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Let $(X, \mathcal X, \mu)$ be a measure space. Let $\mu^*$ be the outer measure induced from $\mu$, i.e., $$ \mu^*(A):=\inf \left\{ \sum_{n=1}^{\infty} \mu (B_n) : (B_n) \subset \mathcal X , A \subset \bigcup_n B_n \right\} \quad \forall A \subset X. $$

Let $\mathcal{M}$ be the collection of all $\mu^*$-measurable sets, i.e., those sets $A \subseteq X$ such that $$ \mu^*(E)=\mu^*(E\cap A) +\mu^*(E\cap A^c) \quad \forall E\subset X. $$

Let $(X, \overline{\mathcal X}, \overline \mu)$ be the completion of $(X, \mathcal X, \mu)$. Then

  • $\mathcal M$ is a $\sigma$-algebra on $X$,
  • $\mu^*|_{\mathcal M}$ is a complete measure,
  • $\mathcal X \subset \overline{\mathcal X} \subset \mathcal M$,
  • $\overline \mu = \mu^*|_{\overline{\mathcal X}}$, and $\mu = \overline \mu |_{\mathcal X}$.

I would like to prove a result mentioned in this answer, i.e.,

Theorem If $\mu$ is $\sigma$-finite then $\overline{\mathcal X} = \mathcal M$.

Could you have a check on my attempt?


Proof Fix $A \in \mathcal M$. Let's prove that $A \in \overline{\mathcal X}$. Let $\mathcal N$ be the collection of all subsets of $\mu$-null subsets of $X$, i.e., $$ \mathcal N := \{A \subset X :\exists N \in \mathcal X \text{ such that } A \subset N \text{ and } \mu (N)=0\}. $$

Then $A \in \overline{\mathcal X}$ if and only if $A = B \cup C$ for some $B \in \mathcal X$ and $C \in \mathcal N$.

  1. $\mu$ is finite.

Because $\mu^* (A) \le \mu^*(X) = \mu (X) < \infty$, for each $n \in \mathbb N$ there is a sequence $(B_{nm})_m \subset \mathcal X$ such that $A \subset \bigcup_m B_{nm}$ and $\mu^* (A) > \sum_{m=1}^{\infty} \mu (B_{nm}) - \frac{1}{n}$. Let $B := \bigcap_n \bigcup_m B_{nm}$. Then $A \subset B \in \mathcal X$ and $\mu^* (A) = \mu (B)$. It follows from $\mu^* (A) < \infty$ that $\mu^* (A')=0$ with $A' := B \setminus A$. With similar reasoning, there is $B' \in \mathcal X$ such that $A' \subset B'$ and $\mu^* (A') = \mu (B')$. It follows that $A' \in \mathcal N$. We have $$ A^c := X \setminus A = (X \setminus B) \cup A'. $$

It follows that $A^c \in \overline{\mathcal X}$ and thus $A \in \overline{\mathcal X}$.

  1. $\mu$ is $\sigma$-finite.

There is a sequence of $(X_n)$ of pairwise disjoint sets in $\mathcal X$ such that $\bigcup_n X_n = X$ and $\mu (X_n) < \infty$ for all $n$. Let $\mathcal X_n$ be the sub $\sigma$-algebra that $\mathcal X$ induces on $X_n$, i.e., $\mathcal X_n := \{A \cap X_n : A \in \mathcal X\}$. Let $\mu_n$ be the restriction of $\mu$ to $\mathcal X_n$, i.e., $\mu_n (A) := \mu (A)$ for all $A \in \mathcal X_n$. We construct the corresponding objects $(\mu^*_n, \mathcal M_n, \overline{\mathcal X_n}, \overline{\mu_n})$ from the measure space $(X_n, \mathcal X_n, \mu_n)$.

Clearly, $\mu_n$ is finite. We need the following lemmas, i.e.,

  • Lemma 1 $\mathcal M_n = \{A \cap X_n : A \in \mathcal M\}$.

  • Lemma 2 $\overline{\mathcal X_n} = \{A \cap X_n : A \in \mathcal{\overline X}\}$.

Let $A_n := A \cap X_n$. By Lemma 1, $A_n \in \mathcal M_n$. By (1.), $\overline{\mathcal X_n} = \mathcal M_n$. So $A_n \in \overline{\mathcal X_n}$. By Lemma 2, $\overline{\mathcal X_n} \subset \mathcal{\overline X}$. This implies $A_n \in \mathcal{\overline X}$ for all $n$. On the other hand, $A = \bigcup_n A_n$. It follows that $A \in \mathcal{\overline X}$. This completes the proof.

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  • $\begingroup$ Is it necessary that the sequence $(X_n)$ is pairwise disjoint? $\endgroup$
    – ashpool
    Commented Aug 9, 2023 at 2:17
  • $\begingroup$ I checked your proof, and I think it works. Just one comment: Instead of Lemma 2, I think you could use the more general result that completion of measure space preserves inclusion. $\endgroup$
    – ashpool
    Commented Aug 9, 2023 at 2:44

1 Answer 1

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Some simplifications:


$\mu^*(A)=\inf\{\mu(B)\mid A\subseteq B\in\mathcal X\}\tag1$

For this beware that we have $A\subseteq\mu^*(A)\leq\mu(\bigcup_{n=1}^{\infty}B_n)\leq\sum_{n=1}^{\infty}\mu(B_n)$ whenever the $B_n\in\mathcal X$ cover $A$ and that also $\bigcup_{n=1}^{\infty}B_n\in\mathcal X$.


