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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.

Does it follow that all open subsets of $X$ are Caratheodory-measurable by $\phi$ ?


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

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up vote 3 down vote accepted

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.

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