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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?

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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.

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