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Let $K$ be a compact Hausdorff space. Does there exist a finite Borel measure on $K$, assigning positive values to all non-empty open sets of $K$?

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Nate's answer here gives a counterexample (the Stone–Čech compactification of an uncountable discrete set). You have uncountably many open singletons which you can't charge all at the same time. –  t.b. Jun 27 '12 at 8:55
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At least if $K$ is a topological group, the answer is yes: Haar measure on $K$. –  Xabier Domínguez Jun 27 '12 at 10:05
    
@XabierDomínguez, does this work if you take an uncountable power of the two-element group with the product topology? –  user16299 Jul 2 '12 at 9:27
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@YemonChoi: If I don't seriously misunderstand something here, this is straightforward in this case. The measure you get is an extension of the usual coin-flipping measure. The open sets in the product topology are finite intersections of cylinder sets and $\mu\big(\pi_i^{-1}(U_i)\cap\pi_j^{-1}(U_j)\big)=\mu\big(\pi_i^{-1}(U_i))\mu(\pi‌​_i^{-1}(U_i)\big)$. So an open set can be identified with a finite sequence of coin-flips and they have all positive measure. –  Michael Greinecker Jul 5 '12 at 6:56
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@Yemon: Let $U$ be an open null set of a compact group $G$. By compactness we can cover $G$ by finitely many translates of $U$ and deduce that Haar measure is zero. Similarly you can show that no open set in a locally compact group has zero measure: you'd deduce first that Haar measure is zero on compact sets and then deduce that it's zero on every measurable set by tightness. Alternatively use $\tau$-additivity of Haar measure. –  t.b. Jul 8 '12 at 11:40
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