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 $\lim _{n \to \infty} m(A_n)=m(\cup_{n=1}^\infty A_n)$ ?
$(X,m)$ be regular outer measure space . If $\{A_n\}$ is increasing sequence of sets then $\lim _{n \to \infty} m(A_n)=m(\cup_{n=1}^\infty A_n)$?
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measure-theory
outer-measure
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0What have you done so far? – 2017-02-14
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0@Filburt : Nothing much .. for each $A_n$ we get a measurable subset $B_n$ but it seems difficult to make an increasing sequence out of those $ B_n$ in such a way that they relate to $A_n$ – 2017-02-14
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0Review your definition. Is $B\subset A$ or $A\subset B$? – 2017-02-14
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0@Filburt : ah sorry I mean $A \subseteq B$ – 2017-02-14
1 Answers
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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.