Comparing two different principles of premeasure-to-measure extension

Question

It is well-known that a premeasure $\mu_0$ (possibly taking infinite values) on a ring of subsets $\Omega_0$ of a set $X$ can be extended to a complete measure space $(X, \Omega_C, \mu_C)$ ($C$ for Carathéodory) where $\mu_C$ is the restriction of the induced outer measure $\mu^*$ and $\Omega_C$ is the set of all Carathéodory-measurable sets $M$, i.e., $\mu^*(E) = \mu^*(E\cap M) + \mu^*(E\setminus M)$ for all $E\subseteq X$.

Nik Weaver (in Measure Theory and Functional Analysis, Chapter 2, Theorem 2.14) gives an alternate complete extension of the premeasure $(X, \Omega_0, \mu_0)$—let's call it $(X, \Omega_N, \mu_N)$ ($N$ for Nik). I want to compare and hopefully show that the two extensions are the same.

Let me first describe Nik's approach. For him, $\mu_0$ is finite-valued, so let's assume this throughout. He breaks his construction into two cases:

1. $X\in\Omega_0$: Here he declares $\Omega_N$ to be the set of all subsets $M$ of $X$ which can be approximated arbitrarily well by those in $\Omega_0$, i.e., for each $\epsilon > 0$, there exists an $A\in\Omega_0$ such that $\mu^*(M\mathbin{\Delta} A) < \epsilon$. Further, he declares $\mu_N$ to be the restriction of $\mu^*$. Then he shows that $(X, \Omega_N, \mu_N)$ indeed is a complete measure space extending the premeasure.

2. $X\notin\Omega_0$: Here he first defines premeasures $(A, \Omega_0^A, \mu_0^A)$ for each $A\in\Omega_0$ where $\Omega_0^A$ is the set of all those subsets of $\Omega_0$ that are also contained in $A$, and $\mu_0^A$ to simply be the restriction of $\mu_0$. By first case, we then have complete measure spaces $(A, \Omega_N^A, \mu^A_N)$ extending $(A, \Omega_0^A, \mu_0^A)$ for each $A\in\Omega_0$. Finally, he defines \begin{align*} \Omega_N & := \bigcap_{A\in\Omega_0}\{M\subseteq X : M\cap A\in \Omega^A_N\}\text{, and}\\ \mu_N(M) & := \sup_{A\in\Omega_0}\mu_N^A(M\cap A) \end{align*} and shows that $(X, \Omega_N, \mu_N)$ is a complete extension of the original premeasure.

Question: Obviously, one would like to compare Nik's extension to that of Carathéodory. I have successfully shown that the two are equivalent in the first case, namely when $X\in\Omega_0$. However, I have been on the second case for quite some time now, with little progress except showing that $\mu_N\le\mu^*$ on $\Omega_N$. Thus what remains to be shown is the reverse inequality and that $\Omega_N = \Omega_C$. I appreciate any thoughts on this problem.

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For the second case, a nice lemma (from Nik's proof) that might come in handy is the following: If $M\in\Omega_N$ is contained in both $A$ and $B$ of $\Omega_0$, then $\mu^A_N(M) = \mu^B_N(M)$. This helps in proving that $\mu_N$ actually extends $\mu_0$.

Answer 1

$\newcommand\Om\Omega\newcommand\N{\Bbb N}\newcommand\R{\Bbb R}\newcommand\M{\mathscr M}\newcommand\A{\mathscr A}$The equality $\Om_N=\Om_C$ follows immediately from Theorem 1.4 and Remark 1.5 in this paper or its preprint version, where a more general setting is considered -- without requiring what is called the premeasure in the OP to be finite. In that paper, $\M,\M_{\mathsf{Ca}},\A,m,m^*$ correspond to $\Om_N,\Om_C,\Om_0,\mu_0,\mu^*$ in the OP, respectively. (Also, in that paper, the extension of $m$ from $\A$ to $\M$ is defined simply as the restriction of $m^*$ to $\M$.)

