Does Caratheodory measurable imply measurable in general?

Question

In some textbooks, they wrote "Caratheodory measurable, or measurable for short". Does it mean they are equivalent?

I known that Caratheodory measurable implies measurable for Lebesgue measure, but what if we consider a general measure space ? Will this still true?

If not, then why is Caratheodory measurable important? What can this concept do in the general measure space?

what means "Carathéodory measurable"? I just know the Carathéodory extension theorem that defines a complete $\sigma $-algebra from an outer measure — , Jan 02 '20 at 06:44
@Masacroso: "Carathéodory measurable" is probably a term used temporarily (by a book, lecturer, etc.) to denote measurability by Carathéodory's criterion to discuss various equivalences of measurability. Thus, one might say "Lebesgue measurable" to mean inner measure equals outer measure, as Lebesgue originally did, and then discuss the relation (namely, equality) between the two. Regarding the OP's question, possibly of interest is Demystifying the Caratheodory Approach to Measurability. — Dave L. Renfro, Jan 02 '20 at 07:20
Also of possible interest: The blog essay Intuition behind Caratheodory’s Criterion: Think “Sharp Knife” and “Shrink Wrap”! AND the mathoverflow question Definitions of Measurability: Outer-inner measure convergence vs. Caratheodory criterion (and the links given in this last question). — Dave L. Renfro, Jan 02 '20 at 07:28

score 0 · Answer 1 · answered Jan 02 '20 at 17:58

They wrote "measurable" for short since the (outer) measure is fixed. Not because they are equivalent.

For general cases, we can define a measure by Carathéodory's criterion.

Given an outer measure $\mu^*$ on a set $\Omega$, we say that a set $E\subset \Omega$ is $\mu^*$-measurable ("Carathéodory measurable" in my question) if $E$ satisfies Carathéodory's criterion, i.e. $$\mu^*(W) = \mu^*(W\cap E)+\mu^*(W\cap E^c) \text{ for all } W\subset \Omega.$$

The outline of construction is as following:

Show that $\mu^*$ is finite additive.
Show that the measurable sets (here means the sets satisfying Carathéodory's criterion) form an algebra.
Show that the measurable sets form a $\sigma$-algebra.
Define a measure $\mu$ to be the restriction of outer measure $\mu^*$ to the measurable sets.

For details please refer to link.

Does Caratheodory measurable imply measurable in general?

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