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.

  • 1
    $\begingroup$ I think math.stackexchange.com/a/209552/822 answers your question. $\endgroup$ Commented Jun 3, 2020 at 1:49
  • 1
    $\begingroup$ In particular, following the construction given there, we get a non-Lebesgue measurable set $A$ with Lebesgue outer measure $1$ and $\mu(A)=0$. $\endgroup$ Commented Jun 3, 2020 at 1:53
  • $\begingroup$ 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. $\endgroup$ Commented Jun 3, 2020 at 23:36
  • 1
    $\begingroup$ 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). $\endgroup$ Commented Jun 3, 2020 at 23:38
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jun 3, 2020 at 23:46


You must log in to answer this question.

Browse other questions tagged .