How regular are Borel measures (Royden, Real Analysis (3rd ed), Chapter 13, Proposition 14) On page 341 of the 3rd edition of Real Analysis, Royden states the following proposition (and leaves the proof to the reader):
``Let $\mu$ be a measure defined on a $\sigma$-algebra $\mathcal{M}$ containing the Borel sets.  If $\mu$ is outer regular for each compact set or if $\mu$ is inner regular for each bounded open set, then $\mu$ is regular for each $\sigma$-bounded set in $\mathcal{M}$.''
This is Proposition 14 in Chapter 13; it is stated under the assumption that the underlying topological space is a locally compact Hausdorff space, but I would be perfectly happy to assume that we are working on Euclidean space, $\mathbb{R}^d$, and that $\mu(\mathbb{R}^d) < \infty$.  In this case, I know that $\mu$ is regular for any Borel set.  I can also prove the proposition when $\mathcal{M}$ is the completion of the Borel $\sigma$-algebra under $\mu$, but I do not see how to prove it as stated, assuming only that $\mathcal{M}$ contains the Borel $\sigma$-algebra.
Can anyone provide the missing step: that a measure which is regular for every Borel set must be regular for every set in the $\sigma$-algebra $\mathcal{M}$ on which it is defined?
 A: The following example, due to Tim Steger [Personal communication, March 30, 2004], shows that Proposition 14 (page 341 of Royden's Real Analysis, Third edition) is false. The following example has been published in Appendix B of the paper "Accessible points, harmonic measure, and the Riemann mapping", by Fausto di Biase and Tomasz Weiss, published in 2010 in "Seminari di Geometria 2005-2009" by the Universita` di Bologna, Italy. 
Let $A\subset[0,1]$ be a set of Lebesgue outer measure equal to $1$, and such that $[0,1]\setminus A$ has Lebesgue outer measure equal to 1. Let $\nu^*(E)$ be defined as the Lebesgue outer measure of the set $E\cap A$. Then $\nu^*$ is an outer measure on $[0,1]$, in the sense of Royden's Real Analysis, Third edition. Let $\mathcal{M}$ be the $\sigma$-algebra  of $\nu^*$-measurable subsets of $[0,1]$. Let $\nu$ be the restriction of $\nu^*$ to $\mathcal{M}$. Then $\mathcal{M}$ contains all the Borel subsets of $[0,1]$. The measure space $([0,1],\mathcal{M},\nu)$ satisfies the hypothesis in Proposition 14 [loc. cit.] but not its conclusion.  
