Completeness of Borel measure

Question

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?

Answer 1

No, it is not possible for $\mu$ to be complete.

1. There exists a closed subset $K$ of $X$ with $\mu(K)=0$ and a continuous onto map $f\colon K\to2^\omega$.
2. With $K,f$ as above, if $A\subseteq 2^\omega$ is any set not in the universal completion of the Borel sigma-algebra 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 $K_{i_1,\ldots,i_n}\subset K_{i_1,\ldots,i_{n-1}}$. 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 sigma-algebra. Then, there exists a finite Borel measure $\nu$ on $2^\omega$ such that $A$ is not in the completion of the Borel sigma-algebra with respect to $\nu$. The Hahn-Banach theorem gives a regular finite measure $\lambda$ on $X$ such that $\nu=f^\ast\circ\lambda$. If $f^{-1}(A)$ was in the Borel sigma-algebra 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 sigma-algebra wrt $\nu$, contradicting the assumption. QED