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I am reading Foundations of Modern Probability by Kallenberg and stuck on the meaning of increasing limit $A_n\uparrow A$ for sequence of set (pageno 1), I read this question, but I have to confirm that it means

for some n

$A = \bigcap_{m\geq n}\bigcup_{k\geq m} A_k$

zia badar
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1 Answers1

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$\ A_n\uparrow A\ $ means that:

  • The sequence $\ \big\{A_n\big\}\ $ of sets is monotone increasing—that is $\ A_j\subseteq A_k\ $ for all $\ j,k\ $ with $\ k\ge j\ $, and
  • $\ A=\bigcup_\limits{k=1}^\infty A_k\ $

As it happens, in these circumstances, $\ A=\bigcup_\limits{k=m}^\infty A_k\ $ for all $\ m\ ,$ and so \begin{align} \bigcap_{m=n}^\infty\bigcup_\limits{k=m}^\infty A_k&=\bigcap_{m=n}^\infty A\\ &=A \end{align} not just for some $\ n\ ,$ but for all $\ n\ .$ However, if the sequence $\ \big\{A_n\big\}\ $ is not monotone increasing, and $\ B\stackrel{\text{def}}{=}\bigcap_\limits{m=n}^\infty\bigcup_\limits{k=m}^\infty A_k\ ,$ then it's not true that $\ A_n\uparrow B\ $(and $\ B\ $ will generally depend on $\ n\ $).

Thus, while it follows from $\ A_n\uparrow A\ $ that $\ A=\bigcap_\limits{m=n}^\infty\bigcup_\limits{k=m}^\infty A_k\ $ for some $\ n\ ,$ those statements are not equivalent.

lonza leggiera
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  • First apology for modifying my question, I have reverted it to the original question. Actually I was confusing $A_n \uparrow A$ with limit on a new sequence which start from $n$ of the original sequence ${A_n}$. My confusion was cleared after checking monotone class definition from wikipedia which is exactly your answer. – zia badar Feb 20 '24 at 04:35
  • Could you recommend a book containing these basic operations definition? – zia badar Feb 20 '24 at 04:47