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If we are looking over subsets of $\mathbb R$ and considering the outer measure defined exactly as $$\mu^*(A) = \inf\left\{ \sum_{k=1}^\infty \ell(I_k) \text{ where the $I_k$ are open intervals such that } A\subset \bigcup_{k=1}^\infty I_k \right\}$$ and inner measure to be

$$\mu_*(A) = \sup\left\{\mu^*(F) \text{ where the $F$ is a closed subset of $\mathbb R$ such that } F\subset A \right\}$$

then measurability in the sense of Lebesgue:

  • $A$ is measurable if $\color{red}{\mu^*(A)=\mu_*(A)}$

is equivalent to the Caratheodory criterion:

  • $A$ is measurable if $\color{red}{\mu^*(T) = \mu^*(T\cap A) + \mu^*(T\cap A^C)}$ for all test sets $T\subset \mathbb R$

See the proof from page 41-45 of this pdf: http://www.math.harvard.edu/~shlomo/212a/11.pdf; definitions and preceding theorems are found pages 1-40).

First question $\color{red}{\text{STILL OPEN}}$ (perhaps refer to Nate’s comment below)?:

If these two definitions are in fact equivalent, has anyone tried proving the Caratheodory Extension Theorem without the Caratheodory criterion, i.e. from scratch using outer and inner measure, or at least the equivalent definition of $\color{red}{\mu(G\setminus F)<\epsilon \text{ for some } G\supset A\supset F}$ (the "inner-outer" characterization from the Harvard pdf)? If so, could you provide a rundown of the proof (or a link to some article)?

If it doesn't work generally, how about in the specific case where outer measure is defined the way I defined it above (e.g. as in the proof presented here, page 28 of the pdf: https://www.stat.washington.edu/jaw/COURSES/520s/521/bk521reJaw2012.pdf)?

And second question $\color{red}{\text{(resolved I believe by the comment by Daniel Fischer)}}$:

Consider a weaker criterion than Caratheodory's: I just want that $\mu^*(X) = \mu^*(X\cap A) + \mu^*(X\cap A^C)$ for just the "biggest set" $X$ (I'm talking about the set for which $A^C=X\setminus A$ for all $A$, i.e. the universal set; assume also that $\mu(X)$ is not infinite). We are still using outer measure as I defined it above.

Is this an equivalent condition to the notions of measurability I described before? If not, is there an explicit counterexample (one that I can use pedagogically to motivate Caratheodory's stronger criterion)?

Related questions:

$\color{red}{\text{Thanks to Daniel Fischer to a comment for what I believe to be a resolution to Question 2:}}$

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    $\begingroup$ Re your second question, if $\mu(X) < +\infty$, your condition is equivalent to Carathéodory measurability. The two are (generically) not equivalent if $\mu(X) = +\infty$, since then $\mu(X) = \mu^{\ast}(A) + \mu^{\ast}(X\setminus A)$ for all $A \subset X$. Re the first question, I think I have read somewhere that it's possible. But it would be very awkward, I guess. $\endgroup$ Oct 8 '19 at 21:22
  • $\begingroup$ @DanielFischer Thanks for the link. Do you mind writing out (or linking to) an explicit example for the infinite case where my condition says “it’s $\mu^*$-measurable” when in fact the other conditions say it’s not? $\endgroup$
    – D.R.
    Oct 9 '19 at 1:05
  • $\begingroup$ Oy vey. An explicit example of a non-measurable set is hard to come by. If AC doesn't hold (in which particular way it would need to fail would be a question on which Asaf may be able to shed light) it is possible that every subset of $\mathbb{R}$ is Lebesgue-measurable. On different measurable spaces, explicit examples may be easy to construct, however. $\endgroup$ Oct 9 '19 at 11:30
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    $\begingroup$ To my mind, the Caratheodory extension theorem in this context is the statement that "the collection of measurable sets is a $\sigma$-algebra and the outer measure is countably additive on this $\sigma$-algebra". Which is exactly what Sternberg proves in slides 27-38. So if that is not what you are looking for, then perhaps you could precisely state the result you want to see proved. $\endgroup$ Oct 17 '19 at 21:15
  • $\begingroup$ @NateEldredge his proof is so short! In three slides he proved the extension theorem (and then went on to talk about open and closed sets)?! If someone can prove that so quickly why need Carathéodory at all? $\endgroup$
    – D.R.
    Oct 19 '19 at 5:28

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