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.


1 Answer 1


I'll answer my own question since I just came up with a simple proof, in case someone else finds themselves having trouble as I did.

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.


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