completion of a measure space

If the measure space $(X,𝒮,\mu )$ is not complete, then it can be completed in the following way. Let

$$𝒵=\bigcup _{E\in 𝒮,\mu (E)=0}𝒫(E),$$

i.e. the family of all subsets of sets whose $\mu$-measure is zero. Define

$$\overline{𝒮}=\{A\cup B:A\in 𝒮,B\in 𝒵\}.$$

We assert that $\overline{𝒮}$ is a $\sigma$-algebra. In fact, it clearly contains the emptyset, and it is closed under countable unions because both $𝒮$ and $𝒵$ are. We thus need to show that it is closed under complements. Let $A\in 𝒮$, $B\in 𝒵$ and suppose $E\in 𝒮$ is such that $B\subset E$ and $\mu (E)=0$. Then we have

$${(A\cup B)}^c=A^c\cap B^c=A^c\cap {(E-(E-B))}^c=A^c\cap (E^c\cup (E-B))=(A^c\cap E^c)\cup (A^c\cap (E-B)),$$

where $A^c\cap E^c\in 𝒮$ and $A^c\cap (E-B)\in 𝒵$. Hence ${(A\cup B)}^c\in \overline{𝒮}$.

Now we define $\overline{\mu }$ on $\overline{𝒮}$ by $\overline{\mu }(A\cup B)=\mu (A)$, whenever $A\in 𝒮$ and $B\in 𝒵$. It is easily verified that this defines in fact a measure, and that $(X,\overline{𝒮},\overline{\mu })$ is the completion of $(X,𝒮,\mu )$.

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	Derivation
Classification	msc 28A12