# Characterization of dense open subsets of the real numbers

Does the complement of every dense open subset of the real numbers have Lebesgue measure $0$?

This is certainly not a characterization of dense open subsets of reals, since the complement of the irrationals(which is not open) has Lebesgue measure $0$. So, is there any useful characterization of dense open subsets of the reals? Also, please mention what happens if the Axiom of Choice is not assumed.

• There's really nothing to do with the axiom of choice here. Except with the minor point that the axiom of choice is quite essential for assuming that the Lebesgue measure is countably additive, but that has nothing do with your question really. Just a general point to make about measures and choice. – Asaf Karagila Jul 30 '14 at 21:02
• @AsafKaragila I didn't recall choice being used at all to get countable additivity, but only to prove the existence of non-measurable sets (as a consequence of additivity). – user61527 Jul 31 '14 at 3:52
• @T.Bongers: What sigma additive measures without atoms exist on a set which is a countable union of countable sets? (And recall that the real numbers are consistently such a set in some models where choice fails badly.) – Asaf Karagila Jul 31 '14 at 3:54

No, it does not. Fix an enumeration of the rationals $\{r_k\}_{k = 0}^{\infty}$ and put an open interval
$$\mathcal{O}_k = \left(r_k - \frac{1}{2^{k + 1}}, r_k + \frac{1}{2^{k + 1}}\right)$$
around this point. Define $$\mathcal{O} = \bigcup_k \mathcal{O}_k$$
in which case $\mathcal{O}$ is an open, dense subset of $\mathbb{R}$ with Lebesgue measure at most $2$ (and by lengthening any one of the intervals, we can make it exactly $2$); hence its complement has infinite measure (and as far as I see, choice has nothing to do with this).
• If we choose an arbitrary enumeration of the rationals we can't say that the family of $\mathcal{O}_k$ are disjoint. So, the measure of $\mathcal{O}$ is $\le 2.$ – mfl Jul 30 '14 at 19:09