# completion of a measure space

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 $(X,\mathscr{S},\mu)$.

Title completion of a measure space CompletionOfAMeasureSpace 2013-03-22 14:06:59 2013-03-22 14:06:59 Koro (127) Koro (127) 9 Koro (127) Derivation  msc 28A12