$(X,m)$ be regular outer measure space . If $\{A_n\}$ is increasing sequence of sets then $\lim _{n \to \infty} m(A_n)=m(\cup_{n=1}^\infty A_n)$?

Question

Let $(X,m)$ be an outer measure space which is regular i.e. for any $A \subseteq X$ there is a Caratheodory measurable set (w.r.t. $m$) $A \subseteq B$ such that $m(A)=m(B) $ . Then is it true that $m$ is continuous from below i.e. if $\{A_n\}$ is an increasing sequence of subsets in $X$ then $\lim _{n \to \infty} m(A_n)=m(\cup_{n=1}^\infty A_n)$ ?

Answer 1

For each $A_n$, let $B_n\supset A_n$ a $m$-measurable set with $m(B_n)=m(A_n)$. Define an ascending sequence $\{C_n\}$ of $m$-measurable sets as the following:

$$C_n = \bigcap_{i=n}^\infty B_i,~~n\in\mathbb{N}.$$

Note that $A_n\subset C_n\subset B_n$ for all $n\in\mathbb{N}$. Then

$$m(\bigcup_{n=1}^\infty A_n)\leq m(\bigcup_{n=1}^\infty C_n) = \lim_{n\to\infty}m(C_n)\leq\lim_{n\to\infty}m(B_n)=\lim_{n\to\infty}m(A_n).$$

The opposite inequality

$$m(\bigcup_{n=1}^\infty A_n) \geq \lim_{n\to\infty}m(A_n)$$

follows from $$m(\bigcup_{n=1}^\infty A_n) \geq m(A_n)~~\forall n\in\mathbb{N}.$$

This completes the proof.