# Extension of group valued measures

Let $$G$$ be a topological complete Hausdorff group (written additively with neutral element $$0$$), let $$\mathcal{R}$$ be a ring of sets (i.e. a nonempty set of sets closed under union and relative complement) and let $$\mu:\mathcal{R}\rightarrow G$$ be a measure, i.e. a set function with the following properties:

1. $$\mu(A\cup B) = \mu(A)+\mu(B)$$ whenever $$A,B\in\mathcal{R}$$, $$A\cap B=\emptyset$$.
2. $$\mu(A_n)\rightarrow 0$$ whenever $$(A_n)_{n\in\mathbb{N}}$$ is a decreasing sequence of sets of $$\mathcal{R}$$ with $$\bigcap_{n\in\mathbb{N}}A_n = \emptyset$$.

Assume further that $$\mu$$ has the following property:

1. If $$B\in\mathcal{R}$$ and $$(A_n)_{n\in\mathbb{N}}$$ is a pairwise disjoint sequence of sets of $$\mathcal{R}$$ with $$A_n\subseteq B$$ for all $$n\in\mathbb{N}$$, then $$\mu(A_n)\rightarrow 0$$.

How can I prove that, under the conditions listed, for every increasing sequence of sets $$(A_n)_{n\in\mathbb{N}}$$ of $$\mathcal{R}$$ which is bounded by another set $$B$$ of $$\mathcal{R}$$, the sequence of measures $$(\mu(A_n))_{n\in\mathbb{N}}$$ is a Cauchy sequence in $$G$$?

This is part of Takahashi's proof that such a measure $$\mu$$ can be extended from $$\mathcal{R}$$ to the $$\delta$$-ring generated by $$\mathcal{R}$$, but sadly I cannot find a detailed proof anywhere, and have failed to prove it myself. Any help would be greatly appreciated.

Assume it is false, and take an increasing sequence $$(A_n)_{n\in\mathbb{N}}$$ of elements of $$\mathcal{R}$$ bounded by some set $$B\in\mathcal{R}$$ (i.e. $$A_n\subseteq B$$ for all $$n\in\mathbb{N}$$) such that $$(\mu(A_n))_{n\in\mathbb{N}}$$ is not Cauchy; then there exists a nhood $$U$$ of $$0$$ such that, for all $$N\in\mathbb{N}$$, there exist $$n\geq m\geq N$$ with $$\mu(A_n)-\mu(A_m)\not\in U$$. Define a sequence $$(C_n)_{n\in\mathbb{N}}$$ as follows:
1. Take $$N_0=0$$; then there exist $$n_0\geq m_0\geq 0$$ such that $$\mu(A_{n_0})-\mu(A_{m_0})\not\in U$$, and define $$C_0 := A_{n_0}\setminus A_{m_0}$$; this is nonempty since its measure is not $$0$$.
2. Take $$N_1>n_0$$; then there exist $$n_1\geq m_1\geq N_1>n_0$$ such that $$\mu(A_{n_1})-\mu(A_{m_1})\not\in U$$, and define $$C_1 := A_{n_1}\setminus A_{m_1}$$. Again, this is nonempty since its measure is not $$0$$; further, since $$n_1,m_1>n_0$$, we are guaranteed that $$C_1\cap C_0=\emptyset$$.
3. Proceeding in the same way, define $$C_2,C_3$$... to obtain a pairwise disjoint sequence of nonempty sets which are differences of some of the $$A_n$$ (and thus themselves in $$\mathcal{R}$$), such that $$\mu(C_n)\not\in U$$ for all $$n\in\mathbb{N}$$.
Now, by construction, if $$n>m$$, we must have $$C_n\cap C_m=\emptyset$$, since $$C_n=A_i\setminus A_j$$, $$C_m=A_k\setminus A_\ell$$ and $$A_\ell\subseteq A_k \subseteq A_j \subseteq A_i$$. Thus, $$(C_n)_{n\in\mathbb{N}}$$ should converge to $$0$$ by our hypothesis, but does not, a contradiction. It follows that $$(A_n)_{n\in\mathbb{N}}$$ must be a Cauchy sequence, as desired.