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I need some help solving the following problem from basic measure theory, it can be found on Probability & Measure Theory, Ash, section 1.2 (for reference):

Let $\mu$ be counting measure on $\Omega$, where $\Omega$ is an infinite set. Show that there is a sequence of sets $A_n \downarrow \emptyset$ with $\lim_{n\to \infty} \mu(A_n) \neq 0$.

Where $A_n \downarrow \emptyset$ means that the sequence $A_1, A_2, \ldots$ is such that $A_k \supset A_{k+1}$ and $\bigcap_{k = 1}^\infty A_k = \emptyset$.

My idea is to extract a countably infinite set $C \subset \Omega$, and create a sequence $A_1, A_2, \ldots$ with $A_1 = C$ and $A_{k+1} = A_k - \{a\}$ where $a \in A_k$, so that each subsequent set of the sequence "has" one less element than the last. Now, my intuition tells me that each element of the sequence has an infinite number of elements, thus $\lim_{n\to \infty} \mu(A_n) \neq 0$, and the infinite intersection satisfies $\bigcap_{k = 1}^\infty A_k = \emptyset$, because the set only has a countably infinite number of elements.

But how do I justify this? If $$\lim_{n \to \infty} \bigcap_{k = 1}^n A_k= \bigcap_{k = 1}^\infty A_k$$ why would $\lim_{n\to \infty} \mu(A_n) \neq 0$ be true if I'm taking a limit here as well?

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    $\begingroup$ Because $\mu$ is not "continuous"; you don't have that $\lim_{n \to \infty} \mu(A_n) = \mu(\lim_{n \to \infty} A_n)$, which is what you are proving. $\endgroup$ – Mees de Vries Apr 28 '16 at 22:27
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To be true that $A_i \downarrow A \implies \mu(A_i) \downarrow \mu(A)$, you should have that $\mu(A_1) < \infty$.

Hint: construct an example where $\mu(A_1) = \infty$.

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Let $A_n$ be the natural numbers with the first $n$ natural numbers thrown out. I.e. $A_n=\mathbb N\setminus\{1,2\cdots n \}$. Then each $\mu(A_n)=\infty$ and $\cap A_n=\phi$ so that $\mu(\cap A_n)=0$. Can you adapt this argument to your example?

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Choose $x_k \in \Omega$ such that each $x_k$ is distinct.

Let $A_n = \{x_k\}_{ k \ge n}$.

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There is a set $A \subset \Omega$ that can be identified as $\mathbb{Z}$ (in the sense that there is a bijection $f: \mathbb{Z} \to A$). Let $A_n = \mathbb{Z} \setminus \left\{ 1 ,\dots, n \right\}$. This sequence satisfies the claim.

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Try this, which I think is similar to your strategy. Let $C=\{x_1, x_2, \cdots \}$ be a countably infinite set of distinct points. Let $A_n:= \{x_i \in C : i\geq n\}$. So $A_1=C$, $A_2=\{x_2, x_3,\cdots \}$, etc. Then Clearly $$\bigcap_n A_n =\emptyset$$ and $A_n \downarrow$. However, for each $n$, $\mu(A_n)=\infty$, so $\mu(A_n)$ cannot tend to 0.

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