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[parent] 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)$.



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Cross-references: completion, measure, complements, unions, countable, closed under, contains, subsets, complete, measure space
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This is version 4 of completion of a measure space, born on 2004-01-18, modified 2004-01-18.
Object id is 5521, canonical name is CompletionOfAMeasureSpace.
Accessed 2446 times total.

Classification:
AMS MSC28A12 (Measure and integration :: Classical measure theory :: Contents, measures, outer measures, capacities)

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