However, in general $\mu_N\ne\mu^*|_{\Om_N}$. E.g., let $X=\R$, let $\Om_0$ be the powerset of $\N$, and let $\mu_0$ be any finite measure on $\Om_0$. Then $\Om_N$ is the powerset of $\R$ and for any $M\in\Om_N$ such that $M\not\subseteq\N$ we have $\mu_N(M)\le\mu_N(\N)<\infty=\mu^*(M)$ (because no union of members of $\Om_0$ contains such a set $M$), so that $\mu_N(M)\ne\mu^*(M)$. $\quad\Box$

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On the other hand, if $\mu_0$ is $\sigma$-finite, so that $X=\bigcup_{k=1}^\infty A_k$ for some $A_k$'s in $\Om_0$, then $\mu_N=\mu^*|_{\Om_N}$, by the uniqueness of the measure extension -- see e.g. Theorem 1.3 in the mentioned paper.

Moreover, one has the following characterization of the condition $\mu_N=\mu^*|_{\Om_N}$:

Proposition 1: The following three conditions are equivalent to one another:

1. $\mu_N=\mu^*|_{\Om_N}$.
2. For any $M\in\Om_N$ such that $\mu_N(M)=0$ there exist $U_1,U_2,\dots$ in $\Om_0$ such that $M\subseteq\bigcup_{k=1}^\infty U_k$.
3. For any $M\in\Om_N$ such that $\mu_N(M\cap A)=0$ for all $A\in\Om_0$ there exist $U_1,U_2,\dots$ in $\Om_0$ such that $M\subseteq\bigcup_{k=1}^\infty U_k$.

Proof: That conditions 2 and 3 are equivalent to each other follows immediately from the definition of $\mu_N$.

That condition 1 implies condition 2 follows immediately from the definition of $\mu^*$. Indeed, if this implication were false, then for some $M\in\Om_N$ we would have $\mu_N(M)=0$ and $\mu^*(M)=\infty$.

It remains to show that condition 2 implies condition 1. Assume that condition 2 indeed holds. Take any $M\in\Om_N$. We have to show that $\mu_N(M)=\mu^*(M)$. First here, note that $$\mu_N(M)=\sup_{A\in\Om_0}\mu_N(M\cap A) =\sup_{A\in\Om_0}\mu^*(M\cap A)\le\mu^*(M).$$

It remains to show that $\mu^*(M)\le\mu_N(M)$. Here without loss of generality $\mu_N(M)<\infty$. By the definition of $\mu_N$, there is a sequence $(A_k)_{k=1}^\infty$ in $\Om_0$ such that $\mu_N(M\cap A_k)\to\mu_N(M)$. So, $\mu_N(M\cap A)=\mu_N(M)$, where $$A:=\bigcup_{k=1}^\infty A_k=\bigcup_{k=1}^\infty B_k\in\Om_N$$ and $B_k:=A_k\setminus\bigcup_{j=1}^{k-1}A_j\in\Om_0$, so that the $B_k$'s are pairwise disjoint. Therefore and because $\mu_N(M)<\infty$, we have $\mu_N(R)=0$ for $R:=M\setminus A$. So, by condition 2, there exist $U_1,U_2,\dots$ in $\Om_0$ such that $R\subseteq\bigcup_{k=1}^\infty U_k$. So, $$\mu^*(R)=\mu^*\Big(\bigcup_{k=1}^\infty (R\cap U_k)\Big) \le\sum_{k=1}^\infty\mu^*(R\cap U_k) =\sum_{k=1}^\infty\mu_N(R\cap U_k)=0,$$ since $\mu_N(R)=0$. So, $\mu^*(R)=0$ and hence $$\mu^*(M)=\mu^*((M\cap A)\cup R)\le\mu^*(M\cap A)+\mu^*(R) =\mu^*(M\cap A) =\mu^*\Big(\bigcup_{k=1}^\infty(M\cap B_k)\Big) \le\sum_{k=1}^\infty\mu^*(M\cap B_k) =\sum_{k=1}^\infty\mu_N(M\cap B_k) =\mu_N(M\cap A)\le\mu_N(M),$$ so that $\mu^*(M)\le\mu_N(M)$. $\quad\Box$