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Let $X$ be compact Hausdorff space and $\mu$ a finite Borel measure without atoms which is outer regular with respect to open sets and inner regular with respect to compact sets. Can such measure be complete?

flag	asked Feb 9 2012 at 11:45 arc 197●3

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I understand the question as, whether such a measure can be complete on the $\sigma$ algebra of Borel sets. If $X$ is metrizable, the answer is no for cardinality reasons exactly as for the Lebesgue measure -there is a $\mu$-null perfect set, thus with more subsets than there are Borel sets in $X$. The case of a non-metrizable $X$ is the point of the question. If we want to use the same cardinality argument in the latter case, the question is: can we still find a $\mu$-null Borel set $C\subset X$ of cardinality $\tau(X)$? (here $\tau(X)$ = card. of the open sets = card. of Borel sets) – Pietro Majer Feb 9 2012 at 13:22

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Suppose you have such a measure. The problem then would be to construct a compact set $K$ of measure zero, with a subset that is not a Borel set. Certainly this can be done in compact metric space, but what about in exotic compact Hausdorff spaces? – Gerald Edgar Feb 9 2012 at 13:23

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@Pietro: Is it always (i.e. for any $T_2$ space) true that there are as many Borel sets as open sets? Making a rough estimate, apparently one needs that $|\tau|^{\aleph_0}=|\tau|$ (which is true for compact $X$ by a non-trivial result of Shelah, but not true in general). – Ramiro de la Vega Feb 9 2012 at 15:13

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Wouldn't it be enough to have a continuous function from a closed subset $C$ of $X$ onto the Cantor set $\Delta$? You have a subset $D$ f $\Delta$ s.t. neither $D$ nor its complement contains an uncountable closed subset. And can't you build such a function by doing the usual tree construction of a Cantor set in $X$ and identifying to points the branches in the tree? (Or maybe this is a suggestion I should post under the nom de plume unknown (google).) – Bill Johnson Feb 9 2012 at 16:58

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@Bill: I don´t quite understand why that would be enough. Can you explain a bit more how would you finish the argument? – Ramiro de la Vega Feb 10 2012 at 12:56

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Completness of Borel measure

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