# Bolzano-Weierstrass and measures

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 http://en.wikipedia.org/wiki/Kuratowski_convergence). 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 en.wikipedia.org/wiki/Borel_set#Non-Borel_sets – cactus314 Jan 17 '13 at 16:10

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

Definitions:

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

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