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