$\forall A\in\mathcal P(X)\exists B\in\mathcal X[A\subseteq B\text{ and }\mu^*(A)=\mu(B)]\tag2$

If $\mu^*(A)=\infty$ then this indicates that $\mu(X)=\infty$ and we can go for $B=X\in\mathcal X$. Otherwise let $B_n\in\mathcal X$ with $A\subseteq B_n$ and $\mu^*(A)\leq\mu(B_n)\leq\mu^*(A)+\frac1n$ for every $n$. Then $B:=\bigcap_{n=1}^{\infty}B_n$ does the job.


$\mu^*(E\Delta F)=0\implies\mu^*(E)=\mu^*(F)\tag3$

According to $(1)$ proving this boils down to proving that under the condition on LHS it can be deduced that $E\subseteq A\in\mathcal X\implies\mu^*(F)\leq\mu(A)$. This because from $F\subseteq A\cup(E\Delta F)$ it follows directly that $\mu^*(F)\leq\mu^*(A)+\mu^*(E\Delta F)=\mu(A)$.


$A\in\overline{\mathcal X}\text{ iff }\mu^*(A\Delta B)=0\text{ for some }B\in\mathcal X\tag4$

The condition $A\in\overline{\mathcal X}$ states the existence of sets $B,C\in\mathcal X$ and $N\subseteq C$ such that $A=B\cup N$ and $\mu(C)=0$. Then $A\Delta B\subseteq N\subseteq C$ so that $\mu^*(A\Delta B)\leq\mu(C)=0$. Conversely if $\mu^*(A\Delta B)=0$ then sets $C_n\in\mathcal X$ exist $A\Delta B\subseteq C_n$ and $\mu(C_n)<\frac1n$. Then $C:=\bigcap_{n=1}^{\infty}C_n\in\mathcal X$ with $A\Delta B\subseteq C$ and $\mu(C)=0$ showing that $A\Delta B$ is a $\mu$-nullset.


Let me first mention that $(1)$, $(3)$ and $(4)$ can be applied to prove the second inclusion of your third bullet point (i.e. $\overline{\mathcal X}\subseteq\mathcal M$) on a direct way. For $A\in\overline{\mathcal X}$ according to $(4)$ we have some $B\in\mathcal X$ with $\mu^*(A\Delta B)=0$. For any fixed set $E\subseteq X$ the sets $(E\cap A)\Delta(E\cap B)$ and $(E\cap A^c)\Delta(E\cap B^c)$ are subsets of $A\Delta B$ so that $\mu^*$ sends these sets to $0$. Then according to $(3)$ we have:$$\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\cap A^c)=\mu^*(E\cap B)+\mu^*(E\cap B^c)=\mu^*(E)$$where the last equality is based on inclusion $\mathcal X\subseteq\mathcal M$.

Now let's have a look at the case where $\mu$ is a finite measure. It is our aim to prove that $\mathcal M\subseteq\overline{\mathcal X}$ so let $A\in\mathcal M$. Then - because $\mu$ is a finite measure - $\mu^*(A)<\infty$ and according to $(2)$ we have $\mu^*(A)=\mu(B)$ for some $B\in\mathcal X$ with $A\subseteq B$. Then $A\in\mathcal M$ implies that: $$\mu(B)=\mu^*(B)=\mu^*(A\cap B)+\mu^*(A^c\cap B)=\mu^*(A)+\mu^*(B-A)=\mu(B)+\mu^*(B-A)$$and we conclude that $\mu^*(A\Delta B)=\mu^*(B-A)=0$. Then $(4)$ assures that $A\in\overline{\mathcal X}$.


Edit:

If $A\in\mathcal M$ can be written as $\bigcup_{n=1}^{\infty}A_n$ where the $A_n$ are disjoint elements of $\mathcal M$ with $\mu^*(A_n)<\infty$ then $A\in\overline{\mathcal X}$.

Above it was proved that we have $\mathcal M\subseteq\overline{\mathcal X}$ if $\mu$ is a finite measure. However in that proof we did not actually use the fact that $\mu$ is a finite measure but only that $\mu^*(A)<\infty$ for the $A\in\mathcal M$ that we observed. So actually we proved that $\mu^*(A)<\infty$ is for $A\in\mathcal M$ a sufficient condition for $A\in\overline{\mathcal X}$. So we are allowed to conclude that the $A_n$ are elements of $\overline{\mathcal X}$. From this we conclude that also $A\in\overline{\mathcal X}$.

Final note: if $\mu$ is a $\sigma$-finite measure then every $A\in\mathcal M$ can be written as above hence will be an element of $\overline{\mathcal X}$. So in that situation $\mathcal M=\overline{\mathcal X}$.

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  • $\begingroup$ Thank you so much for your elaboration. Actually, my emphasis is on the case $\mu$ is $\sigma$-finite. In my proof, I appeal to Lemma 1 whose proof is not very short. I hope that you can have a check on it. Have a good time at the very end of this year :) $\endgroup$
    – Akira
    Commented Dec 31, 2022 at 15:14
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    $\begingroup$ Glad to help and a good new year for you allready. Next week I might have a second look. $\endgroup$
    – drhab
    Commented Dec 31, 2022 at 15:46
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    $\begingroup$ I have added something about the case where $\mu$ is $\sigma$-finite. $\endgroup$
    – drhab
    Commented Jan 2, 2023 at 16:43
  • $\begingroup$ Thank you so much again! Your proof is much simpler than mine. $\endgroup$
    – Akira
    Commented Jan 2, 2023 at 17:06
  • $\begingroup$ In (4), isn't it necessary that $B$ is a subset of $A$ ? $\endgroup$
    – ashpool
    Commented Aug 8, 2023 at 7:52

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