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Let $\{\mathcal{A}_n\}$ be an infinite sequence of sets with $\mathcal{A}_n \subset \mathcal{M}$, where $\mathcal{M}$ is a bounded subset of $\mathbb{R}$ (for simplicity). Is there a "nice" limit definition $\lim^{\circ}$ and a "nice" measure $\mu^{\circ}$ such that the following two holds:

1) Every $\{\mathcal{A}_n\}$ (as discussed above) has a converging subsequence.

2) If $\{\mathcal{A}_n\}$ converges, and $\mathcal{A} = \lim^{\circ}\mathcal{A}_n$, then $\mu^{\circ}(\mathcal{A}) = \lim_{n\rightarrow\infty}\mu^{\circ}(\mathcal{A}_n)$ (here the last limit is the usual limit).

What I mean by "nice:" For the measure, it should satisfy e.g. $\mu^{\circ}([a,b]) = b-a$ (e.g. like the Lebesgue measure). For the limit, I do not know exactly what I want, but it should not be something utterly useless and trivial.

Let me try to be more clear. For example, set $\mu^{\circ}$ to be the Lebesgue measure, and the limits defined in the standard manner, i.e. let \begin{align} \liminf_{n\rightarrow\infty} \mathcal{A}_n = \{x:x\in\mathcal{A}_n\mbox{ for all but finitely many }n\}, \end{align} \begin{align} \limsup_{n\rightarrow\infty} \mathcal{A}_n = \{x:x\in\mathcal{A}_n\mbox{ for infinitely many }n\}, \end{align} and say that the limit $\lim\mathcal{A}_n$ exists is these values agree. With this setup, the second condition is satisfied, but the first is not (see my earlier question today: Bolzano-Weierstrass for sequences of sets for which I got great responses thanks to many people)

For the sake of experimenting, let $\mu^{\circ}$ be the Lebesgue measure again, but use the Kuratowski convergence for the limits (see e.g. the Wikipedia page Then, the first condition is satisfied (I read the proof somewhere, but now forgot where), but the second condition is not (it is easy to construct a counterexample).

This is just a thought experiment that has been bothering me for a while, and I would greatly appreciate any responses.

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Your measure is "nice" and your convergence of definition is "nice" maybe you should decide who it should behave on some "bad" sets? E.g. cantor set or worse. – cactus314 Jan 8 '13 at 16:25
@john mangual, Thanks for your comment. I was thinking that is one option, e.g. focusing only on, say, open intervals of $\mathbb{R}$ (without playing with the limit or measure definitions). Then, one wonders how big our collection of "nice" sets would be, would they e.g. contain at least Borel sets? I will think about it. – Anon Jan 10 '13 at 7:39
I'm not an analyst, so I don't know how bad these sets can be. I found a non-Borel set on Wikipedia having to do with continued fractions – cactus314 Jan 17 '13 at 16:10

Let $\lambda$ be the Lebesuge measure on $\mathbb{R}$.


$$\lim\,\inf_{n\rightarrow\infty}A_n:=[0,\lim\,\inf_{n\rightarrow\infty}\lambda(A_n)]$$ $$\lim\,\sup_{n\rightarrow\infty}A_n:=[0,\lim\,\sup_{n\rightarrow\infty}\lambda(A_n)]$$

Finally if $\lim\,\inf_{n\rightarrow\infty}A_n=\lim\,\sup_{n\rightarrow\infty}A_n$, then we define $\lim_{n\rightarrow \infty}A_n$ to be: $$\lim_{n\rightarrow \infty}A_n:=\lim\,\inf_{n\rightarrow\infty}A_n=\lim\,\sup_{n\rightarrow\infty}A_n$$ It is clear that if $\{A_n\}_{n\in \mathbb{Z}^+}$ converges then: $$\lambda(\lim_{n\rightarrow \infty}A_n)=\lambda([0,\lim_{n\rightarrow \infty}\lambda(A_n)])=\lim_{n\rightarrow \infty}\lambda(A_n)$$

Next we verifty the Bolzano-Weierstrass property, If $\{A_n\}_{n\in \mathbb{Z}^+}$ is a sequence of subsets of a bounded set $\scr{M}$, then $\forall n\in Z^+[\lambda(A_n)\leq\lambda(\scr M)<\infty]$. Thus the sequence $\{\lambda(A_n)\}_{n\in \mathbb{Z}^+}$ is bounded. Let $\{\lambda(A_{n_k})\}_{k\in \mathbb{Z}^+}$ be a convergent subsequence. Now it follows that: $$\lim\,\inf_{k\rightarrow\infty}A_{n_k}=[0,\lim\,\inf_{k\rightarrow\infty}\lambda(A_{n_k})]=[0,\lim_{k\rightarrow\infty}\lambda(A_{n_k})]$$ $$\lim\,\sup_{k\rightarrow\infty}A_{n_k}=[0,\lim\,\sup_{k\rightarrow\infty}\lambda(A_{n_k})]=[0,\lim_{k\rightarrow\infty}\lambda(A_{n_k})]$$ Hence, $\{A_{n_k}\}_{k\in \mathbb{Z}^+}$ is a convergent subsequence of $\{A_n\}_{n\in \mathbb{Z}^+}$.

Perhaps you won't like this answer (I don't like it as well). You need to come up with better conditions to avoid trivial solutions.

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Hi Amr, Thanks for your answer!. As you have anticipated, I should say that it is not exactly the type of answer I was looking for, but I still think it is a great shot, i.e. you should like your answer, I myself like it :) At least your limit definitions preserve some "information" about the sequence of sets under consideration (their measure). It might be a good idea to proceed like that by trying to "embed" more information about the sequences to the limit definition. I am not sure though how "natural" this approach will turn out to be. – Anon Jan 10 '13 at 7:10
@Anon Thanks. I suggest that you add conditions like $\lim_{n\rightarrow\infty}(A_n\cap E)=(\lim_{n\rightarrow\infty}A_n)\cap E$ – Amr Jan 10 '13 at 9:21

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