# A $\sigma$-algebra $\mathcal{S}$ on $\mathbb{R}$ that is larger than $\mathcal{L}$ such that outer measure is a measure on $(\mathbb{R},\mathcal{S})$.

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 $$\mathbb{R}$$?

• 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. Commented Apr 14, 2023 at 10:20
• @ElchananSolomon ... this is irrelevant. Math texts like Axler assume the Axiom of Choice. Commented Apr 14, 2023 at 10:33
• @ElchananSolomon Thank you very much for your answer. Commented Apr 14, 2023 at 10:43
• @GEdgar Thank you very much for your comment. Commented Apr 14, 2023 at 10:44
• You may be interested in reading my detailed answer in: math.stackexchange.com/questions/1400503/… Commented Apr 14, 2023 at 16:21

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.