PlanetMath: completion of a measure space

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completion of a measure space

(Derivation)

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

$\displaystyle \mathscr{Z} = \bigcup_{E\in\mathscr{S}\, ,\mu(E)=0} \mathscr{P}(E),$

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

$\displaystyle \overline{\mathscr{S}} = \mathscr{S}\cup\mathscr{Z}.$

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

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

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

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