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Problem: Let $\mu$ be the counting measure on an infinite set $\Omega$. Prove that there is a sequence of sets $A_1 \supset A_2 \supset A_3 \dots$ such that $\bigcap A_n = \varnothing$, but $\lim_{n \to \infty} \mu(A_n) \ne 0$.

Attempt:Choose a countably infinite set $\{x_i : i \in \mathbb{N} \} = A_1 \subseteq \Omega$. Define $A_n = \{x_i : i \ge n\}$ Then we have $A_1 \supseteq A_2 \dots$ with $\bigcap A_n = \varnothing$. However, $\mu(A_n) = \infty$ for all $n$.

Does my proof look correct?

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    $\begingroup$ Yes, it is correct. Finding a countable subset requires (part of) AC, but that is an issue only in a set theory course. $\endgroup$ – André Nicolas Jul 12 '13 at 3:40
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Yes.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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  • $\begingroup$ How can one prove that $\cap A_n = \emptyset$? It is not obvious to me. $\endgroup$ – Apoorv Khurasia Aug 11 '18 at 1:23

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