# Measure where every set is Measurable

For the Lebesgue measure the Vitali set is a famous construction of a nonmeasurable set.

I was wondering if we could define some non-trivial outer measure on $2^{\mathbb{R}}$ such that every set is measurable with respect to this measure, or if there always exist nonmeasurable sets with respect to nontrivial measures. (of course we can define the outer measure that is zero on all subsets of $\mathbb{R}$ but i was wondering if we could say something more interesting. If such a measure exists does it have any interesting properties, i.e. does there exist some measure with this property that is Borel-regular? or even Radon? (this seems highly unlikely to me but I'm interested in) I would also like to assume the axiom of choice here.

• What about taking one point and assigning measure one to the sets that contain it, and measure zero to those which don't? – Emilio Oct 15 '16 at 19:19
• I think one can show that if you require translation invariance and countable additivity then non-measurable sets exist. – 3-in-441 Oct 15 '16 at 19:19
• @3-in-441 If you reject the axiom of choice, it is consistent that every subset of $\mathbb{R}$ is measurable. Rejecting choice makes really really weird things possible, however, so please don't. – Daniel Fischer Oct 15 '16 at 19:49
• What counts as nontrivial? We can take a countable set $D = \{ x_n : n \in \mathbb{N}\}$ and set $\mu(A) = \sum_{x_n \in A} 2^{-n}$. Then all sets are $\mu$-measurable. $\mu$ is outer and inner regular. If $D$ is dense, every nonempty open set has positive measure. – Daniel Fischer Oct 15 '16 at 19:55