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Def. A Radon measure $\mu$ on a compact Hausdorff space $X$ is uniformly regular if there is a countable family $\mathcal{A}$ of compact $G_\delta$-subsets of $X$ such that for every open set $U\subseteq X$, $\mu(U)$ = sup$\{\mu(A): A\in\mathcal{A}$ and $A\subseteq U\}$.

My question is: To what extent the following result will be true?

Q: Let $\{X_\alpha:a\in A\}$ be a family of compact Hausdorff space with uniformly regular Radon measure $\mu_\alpha$. The product $X=\prod_{\alpha\in A}X_\alpha$ is uniformly regular.

I think Q is true when $A$ is countable, but I need a sketch of the prove. The most important case when $A$ is uncountable.

Please share your ideas.

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In your definition, if a subset is compact, then it is closed (right??), so why do you have $G_\delta$? – Jason Rute Oct 15 '13 at 16:43
Also, see the last example in Your answer is there. – Jason Rute Oct 15 '13 at 16:52
We have uniformly regular Radon measure not ONLY Radon, and that example is not a compact, since it is an open interval to power $\kappa$ – Ameen Oct 15 '13 at 18:13
@Jason: Not every closed set has to be a $G_\delta$ in a (non-metrizable) compact space. – Ramiro de la Vega Oct 15 '13 at 22:24

If $\kappa$ is uncountable, no Radon measure $\mu$ on $X=2^\kappa$ can be uniformly regular:

Suppose $\mathcal{A}=\{A_n : n\in \omega\}$ is a countable collection of compact $G_\delta$-subsets of $X$. Then each $A_n$ is a countable intersection of clopens and it is therefore determined by a countable set of coordinates $B_n \subseteq \kappa$. Now fix $\alpha \in \kappa \setminus \bigcup_{n \in \omega}B_n$ and consider the clopen $U=\pi_\alpha^{-1}(0)$. Then either $\mu(U) > 0$ or $\mu(X \setminus U)>0$, but no $A_n$ is contained in $U$ or in $X \setminus U$. Hence $\mathcal{A}$ does not determine $\mu$.

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