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Royden Real-analysis 4th edition p.349

Let $X$ be a set. Let $\mu$ be an outer measure on $P(X)$. Define $\mathfrak{M}$ = $\lbrace A\subset X \mid A \text{ is } \mu- \text{ measurable} \rbrace$. Let $\mu^*$ be the restriction of $\mu$ to $\mathfrak{M}$. Then, $(X,\mathfrak{M},\mu^*)$ is a complete measure space.

I know that it is a measure space, but how do i prove that it is complete?

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Subset $A$ of a measure 0 set $B$ would have outer measure 0. Given any $E$ then $\mu(A\bigcap E)=0$ because $A\bigcap E\subset B$; and $\mu(A^{c}\bigcap E)=\mu(E)$ because $B^{c}\bigcap E\subset A^{c}\bigcap E\subset E$ and $\mu(B^{c}\bigcap E)=\mu(E)$ since $B$ is measurable with measure 0.

Hence $\mu(A\bigcap E)+\mu(A^{c}\bigcap E)=\mu(E)$ so $A$ is measurable.

A more intuitive way to look at it is that a null set basically have no effects whatsoever, and definition of measurable sets depend only on what kind of harm the set could do.

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  • $\begingroup$ I have a question, "because $A\bigcap E\subset B$; and $\mu(A^{c}\bigcap E)=\mu(E)$ because $B^{c}\bigcap E\subset A^{c}\bigcap E\subset E$ and $\mu(B^{c}\bigcap E)=\mu(E)$ since $B$ is measurable with measure 0" In the above quote, why dont you only use the fact that $\mu(A^c\cap E)\leq \mu(E)$ in order to have, $\mu(E)\geq \mu(A\cap E)+\mu(A^c\cap E)$. $\endgroup$ – Eduardo Aug 7 '18 at 17:22
  • $\begingroup$ I think there is some problem in my above questioning, because it do not use the mensurability of $B$... Anyway I will be gratefull with some clarification... $\endgroup$ – Eduardo Aug 7 '18 at 17:31

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