Caratheodory extension theorem: which is the "unique extension"

According to Wikipedia, the constructed measure on $\sigma(R)$ is a unique extension. However, in most situations, the $\sigma$-algebra of Caratheodory-measurable sets $M$ is larger than $\sigma(R)$. So is the constructed measure also a unique extension on $M$ and is there an easy way to see this having done the hard work for $\sigma(R)$?

[Or is the actual theorem uniqueness for $M$ and there is an easy way to see uniqueness of extension for $\sigma(R)$?]

I think, you may find interesting Folland's "Real Analysis", (1.10-1.14) in the "Outer measures" section. Let $\mu_0$ be a pre-measure on $R\subseteq 2^X$, and suppose for simplicity that $\mu_0$ is $\sigma$-finite. Denote a corresponding outer measure on $2^X$ as $\mu^*$. Then (1.10) says that the family of $\mu^*$-measurable $M\supseteq R$ is a $\sigma$-algebra, and $\mu:=\mu^*|_M$ is a complete measure. Furthermore, (1.14) says that the extension $\mu'$ of $\mu_0$ to $\sigma(R)$ is unique. As far as I understand, $M$ is a $\mu'$-completion of $\sigma(R)$ and so $\mu$ is a unique extension of $\mu'$ to $M$.
• So $M$ is the $\mu'$-completion of $\sigma(R)$ in the sense of this? Is the proof in Folland too? (or somewhere?) If that is so, I agree that uniqueness of extension over $\sigma(R)$ also gives uniqueness of extension over $M$. Thanks for resolving my query Jul 20, 2014 at 8:59
• Great answer. To add a reference, this post gives an explicit answer as to the completion of Borel wrt Lebesgue gives back $M$ itself. math.stackexchange.com/questions/1168953/… Mar 11 at 6:11