Let $ \; \langle X,\mathcal{T} \hspace{.06 in} \rangle \; $ be a second-countable Hausdorff space.


Let $ \; \phi : 2^X \to [0,+\infty] \; $ be an outer regular [outer measure](http://en.wikipedia.org/wiki/Outer_measure#Formal_definitions).


Does it follow that all open subsets of $X$ are [Caratheodory-measurable](http://en.wikipedia.org/wiki/Outer_measure#Formal_definitions) by $\phi$ ?
  
  
  

(I already know this holds if $ \; \langle X,\mathcal{T} \hspace{.06 in} \rangle \; $ is regular and $ \; \phi(X) < +\infty \; $ .)



Do outer regular outer measures always measure open sets?
No. Suppose there is a set $E$ that is not measurable. Take the topology that is generated by $E$ and the original topology. This is also second countable and $\phi$ is regular for it.


