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It's a theorem that

Let $(X, M, m)$ be a measure space and let $\{A_k\}$ be a sequence of measurable sets. If $A_1 \supset A_2 \supset\dots$ and $m(A_N)<\infty$ for some $N \in \mathbb{N}$ then $m(\bigcap_{k=1}^{\infty} A_k) = \lim_{k \to \infty} m(A_k)$.

I was wondering if measurablity of $\{A_k\}$ is a must or is there an example of non-measurable $\{A_k\}$ such that $m^{*}(\cap_{k=1}^{\infty} A_k) \ne \lim_{k \to \infty} m^{*}(A_k)$ even if for each $k$, $m^{*}(A_k) < \infty$?

user10354138
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  • @user10354138, No! –  Apr 13 '21 at 09:35
  • Did you actually study David's answer? – user10354138 Apr 13 '21 at 09:37
  • @user10354138, my question doesn't suppose $\cap E_n=\emptyset$ and not necessarily $\lim\limits_{n\mapsto \infty} \mu^*(E_n) =0$ –  Apr 13 '21 at 09:43
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    You asked for an example, and David's answer is one. – user10354138 Apr 13 '21 at 09:44
  • @user10354138, I didn't understand David's answer at all! For starters, is $C$ actually Vitali's set? $\mu^(E_n) =\mu^(h_n+C)$ is true because $H$ is countable? All $\mu^(E_n)>0$ doesn't imply $\lim \mu^(E_n)>0$ so how contradiction? –  Apr 13 '21 at 15:58
  • I have posted a detailed answer to your question. Please, let me know if you have any question regarding my answer. If my answer provides relevant / helpful information regarding your question, please, upvote it. If my answer actualy answers your question, accept it too, please. To upvote, click the triangle pointing upward above the number (of votes) in front of the question. To accept the answer, click on the check mark beside the answer to toggle it from greyed out to filled in. – Ramiro Apr 14 '21 at 00:24
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    @user10354138 There are simpler examples than the one presented by David. Take a look in my answer below. – Ramiro Apr 14 '21 at 04:15
  • @L.G. The set $C$ mentioned by David Ullrich in the link is a Vitali set. –  Apr 14 '21 at 06:06

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is there an example of non-measurable $\{A_k\}$ such that $m^{*}(\cap_{k=1}^{\infty} A_k) \ne \lim_{k \to \infty} m^{*}(A_k)$ even if for each $k$, $m^{*}(A_k) < \infty$?

The answer is YES. Here is a simple example.

Let $(\Bbb N, \Sigma, m)$ be a measure space where $\Sigma =\{\emptyset, \Bbb N\}$ and $m$ is a measure defined as $m(\emptyset)=0$ and $m(\Bbb N)=1$.

It is easy to see that, $m^*(\emptyset)=0$ and, for all $E \subseteq \Bbb N$, if $E\neq \emptyset$, then $m^*(E)=1$.

For each $k \in \Bbb N$, let $A_k=\{k, k+1, k+2, \cdots \} \subseteq \Bbb N$.

We have $\bigcap_k A_k = \emptyset$. So, we have $$m^*\left (\bigcap_k A_k \right) = m^*(\emptyset)= 0 $$

and, for each $k \in \Bbb N$, $A_k \neq \emptyset$. So, $m^*(A_k)=1$. So, we have $$m^*\left (\bigcap_{k=1}^{\infty} A_k \right) =0 \ne 1= \lim_{k \to \infty} m^{*}(A_k)$$

Ramiro
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