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If $m$ is an outer measure defined on a set $X$, we say that a subset $E$ of $X$ is Carathéodory-measurable with respect to $m$ if for all subsets $A$ of $X$, we have $m(A)=m(A\cap E) + m(A\cap E^c)$. And if $M$ is the set of all Carathéodory-measurable sets with respect to $m$, then $M$ is a complete sigma algebra on $X$ and $m$ restricted to $M$ is a complete measure on $X$.

My question is, is $M$ "optimal" in some way? Is it the biggest or smallest subset of $P(X)$ such that $m$ restricted to that subset is a measure? Is it the biggest or smallest subset of $P(X)$ such that $m$ restricted to that subset is a complete measure?

To put it another way, what is it that makes the Carathéodory measurability criterion the "best" criterion for measurability? Or is it not the best, but just an arbitrary choice out of a sea of infinitely many equally good stronger and weaker measurability criteria?

Keshav Srinivasan
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2 Answers2

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The answer is no, the Lebesgue measure can be extended to a translation invariant measure in a $\sigma$-algebra that properly contain the Lebesgue sigma álgebra. However this example is not trival. In fact it is content of a paper of Annals of Mathematics.

See: https://www.jstor.org/stable/1969435

Eduardo
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  • OK, it may not be the largest, but what about the smallest? Is $M$ the smallest subset of $P(X)$ such that $m$ restricted to that subset is a complete measure? – Keshav Srinivasan Sep 17 '18 at 21:21
  • I can think in a pathological example. Start with $\Omega=\mathbb{R}$ and define $\mu:\mathscr{P}(\mathbb{R})\to [0,\infty]$ by $\mu^{\star}(A)=0$ for any $A\in \mathscr{P}(\mathbb{R}).$ Then the $\sigma$-álgebra of the $\mu^\star$-measurable sets is $\mathscr{P}(\mathbb{R})$ and $\mu^\star$ restricted to the Lebesgue sigma algebra is a complete measure. – Eduardo Sep 18 '18 at 02:32
  • Is there some well-behaved class of outer measures for which $M$ is the smallest subset of $P(X)$ such that $m$ restricted to that subset is a complete measure? – Keshav Srinivasan Sep 18 '18 at 16:48
  • I dont know, but I think this is a non trivial question. – Eduardo Sep 18 '18 at 18:02
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Edit: Yes, the Caratheodory condition defines the largest $\sigma$-algebra $\sum$ such that the restriction of $m$ to $\sum$ is a measure and such that we also have $$m(A)=\inf_{A\subset E\in\sum}m(E)$$for all $A\subset X$. That follows from this:

Motivation for C's definition of measurability: Suppose $(X,\sum,\mu)$ is a measure space. Define the outer measure of $A\subset X$ by $\mu^*(A)=\inf_{A\subset E\in\sum}\mu(E)$. If $A\subset X$ and $E\in\sum$ then $\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E)$.

Proof. Suppose $A\subset X$ and $E\in\sum$. Choose $F$ with $A\subset F\in\sum$. Since $\mu$ is a measure, $A\cap E\subset F\cap E\in\sum$ and $A\setminus E\subset F\setminus E\in\sum$ we have $$\mu(F)=\mu(F\cap E)+\mu(F\setminus E)\ge\mu^*(A\cap E)+\mu^*(A\setminus E).$$Taking the inf over $F$ shows that $$\mu^*(A)\ge\mu^*(A\cap E)+\mu^*(A\setminus E).$$

The other inequality is clear, since it's easy to see that $\mu^*$ is subadditive: $\mu^*(B\cup C)\le\mu^*(B)+\mu^*(C)$ for every $B$ and $C$. (If $B\subset E_a\in\sum$ and $C\subset E_2\in\sum$ then $$\mu^*(B\cup C)\le\mu(E_a\cup E_2)\le\mu(E_1)+\mu(E_2);$$now take the inf over $E_1$ and $E_2$ on the right side.)

David C. Ullrich
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  • But there may be other measurability conditions that also imply that that condition holds for measurable sets $A$. So in that sense it might still be arbitrary. – Keshav Srinivasan Sep 12 '18 at 13:48
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    @KeshavSrinivasan Yes. I said that I wasn't claiming the definition was optimal in any sense. – David C. Ullrich Sep 12 '18 at 14:04
  • Are you saying that there are measure spaces that do not satisfy $m(A)=\inf_{A\subset E\in\sum}m(E)$? – Keshav Srinivasan Sep 13 '18 at 03:09
  • @KeshavSrinivasan No, I didn't say that. If we start with a general outer measure $m$ and then define $\sum$ by the C condition it's not clear to me whether it follows that $m(A)=\inf_{A\subset E\in\sum}m(E)$ for every $A$. Of course that's true if $m$ was defined in the way outer measures are typpically defined, but I don't see why it's clear in general. What I'm saying is what I said: If we add that condition then the $C$ definition of meaurability gives the largest possible $\sigma$-algebra. – David C. Ullrich Sep 13 '18 at 13:02
  • Could you please clarify how you answer squares with Eduardo's? As far as I could understand, Lebesgue $\sigma$-algebra is the largest $\sigma$-algebra $\Sigma$ that allows a given general outer measure $m$ to be recovered from its restriction on $\Sigma$ via $m(A)=\inf_{A\subset E\in\Sigma}m(E)$. At the same time, this resulting Lebesgue measure space can be extended (according to Kakutani&Oxtoby 1950), but this extension may disagree with the outer measure $m$ outside the Lebesgue $\sigma$-algebra. Is this correct? – paperskilltrees Jun 23 '22 at 13:47
  • Should I have replaced "a given general outer measure $m$" with "a given regular outer measure $m$"? – paperskilltrees Jun 23 '22 at 14:21