proof of dominated convergence theorem


Define the functions hn+ and hn- as follows:

hn+(x)=sup{fm(x):mn}
hn-(x)=inf{fm(x):mn}

These suprema and infima exist because, for every x, |fn(x)|g(x). These functions enjoy the following properties:

For every n, |hn±|g

The sequenceMathworldPlanetmath hn+ is decreasing and the sequence hn- is increasing.

For every x, limnhn±(x)=f(x)

Each hn± is measurable.

The first property follows from immediately from the definition of supremum. The second property follows from the fact that the supremum or infimumMathworldPlanetmath is being taken over a larger set to define hn±(x) than to define hm±(x) when n>m. The third property is a simple consequence of the fact that, for any sequence of real numbers, if the sequence convergesPlanetmathPlanetmath, then the sequence has an upper limitMathworldPlanetmath 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

{xhn-(x)y} and {xhn+(x)y}

are measurable. However, by the definition of hn±, these sets can be expressed as

mn{xfn(x)y} and mn{xfn(x)y}

respectively. Since each fn is assumed to be measurable, each set in either union is measurable. Since the union of a countableMathworldPlanetmath number of measurable setsMathworldPlanetmath is itself measurable, these unions are measurable, and hence the functions hn± are measurable.

Because of properties 1 and 4 above and the assumptionPlanetmathPlanetmath that g is integrable, it follows that each hn± is integrable. This conclusionMathworldPlanetmath and property 2 mean that the monotone convergence theoremMathworldPlanetmath is applicable so one can conclude that f is integrable and that

limnhn±(x)𝑑μ(x)=limnhn±(x)dμ(x)

By property 3, the right hand side equals f(x)𝑑μ(x).

By construction, hn-fnhn+ and hence

hn-fnhn+

Because the outer two terms in the above inequalityMathworldPlanetmath tend towards the same limit as n, the middle term is squeezed into converging to the same limit. Hence

limnfn(x)𝑑μ(x)=f(x)𝑑μ(x)
Title proof of dominated convergence theoremPlanetmathPlanetmath
Canonical name ProofOfDominatedConvergenceTheorem1
Date of creation 2013-03-22 14:33:58
Last modified on 2013-03-22 14:33:58
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 5
Author rspuzio (6075)
Entry type Proof
Classification msc 28A20
Related topic ProofOfDominatedConvergenceTheorem