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How to prove if a class is monotone, then its limit supremum equals its limit infimum. Example, ${A_{n}}$ is a monotone class with $A_{n} \subset \Omega$, and $A_{1} \subset A_{2} \subset A_{3}... $, or $A_{1} \supset A_{2} \supset A_{3}...$, then $\limsup_{n}A_n =\liminf_{n}A_n$.

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What’s your definition of monotone in this context? –  Brian M. Scott Mar 11 '13 at 21:05
    
@BrianM.Scott yeah, i updated it –  Dylan Zhu Mar 11 '13 at 21:13

1 Answer 1

I’ll do the monotone increasing case and leave the other case to you; feel free to ask for help with it if you get stuck.

Suppose that $A_n\subseteq A_{n+1}$ for each $n\in\Bbb N$. Then for any $x$ and $n$, $x\in A_n$ iff $x\in A_k$ for all $k\ge n$. Let $N(x)=\{n\in\Bbb N:x\in A_n\}$; then for each $x$, either $N(x)=\varnothing$, or $N(x)=\{k\in\Bbb N:k\ge n\}$, where $n=\min N(x)$. From this it’s easy to see that

$$\begin{align*} x\in\liminf_nA_n\quad&\text{iff }\quad\Bbb N\setminus N(x)\text{ is finite}&&\text{by definition}\\ &\text{iff }\quad N(x)\text{is infinite}&&\text{by the observation above}\\ &\text{iff }\quad x\in\limsup_nA_n&&\text{by definition}\;, \end{align*}$$

and it’s not hard to check that in fact $\liminf_nA_n=\limsup_nA_n=\bigcup_{n\in\Bbb N}A_n$.

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