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Carathéodory's construction assigns to each outer measure $$\phi^+:\mathcal{P}(S)\to[0,\infty]$$ a $\sigma$-algebra $\Sigma$ where the restriction $\phi$ of $\phi^+$ is a measure. $\Sigma$ is easy to define: It is the collection of all those $X\subseteq S$ such that for any $A\subseteq S$, we have $$\phi^+(A)=\phi^+(A\cap X)+\phi^+(A\setminus X).$$

Riesz representation theorem also considers an outer measure and extracts from it a $\sigma$-algebra where the outer measure is a (nice, essentially regular) measure. In this case, $S$ is assumed to be Hausdorff and locally compact. We are given a positive linear functional $\Lambda\in C_c(S)$ (the collection of continuous complex-valued functions on $S$ with compact support). We start by defining $$\phi^+(V)=\sup\{\Lambda f\mid {\rm supp}(f)\subset V\}$$ for $V$ open, and then extend $\phi^+$ to all ${\mathcal P}(S)$ by $$\phi^+(E)=\inf\{\phi^+(V)\mid V\text{ is open, and }E\subseteq V\}.$$ This is an outer measure. The $\sigma$-algebra one associates to $\phi^+$ is $\mathcal{M}$, the collection of all $X\subseteq S$ such that for any compact $K\subseteq S$ we have that $\phi^+(X\cap K)<\infty$ and $$\phi^+(X\cap K)=\sup\{\phi^+(C)\mid C\text{ is compact, and }C\subseteq X\cap K\}.$$ Again, the restriction $\phi$ of $\phi^+$ to $\mathcal{M}$ is a measure.

Are there any reasonable circumstances under which we should expect that $\Sigma=\mathcal{M}$, or at least that one of the two containments holds?

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The answer is yes. The inclusion is easy in one direction and, in the other direction you can reduce to the case where $A$ is open and, then, to the case where $A$ is compact, so that the two definitions reduce to the same thing. –  George Lowther Jun 22 '12 at 14:38
    
Oh, you are right! Thanks. –  Bruce Jun 22 '12 at 16:42
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I might misunderstand your question, but it seems that you are asking about the distinction between the measure and "essential measure" in Bourbaki measure theory ("presque partut" vs. "localment presque partut"). In locally compact paracompact spaces there is not such a "byzantine" distinction (where the term "byzantine" is used by A.Weil is the comments in his collected papers). However, the collections of measurable sets is the same for the two measures; essentially, the "local" version is a "smaller" measure with some sets with infinite (total variation) measure for the initial measure which became of (total variation) measure zero for the local version.

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I should have checked Bourbaki's book. Many thanks. –  Bruce Jun 22 '12 at 16:43
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