# Lebesgue Measure: Is it Optimal?

Consider a collection $$\Sigma$$ of subsets of $$\mathbb{R}^d$$ such that

1. $$A, B \in \Sigma \implies (A \cup B \in \Sigma \hspace{0.3mm} \text{ and } A \cap B \in \Sigma \hspace{0.3mm} \text{ and } A \setminus B \in \Sigma).$$
2. $$\Sigma$$ is closed under countable unions and countable intersections.
3. If $$A$$ is either open or closed in the usual topology of $$\mathbb{R}^d$$, then $$A \in \Sigma$$.
4. If $$A \in \Sigma$$ and $$c \in \mathbb{R}^d$$, then $$c + A \in \Sigma$$.

where $$x \in c + A \iff x-c \in A$$. Also, consider a function $$\mu : \Sigma \to [0,+\infty]$$ such that

1. $$\mu([0,1]^d) = 1$$.
2. $$\mu(\varnothing) = 0$$.
3. If $$A \in \Sigma$$ and $$c \in \mathbb{R}^d$$, then $$\mu(c+A) = \mu(A)$$.
4. $$\mu$$ is countably additive on $$\Sigma$$.

Then, if I'm not mistaken, it can be shown that

If $$A \in \Sigma$$ and $$A$$ is Lebesgue measurable, then $$\mu(A) = m(A)$$ where $$m$$ is the Lebesgue measure.

Loosely speaking, my question is what if $$A$$ is not Lebesgue measurable?

Is it possible that a non- Lebesgue measurable set $$A$$ can be in $$\Sigma$$, or would this automatically contradict disjoint additivity?

Is it possible that $$\mu(A)$$ is not equal to the Lebesgue outer measure when $$A \in \Sigma$$ and $$A$$ is not Lebesgue measurable?

An outline of a proof would be beautiful. If that's not feasible, then a reference would also be greatly appreciated. Thank you for any insight on the matter.

• I think math.stackexchange.com/a/209552/822 answers your question. Jun 3, 2020 at 1:49
• In particular, following the construction given there, we get a non-Lebesgue measurable set $A$ with Lebesgue outer measure $1$ and $\mu(A)=0$. Jun 3, 2020 at 1:53
• I see, and if $\mu(A)=0$ then additivity is not disturbed. That said, I can't see I particularly understand the construction ("enumerate the countable subsets of $\mathbb{R}$" the author says). I also apologize if my question was a duplicate. Jun 3, 2020 at 23:36
• It's transfinite induction. The indices $\xi$ are not integers, but all the ordinals less than $\mathfrak{c}$ (of which there are exactly $\mathfrak{c}$ many, in particular uncountably many). Jun 3, 2020 at 23:38
• The theorem from Fremlin seems to say that the new measure space $(X, \Sigma', \mu')$ is again complete (417A (ii)). I don't think the answer from Rookatu is correct in that sense. Jun 3, 2020 at 23:46