$\tau$-additive measures on a complete metric space are tight

Question

Let $X$ be a complete metric space. Are all $\tau$-additive Borel measures on $X$ tight?

In Bogachev's "Measure Theory", vol. 2, in the proof of Theorem 8.9.4 (end of page 213) it says:

Note that if $X$ is a complete metric space, then $\mathcal{M_\tau}(X)=\mathcal{M}_t(X)$, and $\mathcal{M}_t(X) \subset \mathcal{M_\tau}(X)$ for any metric space.

There:

• $\mathcal{M}_t(X)$ is the set of tight Baire measures, which are the same as tight Borel measures since we are in a metric space;
• $\mathcal{M}_\tau(X)$ is the set of $\tau$-additive Borel measures.

Now the second part of the statement is basically implied (at least for locally finite measures) by Proposition 7.2.2 (page 74), which says that every Radon measure is $\tau$-additive. However, there is no proof for the first part of the statement.

Does anyone have a reference of the proof, or directly the proof?

Answer 1

The equality $\mathcal M_\tau(X)=\mathcal M_t(X)$ for a complete metric space $X$ follows from three facts:

1) For any finitely additive measure $\mu$ on $X$ its support $supp(\mu)$ (i.e., the set of points $x\in X$ whose any neighborhood $O_x$ has positive measure $\mu(O_x)$) has countable cellularity and hence is separable.

2) For a $\tau$-additive measure $\mu$ the complement $X\setminus supp(\mu)$ has measure zero, so the measure $\mu$ ``lives'' on its support (in the sense that $\mu(supp(\mu))=1$).

3) Each countably additive Borel measure on a Polish (= separable completely metrizable space) is Radon. This follows from the existence of metrizable compactifications of Polish spaces.

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The same argument shows that the equality $\mathcal M_\tau(X)=\mathcal M_t(X)$ holds for any metrizable space $X$ whose any closed separable subspace is universally measurable (i.e., measurable with respect to any $\sigma$-additive Borel measure on some metrizable compactification of $X$).