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.
 A: 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.
