# Are the Caratheodory measurable sets the smallest completion of Borel sets in general?

It is well known that,

1. The Caratheodory-measurable sets and the corresponding (extended) measure form a complete measurable space.

2. Measurable space with the Borel sets as the sigma algebra may not be complete.

So, my question is, if we extend the measure from the pre-measure defined on the smallest algebra containing all the open sets, will the result Caratheodory-measurable sets lead to the smallest completion of that with Borel sets? This should be true in $\mathbb{R}^n$ with the canonical topology. But is it true in general?

Consider $\mathbb{R}^n$ with the Borel algebra and the counting measure. Then the Carathéodory-measurable sets are all the subsets of $\mathbb{R}^n$, i.e. the discrete $\sigma$-algebra. But this is not the completion of the Borel algebra w.r.t. the counting measure - in fact it is already complete w.r.t. this measure.
Edit: In answer to your comment, your claim is indeed true if you start with a sigma-finite pre-measure. In general, the sigma-algebra of Carathéodory-measurable sets is the saturation of the completion of the generated sigma-algebra (the Borel algebra in this particular case). The saturation of a measure space $\left(X,\mathcal{M},\mu\right)$ is the measure space $\left(X,\tilde{\mathcal{M}},\tilde{\mu}\right)$ where:
1. $\tilde{\mathcal{M}}$ is the sigma-algebra of locally measurable sets in the original measure space: A subset of $X$ is locally measurable (w.r.t. to $\left(X,\mathcal{M},\mu\right)$) if its intersection with every measurable set of finite measure ($A \in \mathcal{M}$ with $\mu (A) < \infty$) is measurable. In a sigma-finite measure space, every locally measurable set is measurable, so $\tilde{\mathcal{M}} = \mathcal{M}$.
2. $\tilde{\mu}\left(E\right)$ is defined as $\mu(E)$ if $E \in \mathcal{M}$ and as $\infty$ if $E \in \tilde{\mathcal{M}}\backslash\mathcal{M}$.