑μ.

First of all by the monotonicity of the sequence we have

f(x)=supkfk(x)

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

supkXfk𝑑μXf(x)𝑑μ.

So take any simple measurable function s such that 0sf. Given also α<1 define

Ek={xX:fk(x)αs(x)}.

The sequence Ek is an increasing sequence of measurable sets. Moreover the union of all Ek is the whole space X since limkfk(x)=f(x)s(x)>αs(x). Moreover it holds

Xfk𝑑μEkfk𝑑μαEks𝑑μ.

Since s is a simple measurable function it is easy to check that EEs𝑑μ is a measureMathworldPlanetmath and hence

supkXfk𝑑μαXs𝑑μ.

But this last inequalityMathworldPlanetmath holds for every α<1 and for all simple measurable functions s with sf. Hence by the definition of Lebesgue integral

supkXfk𝑑μXf𝑑μ

which completesPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Title proof of monotone convergence theorem
Canonical name ProofOfMonotoneConvergenceTheorem
Date of creation 2013-03-22 13:29:56
Last modified on 2013-03-22 13:29:56
Owner paolini (1187)
Last modified by paolini (1187)
Numerical id 7
Author paolini (1187)
Entry type Proof
Classification msc 28A20
Classification msc 26A42