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I'm self-learning Measure Theory using Real Analysis book of Folland. Unfortunately, I got stuck in this problem and couldn't find any clue to solve this. Can someone help me, or give me some hint so I can solve it. Thanks so much:

Let ($X$, $M$, $\mu$) be a measure space, $\mu^*$ the outer space induced by $\mu$ according to the formula: $$\mu^*(E) = \inf\{\sum_1^\infty \mu_0(A_j): A_j \in M, E \subset \bigcup_1^\infty A_j\}$$ $M^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, and $\overline{\mu} = \mu^*|M^*$. Prove that $\overline{\mu}$ is the saturation of the completion of $\mu$

Although I can solve the first problem(which states that if $\mu$ is $\sigma$-finite, then $\overline{\mu}$ is the completion of $\mu$), but I still got no clue to solve this general problem. Please help me, I really appreciate.

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  • $\begingroup$ $u^*(E)$ was supposed to be $\mu^*(E)$, and "$\sigma$-measurable" was something else, probably "$\mu^*$-measurable", yes? Anyway, a valid non-answer to your question is you should worry about non-$\sigma$-finite measures last; the measures that come up in most contexts are $\sigma$-finite. $\endgroup$ – David C. Ullrich Jun 27 '15 at 17:51
  • $\begingroup$ I'm so sorry for that. I edited the question to be correct. I'm so careless, sorry! $\endgroup$ – le duc quang Jun 28 '15 at 1:21
  • $\begingroup$ If you have had time to think about my purported solution, I would appreciate some feedback. $\endgroup$ – Matt Rosenzweig Jul 4 '15 at 16:40
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To avoid confusion, let's clarify some notation. Let $\mathcal{M}^{*}$ denote the $\sigma$-algebra of $\mu^{*}$-measurable subsets of $X$. Let $\bar{\mathcal{M}}$ denote the completion of $\mathcal{M}$ with respect to $\mu$. Let $\widetilde{\bar{\mathcal{M}}}$ denote the $\sigma$-algebra of locally measurable subsets of $(X,\bar{\mathcal{M}},\bar{\mu})$. Since $\mu^{*}$ restricts to a complete measure on $\mathcal{M}^{*}$, we know that $\mu^{*}|{\mathcal{M}}^{*}$ coincides with the completion of $\mu$ on the completion $\bar{\mathcal{M}}$ of $\mathcal{M}$ with respect to $\mu$, so there is no ambiguity in using the notation $\bar{\mu}$.


Lemma. Let $(X,\mathcal{M},\mu)$ be a measure space, and let $\mu^{*}$ be the outer measure induced by $\mu$. For $E\in\mathcal{M}^{*}$, with $\mu^{*}(E)<\infty$, there exists $A\in\mathcal{M}$ such that \begin{align*} E\subset A, \quad \mu^{*}(E)=\mu(A) \end{align*} In particular, if $\mu^{*}(E)=0$, then $E$ is contained in a $\mu$-null set $N\in\mathcal{M}$.

Proof. By definition of infimum, there exists a countable collection of sets $\left\{A_{j}\right\}\subset\mathcal{M}$ such that \begin{align*} E\subset A:=\bigcup_{j=1}^{\infty}A_{j}, \quad \sum_{j=1}^{\infty}\mu(A_{j})<\mu^{*}(E)+\epsilon \end{align*} Since $\mathcal{M}$ is a $\sigma$-algebra, $A\in\mathcal{M}$, and by subadditivity, $\mu(A)\leq\mu^{*}(E)+\epsilon$.

We can apply the preceding result for each $\epsilon_{n}=1/n$, to obtain a collection of sets $\left\{A^{n}\right\}\subset\mathcal{M}$ such that $E\subset A^{n}$ and $\mu(A^{n})\leq\mu^{*}(E)+1/n$. If we set $A:=\bigcap_{n}A^{n}$, then $E\subset A$ and by monotonicity, \begin{align*} \mu(A)\leq\mu(A^{n})\leq\mu^{*}(E)+\dfrac{1}{n},\qquad\forall n \end{align*} Letting $n\rightarrow\infty$, we obtain $\mu(A)\leq\mu^{*}(E)$. The reverse inequality holds by monotonicity. $\Box$


