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I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The author wrote as follows on p.52 in this book:

We have accomplished the major goal of this section, which was to show that outer measure restricted to Borel sets is a measure. As we will see in this subsection, outer measure is actually a measure on a somewhat larger class of sets called the Lebesgue measurable sets.

Is there a $\sigma$-algebra $\mathcal{S}$ on $\mathbb{R}$ such that outer measure is a measure on $(\mathbb{R},\mathcal{S})$ and $\mathcal{L}\subsetneq\mathcal{S}$, where $\mathcal{L}$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}$?

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    $\begingroup$ Under certain set-theoretic assumptions, all sets are Lebesgue measurable, and you can't get any bigger than that. See: en.wikipedia.org/wiki/Solovay_model. $\endgroup$ Commented Apr 14, 2023 at 10:20
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    $\begingroup$ @ElchananSolomon ... this is irrelevant. Math texts like Axler assume the Axiom of Choice. $\endgroup$
    – GEdgar
    Commented Apr 14, 2023 at 10:33
  • $\begingroup$ @ElchananSolomon Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Commented Apr 14, 2023 at 10:43
  • $\begingroup$ @GEdgar Thank you very much for your comment. $\endgroup$
    – tchappy ha
    Commented Apr 14, 2023 at 10:44
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    $\begingroup$ You may be interested in reading my detailed answer in: math.stackexchange.com/questions/1400503/… $\endgroup$
    – Ramiro
    Commented Apr 14, 2023 at 16:21

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In a certain sense, no.

Write $\lambda^*$ for Lebesgue outer measure on $\mathbb R$. Suppose $\lambda^*$ is a measure on $\mathcal S \supsetneq \mathcal L$. Let $E \in \mathcal S \setminus \mathcal L$. Now $$ E = \bigcup_{M=1}^\infty \big([-M,M]\cap E\big) . $$ So there is $M > 0$ so that $E \cap [-M,M]$ is not Lebesgue measurable. But then $$ \lambda^*\big([-M,M]\cap E\big) + \lambda^*\big([-M,M]\setminus E\big) < \lambda^*\big([-M,M]\big) . $$ All three of these sets belong to $\mathcal S$, and thus $\lambda^*$ is not a measure on $\mathcal S$.


There is still the possibility of $\mathcal S \supsetneq \mathcal L$ and a measure $\mu$ on $\mathcal S$ that agrees with Lebesgue measure on $\mathcal L$. It is just that $\mu$ isn't Lebesgue outer measure.

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  • $\begingroup$ GEdgar, Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Commented Apr 14, 2023 at 10:43
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    $\begingroup$ Note: "thanks" comments are discouraged here, and on all stackexchange forums. $\endgroup$
    – GEdgar
    Commented Apr 14, 2023 at 15:30

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