This is a homework problem and I need some guidance on a proof.

Let $(X,\mathcal{M},\mu)$ be a measure space, $\mu^*$ the outer measure induced by $\mu$ according to (1.12), $\mathcal{M}^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, and $\overline{\mu}=\mu^*\mid_{\mathcal{M}^*}$.

(a) If $\mu$ is $\sigma$-finite, then $\overline{\mu}$ is the completion of $\mu$.


(1.12) $\mu^*(E) = \inf\left\{ \sum_{1}^\infty \mu(A_j)\ \mid\ A_j \in \mathcal{M},\ E \subset \bigcup_1^\infty A_j \right\}.$

Definition: $A \subset X$ is $\mu^*$-measurable for an outer measure $\mu^*$ if $$\mu^*(E) = \mu^*(E \cap B) + \mu^*(E \cap B^C)$$ for all $E \subset X$.

We have the following Lemma:

For an measure $\mu^*$ induced by a premeasure, if $\mu^*(E) < \infty$, then $E$ is $\mu^*$-measurable iff there exists $B \in \mathcal{A}_{\sigma\delta}$ with $E \subset B$ and $\mu^*(B\backslash E) =0$, where $\mathcal{A}_\sigma$ is the collection of countable unions of sets in $\mathcal{M}$, and $A_{\sigma\delta}$ is the collection of countable intersections of sets in $\mathcal{A}_\sigma$.

with the following Corollary:

If $\mu$ is $\sigma$-finite, then the condition $\mu^*(E) < \infty$ is superfluous.

What I started with:

Suppose there is some $E\in \mathcal{M}$ such that $E$ is not $\mu^*$-measurable but $\mu(E) = 0$. Then, since $E \in \mathcal{M}$, we have \begin{equation} \mu^*(E) = \inf \left\{ \sum \mu(A_j)\ \mid\ A_j \in \mathcal{M},\ E \subset \bigcup A_j\right\} = \mu(E) = 0. \end{equation} Consider that $\mu$ is $\sigma$-finite. Then $X = \bigcup_{j=1}^\infty F_j$, so $X \in \mathcal{M}_\sigma$, the collection of countable unions of sets in $X$. Likewise, $X \in \mathcal{M}_{\sigma\delta}$, the collection of countable intersections of sets in $\mathcal{M}_\sigma$. Then, $\mu^*(X\backslash\ E^C) = 0$, so $E^C$ is $\mu^*$-measurable, and hence $E$ is $\mu^*$-measurable. Therefore, no such $E$ can exist, and every $\mu$-null set is a $\mu^*$-null set in $\mathcal{M}^*$. Therefore, $\overline{\mu}$ is the unique completion of $\mu$.

Is this the right track?

From Daniel Fischer's comments, which I do not understand:

By the proposition, every $E \in \mathcal{M}$ is $\mu^*$ measurable, since $\mathcal{M} = \mathcal{M}_\sigma = \mathcal{M}_{\sigma\delta}$. Since every $\mu^*$-measurable set can be approximated from within, assume $\mu^*(E) = \mu(E) = 0$. Then, $A \subset E \subset B \in \mathcal{M}$ and $\mu(B\setminus A) = 0$, so again by the lemma, $A \in \mathcal{M}^*$. Thus every subset of a null set in $\mathcal{M}$ is in $\mathcal{M}^*$.

  • $\begingroup$ Which construction is "according to (1.12)"? Which is the definition of $\mu^\ast$-measurability, the Carathéodory criterion, for all $B$ we have $\mu^\ast(B) = \mu^\ast(B\cap E) + \mu^\ast(B\setminus E)$? $\endgroup$ – Daniel Fischer Feb 8 '15 at 21:51
  • $\begingroup$ @DanielFischer Whoops, thought I removed all those references. Editing. And yes. $\endgroup$ – Emily Feb 8 '15 at 22:09
  • $\begingroup$ Unless you're using a different definition of a measure space, $\mathcal{M}$ is a $\sigma$-algebra, and hence $\mathcal{M}_{\sigma\delta} = \mathcal{M}_\sigma = \mathcal{M}$. Using that simplifies matters considerably. It is then immediate from the lemma and its corollary that every $E\in \mathcal{M}$ is $\mu^\ast$-measurable [assuming $\sigma$-finiteness], and you can easily see that you can also approximate every $\mu^\ast$-measurable set from within, so you get $A\subset E \subset B$ with $\mu(B\setminus A) = 0$. $\endgroup$ – Daniel Fischer Feb 8 '15 at 22:23
  • $\begingroup$ @DanielFischer That's what I'm not understanding. Why do we have that $\mu(B\backslash A) = 0$? $\endgroup$ – Emily Feb 8 '15 at 22:26
  • $\begingroup$ Wait, maybe I'm understanding.... $\endgroup$ – Emily Feb 8 '15 at 22:31

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