Uniformly countably additivity of measures.

Let $\mathcal{K}$ be a family of finitely additive scalar-valued measures defined on some $\sigma$-field $\Sigma$.

We say that $\mathcal{K}$ is uniformly countably additive provided for each decreasing sequence $(E_n)$ of members of $\Sigma$ with $\cap_n E_n = \emptyset$ and each $\epsilon >0$ there is an $N_{\epsilon} >0$ such that $| \mu (E_n) | \leq \epsilon$ for $n$ beyond $N_{\epsilon}$ and all all $\mu \in \mathcal{K}$.

On the other hand, assume that $\mathcal{K}$ is NOT uniformly countably additive.

The textbook says then we can choose a sequence $(\mu_n)$ in $\mathcal{K}$, a disjoint sequence $(E_n)$ in $\Sigma$ and $\delta >0$ such that $|\mu_n (E_n)| > \delta$ for every $n$.

But, I couldn't choose such disjoint sequence $(E_n)$ in $\Sigma$. Only one I could choose is a decreasing sequence whose intersection is the empty set.

Could you help me to choose such disjoint sequence?

Let $F_n$ be a decreasing sequence in $\Sigma$ (that is, $F_{n+1}\subset F_n$) with empty intersection such that $\lim(\mu(F_n))_n$ does not converge uniformly to $0$ for $\mu\in K.$

(I). Suppose $\lim_{n\to \infty}\mu(F_n)=0$ for each $\mu \in K.$

Let $r>0$ be such that $S=\{n\in \Bbb N: \exists \mu\in K\;(\mu(F_n)>r)\}$ is infinite.

Take $n(1)\in S$ and $\mu_1\in K$ with $\mu_1(F_{n(1)})>r.$

Recursively for $j\in \Bbb N$ let $n'(j) >n(j)$ such that $n\geq n'(j)\implies \mu_j(F_n)<r/2.$ Let $E_j=F(n(j))\setminus F(n'(j)).$

Recursively let $n'(j)<n(j+1)\in S$ and let $\mu_{j+1}\in K$ such that $\mu_{j+1}(F_{n(j+1)})>r.$

If $j<k$ then $E_j\cap E_k=\phi$. And $\mu_j(E_j)>r/2$ for all $j$.

(II). (I am presently too sleepy to consider the case $\neg (\forall \mu\in K\;(\lim_{n\to \infty}\mu(F_n)=0)$. )

• Thanks! I used your argument to prove the claim. On the other hand, we can choose a decreasing sequence $(F_n)$ with empty interior and $\epsilon >0$ such that $S = \{ n \in \mathbb{N} | \exists \mu \in \mathcal{K} : | \mu(F_n)|> \epsilon\}$ is infinite. Also, since $F_n$ is decreasing with empty intersection, $\lim_n \mu (F_n) =0$ for $\mu \in \mathcal{K}$ (sorry to forget to mention that we may assume that all $\mu \in \mathcal{K}$ are countably additive); hence we can choose such $n'(j) > n(j)$. Then by using your argument, we can find such disjoint sequence and positive $\epsilon$. Thanks! – cdamle Oct 30 '17 at 6:06
• I'm uncertain how general the definition of "finitely additive scalar-valued measure" is here. Some def'ns would include the case $\Sigma=\{\phi,S\}$ for some $S\ne \phi,$ with $\mu (S)=\infty$. – DanielWainfleet Oct 30 '17 at 10:32

I think you loss some property on the sequence of $\mu_k$... Since you can choose $\mu_k$ to be delta function central at the same point and then they satisfied they are not uniformly countable additive but we can not choose a sequence ($\mu_n$) in $K$ such that there is a disjoint sequence ($E_n$) in $\Sigma$ and $\delta>0$ such that $|\mu_n(E_n)|>\delta$ for every $n$.