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Let $(X,M, \mu)$ be a measure space. Let $\overline{M}$ be collection of $E \cup Z$ such that $E \in M$ and $Z \subset F$, where $F \in M,$ and $\mu(F) = 0.$ We also know $\bar{\mu}(E) = \mu(E).$

a) Show $\overline{M}$ is smallest sigma algebra containing $M$ and all subsets of elements of $M$ that have measure $0$.

Attempt: Notice if $E_{1} \subset \overline{M}$ then $E_{1} = E \cup Z.$ Since $E, Z \in M$ then $E_{1}^c = (E \cup F)^c \cup (F-Z)$ is in $M$ since $(E \cup F),(F-Z) \in M.$ Hence $\overline{M}$ is sigma algebra. If we let $Z = \emptyset$,then we see $M \subset \overline{M}.$ And since $\emptyset \in M,$ then let $E = \emptyset$, and then all subsets of $M$ with measure $0$ is in $M.$ Does this look okay? I am not sure how to show $\overline{M}$ is the smallest sigma algebra containing $M$

b) Show $\bar{\mu}$ is a measure of $\overline{M},$ and this measure is a completion.

Attempt: First check $\bar{\mu}$ is a measure of $\overline{M}.$ Notice $\bar{\mu}(\emptyset) = \mu(\emptyset) = 0.$ If $A_{1} \subset A_{2}$, and $A_{1} = B_{1} \cup C_{1}, A_{2} = B_{2} \cup C_{2},$ then $\bar{\mu}(A_1) = \mu(B_1) \le \mu(B_2) = \bar{\mu}(A_2).$ Lastly $\bar{\mu}(\cup_{i=1}^{n}A_{i}) = {\mu}(\cup_{i=1}^{n}B_{i}) \le \sum_{i=1}^{n}\mu(B_i).$ Next for completion, we need to show if $F \in \overline{M}$ and $\bar{\mu}(F) = 0,$ if $E\subset F$ then $E \in \overline{M}.$ I am not sure how to this.

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  • $\begingroup$ I don't see where you proved closure under countable unions. $\endgroup$ – bob Sep 12 '19 at 19:12
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Attempt of (a):

What you did here seems okay. Make sure to realize $E_1 \in \overline{M}$, not $E_1 \subset \overline{M}$.

To show it's the smallest $\sigma$-algebra, you have to show $$\overline{M} = \bigcap N$$ where the intersection is over all complete $\sigma$-algebras on $X$ that contain $M$.

Your question on (b) follows from monotonicity, or write $E$ as $\emptyset \cup E$.

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  • $\begingroup$ I am trying the subset approach to show the equality for part a), but its not working out. Any hints on how to approach showing its the smallest sigma algebra containing M? $\endgroup$ – user77404 Aug 11 '14 at 6:11
  • $\begingroup$ @user77404 To show $\overline{M} \subset \cap N$, it suffices to show that $\overline{M} \subset N$ whenever $N$ is a complete $\sigma$-algebra containing $M$. For the opposite inclusion, just note that $\overline{M}$ is a complete $\sigma$-algebra containing $M$, so $$\bigcap N = \overline{M} \bigcap (\bigcap_{N \neq \overline{M}} N) \subset \overline{M}$$ $\endgroup$ – Andrew Maurer Aug 11 '14 at 21:24
  • $\begingroup$ @user77404 your first statement is proving it is the smallest, since you claim $\bar{M}⊂N$ for all complete σ-algebra containing $M$. If it is proved, nothing more is needed to prove. But, how we can show that? $\endgroup$ – Susan_Math123 Oct 23 '15 at 3:56
  • $\begingroup$ @andybenji can you continue your solution part (b) further? $\endgroup$ – Susan_Math123 Oct 23 '15 at 4:51
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The argument "since $E, Z \in M$" is incorrect, since $Z$ needn't be in $M$. It is true, though, that $Z \subset F$ for some null measurable set $F$.

Here is a suggestion:

Without loss of generality, if $E \cup Z \in \overline{M}$, while $Z \subset F \in Null(M)$, one can assume that $E \cap F = \phi$. Indeed, otherwise one could write $E \cup Z = (E \setminus F) \cup [(F \cap E) \cup Z]$, and have $E \setminus F \in M, (F \cap E) \cup Z \subset F \in Null(M)$

So, $E^c = E^c \setminus F \cup F \in \overline{M}$, since $E^c \setminus F \in M, F \in Null(M)$

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