It is well known that,
The Caratheodory-measurable sets and the corresponding (extended) measure form a complete measurable space.
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?
In general, no.
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:
You can find more details about this, as well as a guided proof of your claim, in Folland's "Real Analysis", ex. 16, 22 in chapter 1.