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Is that true that if $\{E_k\}_1^\infty$ is a decreasing countable colection of sets and $m^*(E_1)<\infty$, then $$m^*\left(\displaystyle\cap_{k=1}^{\infty}E_k\right)=\lim_{k\to\infty}[m^*(E_k)]?$$

I think it's not true, because we don't have that each $E_k$ is measurable.

Does someone know some prove or some counterexample?

PS: m*(A) is the Lebesgue outer measure of A.

Manatee
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    If $E_k$ are not assumed to be measurable sets then the equality does not hold. – Ramiro May 29 '21 at 12:48
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    See https://math.stackexchange.com/questions/1764727/general-property-regarding-outer-measure-for-a-nested-sequence-of-sets-measurab – Ramiro May 29 '21 at 12:48
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    If $m^*$ is an arbitrary outer measure, the there are trivial counter-examples: – Ramiro May 29 '21 at 12:52
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    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 $\bigcap_k A_k = \emptyset$. 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 May 29 '21 at 12:55

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I don't think the measurable thing is really the point of the question, you're getting into some weird AOC stuff otherwise. I think (assuming the sets are measurable) that the statement is true.

It might be helpful to consider the (increasing) sequence of sets $A_{1} = E_{1}, A_{k} = E_{1}\setminus E_{k}$ for $k=2,3,...$

Their union is $E_{1} \setminus\bigcap\limits_{k=1}^{\infty}E_{k}$

Try to continue from here :)

platypus22
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    That the statement is true if the sets are measurable is elementary. The whole point of the question is to check the case where the sets are not measurable. – Jason May 28 '21 at 20:07
  • @Jason Do you have some trivial counterexample that shows that my statement is false? – Manatee May 28 '21 at 20:13
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    @Manatee It's by no means trivial, but check the top answer to the the question Good Morning Captain linked. – Jason May 28 '21 at 20:35
  • @Jason That's hard to adapt it to my context. I’ve never seen these sets before. – Manatee May 28 '21 at 20:40
  • @Manatee Any counterexamples have to be non-measurable, and non-measurable sets are inherently difficult to describe. – Jason May 30 '21 at 03:45