Let $E$ be a locally measurable subset of $(X,\bar{\mathcal{M}},\bar{\mu})$. I claim that $E\in\mathcal{M}^{*}$. It suffices to show that for any $F\subset X$ with $\mu^{*}(F)<\infty$, we have \begin{align*} \mu^{*}(F)\geq\mu^{*}(E\cap F)+\mu^{*}(E^{c}\cap F) \end{align*} By the lemma, there exists a set $A\in\mathcal{M}$ such that \begin{align*} F\subset A, \quad \mu^{*}(F)=\mu(A) \end{align*} So $E\cap A$ is measurable, whence $E^{c}\cup A^{c}\cap A=E^{c}\cap A$ is measurable. By monotonicity and additivity, we see that \begin{align*} \mu^{*}(E\cap F)+\mu^{*}(E^{c}\cap F)\leq\mu^{*}(E\cap A)+\mu^{*}(E^{c}\cap A)=\mu(A)=\mu^{*}(F) \end{align*}

The reverse inclusion also holds: $E\in\mathcal{M}^{*}$ implies $E\in\widetilde{\bar{\mathcal{M}}}$. For any set $A\in\bar{\mathcal{M}}$ with $\bar{\mu}(A)<\infty$, $\mu^{*}(E\cap A)<\infty$, whence there exists a set $B\in\mathcal{M}$ such that $E\cap A\subset B$ and $\mu^{*}(E\cap A)=\mu(B)$. Since $E\cap A, B\in\mathcal{M}^{*}$, $\mu^{*}(B\setminus (E\cap A))=0$. But then there exists $N\in\mathcal{M}$, such that \begin{align*} B\setminus (E\cap A)\subset N, \quad \mu^{*}(B\setminus (E\cap A))=\mu(N)=0 \end{align*} We conclude that $B\setminus (E\cap A)\in\bar{\mathcal{M}}$, whence \begin{align*} E\cap A=B\setminus (B\setminus (E\cap A))\in\bar{\mathcal{M}} \end{align*}

With $E$ as above, suppose $\mu^{*}(E)<\infty$. Then, as asserted before, there exists a set $A\in\mathcal{M}$ such that $E\subset A$ and $\bar{\mu}(E)=\mu(A)$. But then $E=E\cap A\in\bar{\mathcal{M}}$. We conclude that

\begin{align*} \bar{\mu}(E)=\begin{cases}\bar{\mu}(E) & {E\in\bar{\mathcal{M}}}\\ \infty & {E\in\widetilde{\bar{\mathcal{M}}}\setminus\bar{\mathcal{M}}} \end{cases}, \end{align*} which is the definition of the $\widetilde{\bar{\mu}}$, the saturation of $\bar{\mu}$.

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  • $\begingroup$ Sorry for not replying you soon @matt. I will see your answer carefully tomorrow and ask you if I have any question. Thanks so much for answering this problem. I really appreciate $\endgroup$ – le duc quang Jul 4 '15 at 17:14
  • $\begingroup$ Hi @matt, I took a look at your proof. It's all very convincing, except the last point. In the book, if $E$ is in completion of $\mathcal{M}$, we write $E = M \cup N$, here $M \in \mathcal{M}, N \subset P, P \in \mathcal{M}, \mu(P) = 0$, then completion measure of $E$ is equal to $\mu(M)$. So I don't see why your last point can show that $\overline{ \mu}$ is the saturation of completion of $\mu$. Can you please clarify it clearer? $\endgroup$ – le duc quang Jul 5 '15 at 7:41
  • $\begingroup$ And another thing is that, the condition that $E$ be a locally measurable set in $(X, \overline{\mathcal{M}}, \overline{\mu})$ mean that $E \cap A \in \overline{\mathcal{M}}$ for all $A \in \overline{\mathcal{M}}$ and $\overline{\mu}(A) < \infty$. So when you have $\mu(A) < \infty$, I think you can only conclude that $E \cap A \in \overline{\mathcal{M}}$, not $\mathcal{M}$, so I don't think you can conclude that $E \cap A$ is measurable. $\endgroup$ – le duc quang Jul 5 '15 at 16:37
  • $\begingroup$ @leducquang: So the measure on the completion $\sigma$-algebra $\overline{\mathcal{M}}$ is uniquely determined by $\mu$; I believe Folland at least states this. So $\overline{\mu}:=\mu^{*}|\overline{\mathcal{M}}$ agrees with the definition you gave in your comment. Regarding the second comment, so then $E\cap A \in\overline{\mathcal{M}}$ and $E^{c}\cup A^{c}\cap A=E^{c}\cap A\in\overline{\mathcal{M}}$. As stated before, $\mu^{*}$ restricts to a measure on $\overline{\mathcal{M}}$, so disjoint additivity holds. Replacing $\mu(A)$ with $\overline{\mu}(A)$, but these two quantities are equal. $\endgroup$ – Matt Rosenzweig Jul 5 '15 at 16:47
  • $\begingroup$ Ok, I understand your point now. Thanks so much for your clarification, @matt $\endgroup$ – le duc quang Jul 6 '15 at 1:50

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