4
$\begingroup$

Let $(X,\mathscr{A},\mu)$ be a measure space. The completion of $\mathscr{A}$ with respect to $\mu$ is defined as $$\mathscr{A}_\mu :=\left\{A\subseteq X\mid \exists E,F\in \mathscr{A} : E\subseteq A\subseteq F, \mu(F\setminus E)=0 \right\}.$$ The completion of $\mu$ is defined as a function $\overline{\mu}:\mathscr{A}_\mu\to[0,\infty]$ by $$\overline{\mu}(A) :=\sup\left\{\mu(B)\mid B\in\mathscr{A}:B\subseteq A\right\}.$$ The triple $(X,\mathscr{A}_\mu,\overline{\mu})$ now forms a complete measure space.

I am trying to show that if we again complete this completion, we obtain the same measure space. Specifically, I am trying to show that $(\mathscr{A}_\mu)_{\overline{\mu}}=\mathscr{A}_\mu$ (and $\overline{\overline{\mu}}=\overline{\mu}$).

My observations so far: The inclusion $\mathscr{A}_\mu\subseteq(\mathscr{A}_\mu)_{\overline{\mu}}$ follows easily. So, suppose that $A\in (\mathscr{A}_\mu)_{\overline{\mu}}$. Then we can find $E,F\in\mathscr{A}_\mu$ so that $E\subseteq A\subseteq F$ and $\overline{\mu}(F\setminus E)=0$. We need to find two sets $E', F'\in\mathscr{A}$ so that $E'\subseteq A\subseteq F'$ and $\mu(F'\setminus E')=0$. Since $E,F\in \mathscr{A}_\mu$ we can find $E_1,E_2,F_1,F_2\in\mathscr{A}$ so that $$E_1\subseteq E\subseteq E_2 \quad \text{and}\quad\mu(E_2\setminus E_1)=0,$$ and $$F_1\subseteq F\subseteq F_2 \quad\text{and}\quad\mu(F_2\setminus F_1)=0.$$ We now have the chain of inclusions $E_1\subseteq E\subseteq A\subseteq F\subseteq F_2$. Setting $E'=E_1$ and $F'=F_2$ seems the obvious next step, but then we need to show that $\mu(F_2\setminus E_1)=0$. However, I don't see why this should be true. On the other hand, we may set $E'=E_1\cup F_1$ and $F'=E_2\cup F_2$, so that $$0\leq \mu((E_2\cup F_2)\setminus(E_1\cup F_1))\leq \mu((E_2\setminus E_1)\cup (F_2\setminus F_1)) \leq 0.$$ But now $E_1\cup F_1$ may not be contained in $A$, because $F_1$ may not be contained in $A$. We may modify our choice to $E'=E_1\cap F_1$, which is contained in $A$. However, now: $$\mu((E_2\cup F_2)\setminus (E_1\cap F_1))\leq \mu(E_2\setminus F_1) + \mu(F_2\setminus E_1),$$ which does not tell us anything because the terms on the right hand side are unknown. I am a bit stuck.

Perhaps another possible approach would be using the fact that $B\in\mathscr{A}_\mu$ if and only if $\mu_\ast(B)=\mu^\ast(B)$, where $\mu_\ast$ and $\mu^\ast$ are the inner- and outer measures induced by $\mu$, respectively.

Any hints would be greatly appreciated!

$\endgroup$
1
  • $\begingroup$ This is exercise 1 of section 1.5 in D.L. Cohn's Measure Theory, second edition. $\endgroup$
    – Nesta
    Commented Oct 5, 2016 at 15:25

1 Answer 1

0
$\begingroup$

DISCLAIMER

I know this is a very old thread. However, after posting about this and working on it, I think I found a nice way to do it.

Proof. By constuction, if $A\in \mathscr A_\mu$, then $A\subset A\subset A$ and $\bar\mu(A-A)=0$. Therefore $A\in(\mathscr A_\mu)_{\bar\mu}$ and hence $\mathscr A_\mu\subset (\mathscr A_\mu)_{\bar\mu}$.

Now let $A\in (\mathscr{A}_\mu)_{\bar\mu}$. Then there exists $E,F\in \mathscr A_\mu$ such that $E\subset A\subset F$ where $\bar\mu(F-E)=0$ so that $\bar\mu(E)=\bar\mu(F)$. It follows that $A-E\subset F-E$. Moreover $E,F\in \mathscr A_\mu$ implies that $F- E=F\bigcap E^c\in\mathscr A_\mu$ since $\mathscr A_\mu$ is a sigma algebra. But then there exists $E_1,F_1\in \mathscr A$ such that $E_1\subset F-E\subset F_1$ and $0=\bar\mu(F-E)=\mu(F_1)=\mu(E_1)$. But taking $E_1=\emptyset$, we find that $E_1\in \mathscr A$ and satisfies $\mu(F_1-E_1)=\mu(F_1)=0$. We now have that $E_1=\emptyset\subset A-E=A\bigcap E^c\subset F-E\subset F_1$ and thus $E_1\subset A-E\subset F_1$. Thus, both $A-E,E\in \mathscr A_\mu$ and hence $A=(A-E)\bigcup E\in \mathscr A_\mu$. Therefore $(\mathscr A_\mu)_{\bar\mu}\subset \mathscr A_\mu$.

We now have that $(\mathscr A_\mu)_{\bar\mu}=\mathscr A_\mu$. Finally, since $A\in \mathscr A\mu$, there exists $E_2,F_2\in \mathscr A\subset \mathscr A_\mu$ such that $E_2\subset A\subset F_2$ and $\mu(F_2-E_2)=0$. Furthermore, by definition $\mu(E_2)=\bar\mu(A)=\mu(F_2)$. Since $\bar\mu=\mu$ on $\mathscr A$, we have automatically $\mu(E_2)=\mu(F_2)=\bar\mu(E_2)=\bar\mu(F_2)$. Therefore, $\bar{\bar\mu}(A)=\bar\mu(E_2)=\mu(E_2)=\bar\mu(A)$ so that $\bar{\bar\mu}=\bar\mu$.

If anyone wants to critique this, I would much appreciate it.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .