completion of a measure space

If the measure space (X,𝒮,μ) is not completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, then it can be completed in the following way. Let


i.e. the family of all subsets of sets whose μ-measure is zero. Define


We assert that 𝒮¯ is a σ-algebra. In fact, it clearly contains the emptyset, and it is closed under countableMathworldPlanetmath unions because both 𝒮 and 𝒵 are. We thus need to show that it is closed under complements. Let A𝒮, B𝒵 and suppose E𝒮 is such that BE and μ(E)=0. Then we have


where AcEc𝒮 and Ac(E-B)𝒵. Hence (AB)c𝒮¯.

Now we define μ¯ on 𝒮¯ by μ¯(AB)=μ(A), whenever A𝒮 and B𝒵. It is easily verified that this defines in fact a measure, and that (X,𝒮¯,μ¯) is the completion of (X,𝒮,μ).

Title completion of a measure space
Canonical name CompletionOfAMeasureSpace
Date of creation 2013-03-22 14:06:59
Last modified on 2013-03-22 14:06:59
Owner Koro (127)
Last modified by Koro (127)
Numerical id 9
Author Koro (127)
Entry type DerivationPlanetmathPlanetmath
Classification msc 28A12