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Definition of regular according to Halmos: Let $\textbf{H}$ be a hereditary $\sigma$-ring (i.e. a $\sigma$-ring such that if $F \in \textbf{H}$ and $E \subset F$ then $E \in \textbf{H}$) and $\mu^*$ an outer measure on $\textbf{H}$. Let $\bar{\textbf{S}}$ be the $\sigma$-ring of all $\mu^*$-measurable sets and $\bar{\mu}$ the restricted measure on $\bar{\textbf{S}}$. $\mu^*$ is said to be regular if $\mu^* = \bar{\mu}^*$, where $\bar{\mu}^*$ is the usual induced outer measure from measure $\bar{\mu}$.

Now to the question; in an exercise an outer measure $\mu^*$ is constructed on the power set of a space $X$ as follows: \begin{equation} \mu^*(E) = 0\ \text{ if }\ E\ \text{ is empty},\quad \mu^*(E) = 1\ \text{ otherwise}. \end{equation} This measure is said to be a finite regular outer measure on $\mathcal{P}(X)$. However I cannot see why it is regular. First lets construct $\bar{\textbf{S}}$: it is clear that it is only the empty set? Since given any non-empty $E$ we have that \begin{equation} 1 = \mu^*(X) \neq \mu^*(E) + \mu^*(E^c) = 2. \end{equation} However, the induced outer measure is defined as \begin{equation} \bar{\mu}^*(E) = \inf \left\{\sum F: E \subseteq F, F \in \bar{\textbf{S}} \right\} \end{equation} and clearly it does not equal $\mu^*$? Am I missing something?

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    $\begingroup$ What you have showed is that $\overline{\mathbb{S}}=\{\emptyset,X\}$. Does this violate outer-regularity? $\endgroup$ – T. Eskin Apr 13 '16 at 16:48
  • $\begingroup$ Yes of course $X$ is also measurable..... How stupid that I did not think of this. Thank you. $\endgroup$ – user128779 Apr 13 '16 at 16:51
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What you have showed is that $\overline{\mathbb{S}}=\{\emptyset,X\}$. Now $\overline\mu$ is the true measure on this set that gives $0$ for empty set and $1$ for $X$. When constructing the outer-measure $\overline\mu^*$, you approximate any non-empty subset of $X$ by covering it with the only possible option in $\overline{\mathbb{S}}$, namely $X$, and taking the $\overline\mu$-measure of that, which is $1$. So the resulting outer measure $\overline\mu^*$ is indeed the same as $\mu^*$ since now every non-empty set has $\overline\mu^*$-measure $1$, and the emptyset has measure $0$. So this outer-measure is indeed regular.

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