# The outer measure on $X$ has a collection $M$ that is a $\sigma$-algebra

This is part of Caratheodory's Theorem taken by Real Analysis, Folland

If $\mu^*$ is an outer measure on $X$, the collection $M$ of $\mu^*$-measurable sets is a $\sigma$-algebra.

We first need to prove that $M$ is closed under complements and closed under finite unions to show that that $M$ is an algebra. I have no problem proving this I just want to know how to prove that $M$ is a $\sigma$-algebra. I know that it will suffice to prove that $M$ is closed under countable disjoint unions. If anyone could give me some advice on how to prove this, please let me know.

• Why are you asking this? There's a proof of the theorem in the book, including this. – David C. Ullrich Aug 18 '15 at 18:36
• I understand that but the proof skips some details... – Wolfy Aug 18 '15 at 18:38
• If you say so - looks fairly complete to me. Regardless, you want someone to write out that whole proof, just hoping to clarify whatever part you don't get? Not likely. You might try asking a more specific question. Or you might spend some time thinking about whatever details you think is missing! What students don't like about Folland is the "terseness". That proof is about a half page in Folland - that means if I were to write it out, including a detailed explanation of every possible thing anyone could miss, it would take three or four pages. Not gonna happen. What details? – David C. Ullrich Aug 18 '15 at 18:50
• Seriously: When you're reading Folland you need to expect to spend an hour or so on each page. That proof took me a few minutes just now, and I'm sort of somewhat moderately expert at that stuff... – David C. Ullrich Aug 18 '15 at 18:51
• The part I am confused about is the induction part if you are looking at the book – Wolfy Aug 18 '15 at 19:20

I don't how it is proved in Folland's book. Normally it is proved as follow.

We only need to prove that for pairwise disjoint $A_n\in M$ $$\mu^*(\bigcup_{n=1}^\infty A_n) \geqslant \sum_{n=1}^\infty \mu^*(A_n)\tag{1}$$ as other half is implied in subadditivity of outer measure.

First note that for outer measure, there is $$A_1\subset A_2\implies \mu^*(A_1)\leqslant \mu^*(A_2)\tag{2}$$ Then for any disjoint $A_n$, since $$\bigcup_{n\leqslant m}A_n \subset \bigcup_{n=1}^{\infty} A_n$$ By additivity for finite unions and (2), there is $$\sum_{n\leqslant m}\mu^*(A_n)=\mu^*(\bigcup_{n\leqslant m} A_n)\leqslant \mu^*(\bigcup_{n=1}^{\infty}A_n)\tag{3}$$ Let $m\to\infty$, (1) is proved.

EDIT: We need to prove that countable union of measurable sets is measurable in order to prove $M$ forms $\sigma$-algebra.

Given $A_n\in M$. Let $$A=\bigcup_{n=1}^\infty A_n=\bigcup_{n=1}^\infty (A_n-\bigcup_{k=1}^{n-1}A_k)=\bigcup_{n=1}^\infty A_n'$$ where $A_1'=A_1$. Clearly $A_n'$ are pairwise disjoint sets. Since $A_n'$ are formed from finite union and complement of measurable sets, they are all measurable. So by (3) $$\sum_{n =1}^{m}\mu^*(A_n')=\mu^*(\bigcup_{n =1}^{m} A_n')\leqslant \mu^*(\bigcup_{n=1}^{\infty}A_n')$$ So $\sum_{n =1}^{\infty}\mu^*(A_n')$ converges. Thus for any $\epsilon>0$, there exists $N>0$ such that $$\sum_{n=N+1}^{\infty}\mu^*(A_n')<\epsilon$$ Let $$A=\bigcup_{n=1}^{\infty}A_n'=\bigcup_{n=1}^{N}A_n'\cup \bigcup_{n=N+1}^{\infty}A_n'=C\cup D$$ Then $$\mu^*(D)=\mu^*(\bigcup_{n=N+1}^{\infty}A_n')\leqslant\sum_{n=N+1}^{\infty}\mu^*(A_n')<\epsilon$$ Since $C$ is measurable, by theorem $12$ on p $41$ in Real Analysis by Royden 4th or by theorem $6$ on p $261$ in Introductory Real Analysis by Kolmogorov and Fomin, there is an elementary set $B$ such that $$\mu^*(B\bigtriangleup C)<\epsilon$$ Also since $$A\bigtriangleup B\subset (B\bigtriangleup C)\cup D$$ There is $$\mu^*(A\bigtriangleup B)\leqslant\mu^*(B\bigtriangleup C)+\mu^*(D)<2\epsilon$$ By the above theorems again, $A=\bigcup_{n=1}^{\infty}A_n$ is measurable.

• First, this is all still irrelevant. Much closer to relevant than previously, but irrelevant. There are no elementary sets in the setting of the theorem the OP was asking about! Those books you cite are talking about a different theorem. A closely related theorem. The theorem we're talking about deals with any outer measure, not just outer measures defined in terms of a collection of elementary sets. Given an outer measure $\mu^*$ on $X$, here's the definition of measurable: $A$ is measurable if for every set $E$ we have $\mu^*(E)=\mu^*(A\cap E)+\mu^*(E\setminus A)$. – David C. Ullrich Aug 19 '15 at 13:32
• And second, even in the case where the outer measure is given by a collection of elementary sets, the proof about is not quite right. You're assuming that $\mu^*(\bigcup A_n')<\infty$. That need not be so. – David C. Ullrich Aug 19 '15 at 13:33