# If $\mu$ is $\sigma$-finite, then the $\sigma$-algebra of all $\mu^*$-measurable sets coincides with the completed $\sigma$-algebra

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.

• Is it necessary that the sequence $(X_n)$ is pairwise disjoint? Commented Aug 9, 2023 at 2:17
• 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. Commented Aug 9, 2023 at 2:44

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 $$\mathcal M=\overline{\mathcal X}$$.

• 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 :) Commented Dec 31, 2022 at 15:14
• Glad to help and a good new year for you allready. Next week I might have a second look. Commented Dec 31, 2022 at 15:46
• I have added something about the case where $\mu$ is $\sigma$-finite. Commented Jan 2, 2023 at 16:43
• Thank you so much again! Your proof is much simpler than mine. Commented Jan 2, 2023 at 17:06
• In (4), isn't it necessary that $B$ is a subset of $A$ ? Commented Aug 8, 2023 at 7:52