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Let $X$ be a Hausdorff compact space, and let $\mathrm {Ba}$, $\mathrm {Bo}$ be its Baire, respectively, Borel, $\sigma$-algebras. Let $\mu:\mathrm {Ba}\to[0,+\infty)$ be a finite Baire measure: it is well-known that it extends uniquely to a regular Borel measure $\tilde\mu:\mathrm {Bo}\to[0,+\infty)$ (since $\mu$ induces a positive linear functional on $C(X)$, this follows from the Riesz-Markov representation theorem, as e.g. stated in Rudin's Real and Complex analysis, thm 2.14). It follows from the regularity that the measure $\tilde\mu:\mathrm {Bo}\to[0,+\infty)$ also satisfies:

1. For any $B\in \mathrm {Bo}$ there exists $A\in \mathrm {Ba}$ such that $\tilde\mu(A\Delta B)=0$.

Question. Is the following stronger statement true, i.e.

2. For any $B\in \mathrm {Bo}$ there exists $A\in \mathrm {Ba}$ such that $B\subset A$ and $\tilde\mu(A\setminus B)=0$.

In other words: Is it the case that the completion of the measure space $(X,\mathrm {Ba},\mu)$ is an extension of $(X,\mathrm {Bo},\tilde\mu)$?

In case the answer is yes, I would be especially happy with an elementary proof independent from the R-M theorem of the fact that $\mathrm {Bo}$ is included in the $\mu$-completion of $\mathrm {Ba}$.

In case the answer is no: Is there an abstract measure-theoretic construction, for abstract measure spaces $(X,\mathcal A,\mu)$, producing an extension into a complete measure space $(X,\mathcal{\tilde A},\tilde\mu)$ verifying the analog of (1), that is: For any $B\in\mathcal{\tilde A}$ there exists $A\in \mathcal A$ such that $\tilde\mu(A\Delta B)=0$, and such that in the case of a finite Baire measure $\mu$ it gives $\tilde{\mathrm{Ba}}\supset{\mathrm{Bo}} $?

(Edit June 9, 2022) Since the answer is no, let me elaborate on point 2. What if one does the Carathéodory construction starting from the measure space $(X,\text{Ba},\mu)$? That is, consider $\mu^*:P(X)\to[0,+\infty)$, the maximum exterior measure which is below $\mu$ on $\text{Ba}$, i.e. $\mu^*(E)=\inf\{\mu(B):E\subset B\in\text{Ba}\}$, and the $\sigma$-algebra $\mathcal C$ of all Carathéodory $\mu^*$-measurable sets. At least for the given examples, $\mathcal C\supset \text{Ba}$. Is this always true? By topological reasons (normality of $X$), $\mu^*$ is $\sigma$-additive on the family of closed sets; moreover (by regularity of $\tilde\mu$), it must coincide with the regular Borel extension $\tilde\mu$ on closed sets.

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Let $X=\{0,1\}^{\aleph_1}$ and $x\in X$, and let $\mu$ be the Borel probability measure such that $\mu(\{x\})=1$. Since $\{x\}$ itself is not a Baire set, there is no Baire set $A\subseteq \{x\}$ such that $\mu(A)=1$. By taking the complements, we get answer no for the question.

For the second question: The usual way of extending a Baire measure to the Borel sets relies on the measure being $\tau$-additive, which is a topological concept. So it is likely that an abstract version would need some kind of property with a topological flavour.

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    $\begingroup$ It looks like all you need is a point $x$ such that $\{x\}$ is not a Baire set. So e.g. you could take $x = \infty$ in the one point compactification of any uncountable discrete set. $\endgroup$
    – Nik Weaver
    Commented Jun 9, 2022 at 5:21
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    $\begingroup$ So here the Baire sets are all countable sets $A\subset X\setminus\{\infty\}$ and all co-countable sets $A\ni\infty$, so that $(X,\text{Ba},\mu)$ is already a complete measure space, but the Borel σ-algebra is the whole $P(X)$. $\endgroup$
    – Pietro Majer
    Commented Jun 9, 2022 at 8:42

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