Let
- $\Omega$ be a set
- $\mathcal A$ be an algebra on $\Omega$
- $E$ be a $\mathbb R$-vector space
- $\mu:\sigma(\mathcal A)\to E$ be $\sigma$-additive
I want to show that for all $A\in\sigma(\mathcal A)$ and $\varepsilon>0$ there is a $B\in\mathcal A$ with $B\subseteq A$ and $$\left\|\mu(A)\right\|_E<\left\|\mu(B)\right\|_E+\varepsilon\;.$$
Let $$\mathcal M:=\left\{A\in\sigma(\mathcal A):\forall\varepsilon>0:\exists B\in\mathcal A:B\subseteq A\text{ and }\left\|\mu(A)\right\|_E<\left\|\mu(B)\right\|_E+\varepsilon\right\}\;.$$ In order to conclude, it would be sufficient to show that $\mathcal M$ is a monotone class. It's clear to me that $\mathcal M$ is closed under countable nondecreasing unions:
- Let $(A_n)_{n\in\mathbb N}\subseteq\mathcal M$ be nondecreasing, $$A:=\bigcup_{n\in\mathbb N}A_n$$ and $\varepsilon>0$
- As $\mu$ is continuous from below, $$\mu(A_n)\xrightarrow{n\to\infty}\mu(A)\tag1$$ and hence $\exists N\in\mathbb N$ with $$\left\|\mu(A)-\mu(A_n)\right\|_E<\frac\varepsilon2\;\;\;\text{for all }n\ge N\tag2$$
- $A_N\in\mathcal M$ $\Rightarrow$ $\exists B\in\mathcal A$ with $$B\subseteq A_N\subseteq A\tag3$$ and $$\left\|\mu(A_N)\right\|_E<\left\|\mu(B)\right\|_E+\frac\varepsilon2\tag4$$ and hence $$\left\|\mu(A)\right\|_E\le\left\|\mu(A)-\mu(A_N)\right\|_E+\left\|\mu(A_N)\right\|_E<\left\|\mu(B)\right\|_E+\varepsilon\tag5$$
How can we show that it $\mathcal M$ is closed under countable nonincreasing intersections?