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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?

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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:

  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} $.

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.

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  • $\begingroup$ Thank you! I like this example because it is so simple. What if we assume the measure is, say, sigma-finite? $\endgroup$ – user112758 May 10 '18 at 1:38

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