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

Question

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 $\mathbb{R}$?

Answer 1

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 $\mathcal S$.

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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.