Let $X$ be a 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?

4$\begingroup$ 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 nonmetrizable $X$ 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) $\endgroup$ – Pietro Majer Feb 9 '12 at 13:22

1$\begingroup$ 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? $\endgroup$ – Gerald Edgar Feb 9 '12 at 13:23

1$\begingroup$ @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 nontrivial result of Shelah, but not true in general). $\endgroup$ – Ramiro de la Vega Feb 9 '12 at 15:13

1$\begingroup$ 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).) $\endgroup$ – Bill Johnson Feb 9 '12 at 16:58

1$\begingroup$ @Bill: I don´t quite understand why that would be enough. Can you explain a bit more how would you finish the argument? $\endgroup$ – Ramiro de la Vega Feb 10 '12 at 12:56
No, it is not possible for $\mu$ to be complete.
 There exists a closed subset $K$ of $X$ with $\mu(K)=0$ and a continuous onto map $f\colon K\to2^\omega$.
 With $K,f$ as above, if $A\subseteq 2^\omega$ is any set not in the universal completion of the Borel sigmaalgebra on $2^\omega$, then $f^{1}(A)$ is not Borel measurable.
In particular, taking $A$ to be something like a Vitali set (e.g., let $\sim$ be the equivalence relation on Cantor space $2^\omega$ where $x\sim y$ iff $x_i=y_i$ for all but finitely many $i$ and choose $A$ such that it contains one element from each equivalence class) then $f^{1}(A)$ is a $\mu$null set which is not Borel. This does require the Axiom of choice.
Proof of 1: Start by letting $S$ be the support of $\mu$. That is, $S$ is the smallest closed subset of $X$ for which $\mu(S)=\mu(X)$. The existence of $S$ follows from compactness of $X$ and regularity of $\mu$; if $S$ is taken to be the intersection of all closed sets of full measure then, for any open $U$ containing $S$, compactness implies that $X$ is the union of $U$ and finitely many open $\mu$null sets, so $\mu(U)=\mu(X)$. Outer regularity gives $\mu(S)=\mu(X)$ as required.
As $X$ is not atomic, there exists distinct points $x\not=y$ in $S$ and, choosing disjoint closed neighbourhoods $K_0,K_1$ of $x,y$ we have $\mu(K_0)\gt0$ and $\mu(K_1)\gt0$. Furthermore, we have $\mu(\{x\})=\mu(\{y\})=0$ so, by outer regularity, $\mu(K_0)$ and $\mu(K_1)$ can be made as small as possible.
Applying this process inductively gives nonempty compact sets $K_{i_1,i_2,\ldots,i_n}$ for $(i_1,\ldots,i_n)\in2^n$ of positive measure such that $K_{i_1,\ldots,i_n}\cap K_{j_1,\ldots,j_n}=\emptyset$ whenever $(i_1,\ldots,i_n)\not=(j_1,\ldots,j_n)$ and $K_{i_1,\ldots,i_n}\subset K_{i_1,\ldots,i_{n1}}$. Furthermore, they can be chosen such that $\mu(K_{i_1,\ldots,i_n})\lt4^{n}$. We can now define $K_x=\bigcap_{n\ge1}K_{x_1,\ldots,x_n}$ for each $x\in2^\omega$, which is nonempty by compactness of $X$ with zero measure (by countable additivity of $\mu)$, and $K\equiv\bigcup_{x\in2^\omega}K_x$ is closed. Defining $f\colon K\to2^\omega$ by setting $f(a)=x$ for $a\in K_x$ satisfies the requied properties. QED
Proof of 2: Suppose that $A\subseteq2^\omega$ is not in the universal completion of the Borel sigmaalgebra. Then, there exists a finite Borel measure $\nu$ on $2^\omega$ such that $A$ is not in the completion of the Borel sigmaalgebra with respect to $\nu$. The HahnBanach theorem gives a regular finite measure $\lambda$ on $X$ such that $\nu=f^\ast\circ\lambda$. If $f^{1}(A)$ was in the Borel sigmaalgebra on $X$ then, by regularity, there would exist sequences of compact sets $B_n\subseteq f^{1}(A)$, $C_n\subseteq f^{1}(A^c)$ with $\lambda(B_n)\to\lambda(f^{1}(A))$ and $\lambda(C_n)\to\lambda(f^{1}(A^c))$. It follows that $f(B_n)\subseteq A$ and $f(C_n)\subseteq A^c$ are compact sets with $\nu(f(B_n))\to\nu(A)$ and $\nu(f(C_n))\to\nu(A^c)$ from which it follows that $A$ is in the completion of the Borel sigmaalgebra wrt $\nu$, contradicting the assumption. QED

$\begingroup$ "Compactness implies that $X$ is the union of $U$ and finitely many open $\mu$null sets": I don't quite follow this step. Could you elaborate? $\endgroup$ – Nate Eldredge Sep 10 '13 at 2:23

$\begingroup$ @ Nate Eldredge: I follow. Let $\mathscr A$ be the set complements of all closed sets $B$ with $\mu(B)=\mu(X)$. Then $\mu(A)=0$ for all $A\in\mathscr A$ and $\{U\}\cup\mathscr A$ is an open cover of $X$ which is assumed to be compact. So there is a finite $\mathscr B\subseteq\mathscr A$ with $X\subseteq U\cup\bigcup\mathscr B$. Now, do you too follow? $\endgroup$ – TaQ Sep 10 '13 at 11:26