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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](http://math.stackexchange.com/q/76751) 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.2012-06-27
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    At least if $K$ is a topological group, the answer is yes: Haar measure on $K$.2012-06-27
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    @XabierDomínguez, does this work if you take an uncountable power of the two-element group with the product topology?2012-07-02
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    @YemonChoi: Yes. Kakutani wrote a paper on how to assign independent measures on product spaces: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pja/11955735822012-07-03
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    @MichaelGreinecker: I know you can define Haar measure on arbitrary powers of compact groups. What is not obvious to me, but perhaps is dealt with in the paper you link to, is that Haar measure assigns strictly positive value to every non-empty open subset of this product2012-07-05
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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.2012-07-05
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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.2012-07-08
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    Thanks @MichaelGreinecker and t.b. -- I was being dim2012-07-09

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