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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 en.wikipedia.org/wiki/Radon_measure. Your answer is there. –  Jason Rute Oct 15 '13 at 16:52
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We have uniformly regular Radon measure not ONLY Radon, and that example is not a compact, since it is an open interval to power $\kappa$ –  Z. Ameen Oct 15 '13 at 18:13
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@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

2 Answers 2

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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hmmmm, I do could not follows you after "consider the clopen $U=\pi^{-1}_\alpha(0)$ ..... One more thing, is there any gap in my answer above. –  Z. Ameen Oct 15 '13 at 22:53

Please could someone double check this candidate answer for my question

For if $I$ is countable, without losing of generality, identify $I$ with the set of natural number $\mathbb{N}$. Assume that the measure $\mu_n$ is uniformly regular on $X_n$ for every $n\in\mathbb{N}$. Then, for each $n\in\mathbb{N}$ there is a countable family $\mathcal{A}_n$ of compact $G_\delta$-subsets of $X_n$ (that makes $\mu_n$ uniformly regular). If we set $\mathcal{A}$ be the family of finite unions of sets of the form $\prod_{n\in\Bbb{N}}A_n$, where $A_n\in\mathcal{A}_n$ for every $n$ and $a_n\neq x_n$ for not more than finite $n$ of $\Bbb{N}$, then, MAYBE, $\mathcal{A}$ become a countable compact $G_\delta$-subsets of $X$.

If this construction is true making a countable family of $G_\delta$-subsets of $X$, then can I generalize to uncountable case.

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There are a few issues: I don't think your proof is finished (as indicated by the word "MAYBE"). (Although you have the right idea.) But more importantly, your proof shows the issue with generalizing to the uncountable case: The set $\mathcal{A}$ you construct would no longer be countable. –  Jason Rute Oct 16 '13 at 3:29
    
@JasonRute Yes, it has not finished yet, but I want to confirm that the above steps are true or not until getting a countable family of subsets of $X$. –  Z. Ameen Oct 16 '13 at 12:42
    
Why my answer has downvoted ! please tell me the reason. –  Z. Ameen Oct 16 '13 at 12:43

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