completion of a measure space
CompletionOfAMeasureSpace
2004-01-18 16:46:24
2009-10-05 17:14:31
Derivation
complete measure
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If the measure space $(X,\mathscr{S},\mu)$ is not complete, then it can be completed in the following way. Let
$$\mathscr{Z} = \bigcup_{E\in\mathscr{S}\, ,\mu(E)=0} \mathscr{P}(E),$$
i.e. the family of all subsets of sets whose $\mu$-measure is zero.
Define
$$\overline{\mathscr{S}} = \{A \cup B : A\in \mathscr{S}, \, B\in \mathscr{Z}\}.$$
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 $$(A\cup B)^c = A^c\cap B^c = A^c\cap (E-(E-B))^c = A^c\cap (E^c\cup (E-B)) = (A^c\cap E^c) \cup (A^c\cap(E-B)),$$
where $A^c\cap E^c\in \mathscr{S}$ and $A^c\cap(E-B)\in\mathscr{Z}$. Hence $(A\cup B)^c \in \overline{\mathscr{S}}$.
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 $(X,\mathscr{S},\mu)$.