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?

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

